Explanation:
Okay, let me try to figure out how to solve this problem. The function given is $ f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor $, and we need to find how many distinct values are in its range for $ x $ in $ [0, \infty) $. Alright, let's start by recalling what the floor function does. The floor of a number is the greatest integer less than or equal to that number. So, for example, $ \lfloor 1.2 \rfloor = 1 $, $ \lfloor -0.5 \rfloor = -1 $, etc.

First, I need to understand the behavior of the function $ \left( -\frac{5}{8} \right)^x $ for $ x \geq 0 $. Since the base is negative, and we're raising it to a real exponent, this might not be straightforward. Let me think about how this expression behaves for different values of $ x $.

Let me note that $ -\frac{5}{8} $ is a negative number between -1 and 0. When we raise a negative number to a real exponent, the result can be complex unless the exponent is an integer. But the problem states that $ f(x) $ is defined for all $ x \in [0, \infty) $ such that $ f(x) $ is a real number. So, perhaps the function is only defined when $ \left( -\frac{5}{8} \right)^x $ is real? That would happen when $ x $ is an integer? Wait, but the problem says it's defined for all $ x $ in [0, ∞). Hmm, maybe there's something I'm missing here.

Wait, actually, for real exponents, $ a^x $ is real only if $ a $ is positive or if $ a $ is negative and $ x $ is rational with an odd denominator. But since $ x $ is a real number here, unless the exponent is an integer, the expression might not be real. But the problem says that $ f(x) $ is defined for all $ x \in [0, \infty) $. So, perhaps there's an assumption that $ x $ is an integer? Wait, but the problem says "for all values of $ x $ in [0, ∞)", but maybe they mean that for each real number $ x $, $ \left( -\frac{5}{8} \right)^x $ is real. But that's not true for all real $ x $, unless $ x $ is an integer. So maybe there's a misunderstanding here.

Alternatively, perhaps the problem is considering $ x $ as a real number, and the expression is interpreted in some other way? Wait, maybe the problem is written incorrectly? Or maybe there's a different interpretation. Wait, maybe the problem is actually meant to have $ x $ as an integer? Because otherwise, the expression $ (-5/8)^x $ would not be real for non-integer real numbers. Let me check the original problem again.

It says: "Let $ f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor $ be a function that is defined for all values of $ x $ in $ [0, \infty) $ such that $ f(x) $ is a real number. How many distinct values exist in the range of $ f(x) $?"

So, the function is defined for all $ x \in [0, \infty) $, but only when $ \left( -\frac{5}{8} \right)^x $ is real. So, perhaps the expression is real for some values of $ x $, but not all? But how can that be? If $ x $ is allowed to be any real number in [0, ∞), then for non-integer $ x $, we can't define $ (-5/8)^x $ as a real number. Therefore, maybe the problem is actually intended to have $ x $ as integers? Because for integer $ x $, even if negative, the exponentiation is real. Let me check if the problem mentions anything about $ x $ being an integer. It doesn't. So, perhaps there's a mistake in the problem?

Alternatively, maybe the problem is considering complex numbers? But the floor function is only defined for real numbers. So, if the problem states that $ f(x) $ is a real number, then $ \left( -\frac{5}{8} \right)^x $ must be real, so perhaps the domain is only those $ x $ for which $ (-5/8)^x $ is real. So, that would be when $ x $ is an integer? Or when $ x $ is rational with an odd denominator?

But the problem says "defined for all values of $ x $ in [0, ∞)" which is confusing. Maybe it's a translation issue or a typo? Wait, perhaps the original problem is in Chinese? Or maybe the user made a typo? Alternatively, maybe the base is actually positive? Wait, maybe the user meant $ (-\frac{5}{8})^x $ but actually, for real exponents, that's problematic. Alternatively, maybe it's supposed to be $ (-\frac{5}{8})^x $ but only for integer values of $ x $, and the problem says "for all values of x in [0, ∞)" but actually, it's for all integers x in [0, ∞). But that's speculation.

Alternatively, maybe they consider $ x $ as real numbers and take the principal value? Wait, for example, complex exponentiation? Then, we can write $ (-\frac{5}{8})^x = e^{x \ln(-5/8)} $, but since $ \ln(-5/8) $ is complex, so $ (-\frac{5}{8})^x $ would be a complex number. But since the floor function is for real numbers, the problem must have $ \left( -\frac{5}{8} \right)^x $ real. Therefore, the problem must be considering $ x $ as integers? Otherwise, it's not real. So, perhaps the problem is intended for integer $ x $?

Alternatively, maybe $ x $ is a real number, but the expression is taken as the magnitude? Or maybe there is a different interpretation. Wait, perhaps the problem is written incorrectly, and it's supposed to be $ (-\frac{5}{8})^x $, but with absolute value? Let me check the original problem again.

Original problem says: $ f(x)=\left\lfloor\left(-\frac{5}{8}\right)^x\right\rfloor $. So, the floor of $ (-5/8)^x $. Given that, if the problem is supposed to have real outputs, maybe the domain is restricted to x such that $ (-5/8)^x $ is real. So, for real numbers, when is $ (-5/8)^x $ real?

Well, if $ x $ is rational with an odd denominator, then it's possible. However, since x is in [0, ∞), and the problem says "for all values of x in [0, ∞)", which is confusing. Maybe there's a misunderstanding here. Alternatively, perhaps x is allowed to be any real number, and the expression is defined through complex numbers, but then floor function would not be defined. So, I think there must be something wrong here.

Wait, maybe the problem is from a competition or textbook where they consider exponents with negative bases for real exponents? For example, if x is an integer, then it's defined. Otherwise, they might not. So, perhaps the problem is intended for integer x, and the wording is just off? Maybe "defined for all values of x in [0, ∞)" is supposed to mean that x is an integer in [0, ∞). But the original problem didn't specify. Hmm.

Alternatively, maybe the problem is in the context of a discrete math course where x is an integer? But without more context, it's hard to tell. Since the problem says x is in [0, ∞), which usually is a continuous interval, but given the floor function, maybe x is integer?

Alternatively, perhaps the problem is considering $ (-\frac{5}{8})^x $ as a complex number and then taking the floor of the real part or something? But that's not standard, and the problem says "f(x) is a real number", so perhaps not.

Alternatively, maybe the problem is using a different definition where for non-integer x, the expression is defined? For example, if x is a real number, then $ (-\frac{5}{8})^x $ could be defined as follows: For x rational with odd denominator, but for irrationals, it's not. However, since the problem says "for all values of x in [0, ∞)", which would require the function to be defined for all real numbers x ≥ 0, but since it's not defined for many real numbers, I think that can't be.

Therefore, maybe there's an error in the problem statement, and x is supposed to be an integer? Let's assume that, because otherwise, the problem doesn't make sense. So, let me proceed under that assumption. Let me check if that leads me to a reasonable answer. If x is an integer, then $ (-5/8)^x $ is real, and we can take the floor of it. So, maybe that's the intended problem. Let me try solving it with that assumption.

If x is an integer ≥ 0, then let's analyze $ (-5/8)^x $. For integer x:

- When x is even, $ (-5/8)^x = (5/8)^x $, which is positive.
- When x is odd, $ (-5/8)^x = -(5/8)^x $, which is negative.

So, for each integer x ≥ 0, let's compute $ f(x) $:

First, let me note that $ 5/8 $ is between 0 and 1. So, $ (5/8)^x $ decreases as x increases. For positive x, it's a positive number approaching 0 as x approaches infinity. For negative x, but since x is non-negative, we don't have that here.

So, for even x:

Let x = 2k, where k is a non-negative integer (since x ≥ 0). Then $ (-5/8)^x = (5/8)^{2k} $. So, that's positive, and decreasing as k increases. So, for even x, the value is positive, and approaching zero.

For odd x:

x = 2k + 1. Then $ (-5/8)^x = - (5/8)^{2k + 1} $. So, negative numbers approaching zero from below.

Therefore, the floor function of this would be:

For even x: floor(positive number between 0 and 1). Since $ (5/8)^x $ is between 0 and 1 for x ≥ 1, and for x = 0, it's 1.

Wait, let's compute for x = 0:

x = 0: $ (-5/8)^0 = 1 $, so floor(1) = 1.

x = 1: $ (-5/8)^1 = -5/8 ≈ -0.625 $, floor(-0.625) = -1.

x = 2: $ (-5/8)^2 = 25/64 ≈ 0.390625 $, floor(0.390625) = 0.

x = 3: $ (-5/8)^3 = -125/512 ≈ -0.244140625 $, floor(-0.244) = -1.

x = 4: $ (-5/8)^4 = 625/4096 ≈ 0.1525878906 $, floor(0.152) = 0.

x = 5: $ (-5/8)^5 ≈ -0.09765625 $, floor(-0.0976...) = -1.

x = 6: $ (-5/8)^6 ≈ 0.06103515625 $, floor(0.061) = 0.

x = 7: ≈ -0.03814697265, floor(-0.038) = -1.

x = 8: ≈ 0.0244140625, floor(0.024) = 0.

x = 9: ≈ -0.01525878906, floor(-0.015) = -1.

And so on.

So, for even x ≥ 0:

At x = 0: floor(1) = 1

For x even and ≥ 2: floor( (5/8)^{even} ) which is between 0 and 1, so floor is 0.

For odd x ≥ 1: floor( - (5/8)^{odd} ) which is between -1 and 0, so floor is -1.

Therefore, the possible values of f(x) are 1, 0, and -1. Wait, so does that mean the range has three distinct values? But let me check for x=0: yes, 1. For even x ≥2: 0. For odd x ≥1: -1. So, that would be three distinct values: 1, 0, -1. So, the answer would be 3?

Wait, but wait, let me check if there are more possibilities. Wait, for example, when x=0, the value is 1. For x=1, it's -1. For x=2, 0. For x=3, -1. For x=4, 0. So yes, only 1, 0, -1.

But wait, let me check for x=0: $ (-5/8)^0 = 1 $. So floor(1) = 1. Correct.

For x=1: -5/8 is -0.625, floor is -1.

x=2: 25/64 ≈ 0.390, floor is 0.

x=3: -125/512 ≈ -0.244, floor is -1.

x=4: 625/4096 ≈ 0.152, floor is 0.

x=5: -5^5 / 8^5 = -3125 / 32768 ≈ -0.095, floor is -1.

Similarly, as x increases, the value approaches zero from the positive and negative sides, but since the magnitude is decreasing, the floor function remains 0 or -1, as we saw. So, for even x ≥ 2, floor is 0, for odd x ≥ 1, floor is -1. And for x=0, floor is 1. So, there are three distinct values: 1, 0, -1. Therefore, the number of distinct values is 3. But wait, the problem says "for all values of x in [0, ∞)" but if x is only integer, then the answer is 3. However, maybe there's more?

Wait, but let me think again. What if x is not an integer? Suppose, for example, x is a non-integer real number. But then, as per earlier, $ (-5/8)^x $ is not real. However, if we assume that x is allowed to be real, but the problem says that f(x) is real, then maybe x is restricted to integers? But since the problem says x is in [0, ∞), which is a continuous interval, but maybe the question assumes that x is integer? Or perhaps there is a different interpretation.

Alternatively, maybe the problem considers $ (-5/8)^x $ as a complex number and takes the floor of its real part? But that's not standard. Alternatively, maybe it's absolute value? But the original problem didn't specify that. Alternatively, maybe it's misprinted and the base is positive? For example, $ (5/8)^x $, which would be straightforward. But the original problem says -5/8.

Alternatively, maybe the base is positive, and there's a typo. But assuming the problem is correct, I need to figure out. Wait, maybe the problem actually allows x to be any real number, and for the expression to be real, x must be rational with odd denominator? But then the range might have more values. However, the problem states that f(x) is defined for all x in [0, ∞). But I don't think that's feasible.

Alternatively, maybe the problem is in a context where complex numbers are considered, but floor function is applied to the real part? Or something like that? But without more context, it's hard to tell.

Alternatively, perhaps the problem is considering that for real exponents, $ (-5/8)^x $ is defined by taking the modulus and then applying the exponent, so like $ |(-5/8)|^x \times e^{i\pi x} $, but then taking the real part? But that seems too speculative.

Alternatively, maybe the problem is using the definition of real exponents for negative bases only when x is rational with odd denominator, but then for x in [0, ∞), the function is defined only at those rationals. However, since the problem says "defined for all values of x in [0, ∞)", that's conflicting. Unless the problem is considering x as integers. Since otherwise, it's not defined for all real numbers in [0, ∞).

Given that, perhaps the problem has a typo, and it's supposed to be $ (5/8)^x $ instead of $ (-5/8)^x $. If that were the case, then the function would be $ f(x) = \lfloor (5/8)^x \rfloor $, and we can analyze it. But since the problem has a negative sign, maybe there's a different approach. Wait, maybe the problem is expecting to interpret $ (-\frac{5}{8})^x $ as $ (-1)^x \times (\frac{5}{8})^x $, and then considering the floor function for real x? But then, unless x is an integer, (-1)^x is not real.

But if we assume that x is real, then $ (-1)^x $ can be written as $ e^{i\pi x} $, which is complex. Therefore, multiplying by $ (\frac{5}{8})^x $, which is real, the whole expression is complex. Hence, the floor function is not defined. Therefore, unless the problem is only considering x integer, which is the only way for $ (-\frac{5}{8})^x $ to be real.

So, assuming that the problem is for integer x, then we can proceed as before, getting three distinct values: 1, 0, -1.

But let me confirm again. Let's check x=0: floor(1) = 1. For x=1: floor(-5/8) = -1. For x=2: floor(25/64)=0. For x=3: floor(-125/512)= -1. For x=4: floor(625/4096)=0. Similarly, for higher x, it alternates between -1 and 0. So the range is { -1, 0, 1 }, so three distinct values. Hence, the answer is 3? But let me check if maybe for some x, the floor function gives something else.

Wait, for example, suppose x is a non-integer. If we assume x is a non-integer, but the problem says "defined for all values of x in [0, ∞)", which is conflicting. Therefore, if we take x to be integers, then the answer is 3. However, the problem says "for all values of x in [0, ∞)", which might mean that the function is defined for all real numbers x ≥ 0. But unless there's another way to interpret the expression.

Alternatively, maybe the problem is in a context where the exponent is allowed to be any real number, but the base is interpreted in a different way? For example, maybe the expression is defined as $ (-1)^x \times (5/8)^x $, which is complex, but taking the real part? Then, applying floor function to the real part? But that is highly non-standard. Alternatively, if the problem is in a context where the expression is defined as the magnitude, so $ |(-5/8)^x| $, which would be $ (5/8)^x $, then floor of that. Then, the function would be $ \lfloor (5/8)^x \rfloor $. But then the negative sign is irrelevant. However, the original problem has the negative sign. So, given that, I think this is a stretch.

Alternatively, perhaps the floor function is applied to the absolute value? But again, the problem doesn't specify that. Given that, I think the only way for the problem to make sense is to assume that x is an integer, leading to three distinct values.

However, maybe I need to check if there are more values. Wait, for example, let me suppose that x is a non-integer, but the expression is still real. For example, take x=1/2. Then $ (-5/8)^{1/2} $ is sqrt(-5/8), which is imaginary. Similarly, for x=1/3, $ (-5/8)^{1/3} $ is real because cube root of negative number is negative. So, for x=1/3, it's real. Similarly, for x=2/3, it's $ [(-5/8)^{1/3}]^2 $, which is positive. Similarly, for x= p/q in reduced form, if q is odd, then it's real, else complex.

Therefore, if x is a rational number with odd denominator, then $ (-5/8)^x $ is real. If x is irrational, it's not real. Therefore, the domain of f(x) is the set of real numbers x ≥ 0 for which x can be expressed as a rational number with odd denominator. But the problem says "defined for all values of x in [0, ∞)" such that f(x) is real. Therefore, the domain is not all of [0, ∞), but only those x which are rational numbers with odd denominator. So, for example, x=1/3, x=2/3, x=1, x=1/5, etc. However, the problem says "defined for all values of x in [0, ∞)", which is confusing. Maybe it's a mistranslation or misstatement, and it actually means that f(x) is defined for all x in [0, ∞) where the expression is real? But even so, the function would only be defined at countably many points, which is not the whole interval.

Given that, if we consider x as real numbers where $ (-5/8)^x $ is real, which would be x rational with odd denominator, then the function f(x) is defined at those points. However, analyzing the range would be complicated because there are infinitely many such x. But maybe the floor function takes on only finitely many values? Let me try.

First, consider x positive real numbers. Let me think about x > 0. For x rational with odd denominator, say x = p/q, where q is odd. Then:

If x is rational, then $ (-5/8)^x = (-1)^{p/q} \times (5/8)^{p/q} $. Since q is odd, (-1)^{p/q} = -1^{p/q} = -1^{p} \times (-1)^{something}... Wait, no. Actually, (-1)^{p/q} is equal to the q-th root of (-1)^p. Since p/q is a rational number. If p is even, (-1)^p = 1, so the q-th root of 1 is 1. If p is odd, (-1)^p = -1, so the q-th root of -1 is -1 since q is odd. Therefore, $ (-5/8)^x = (-1)^{p/q} \times (5/8)^{p/q} = (-1)^{p} \times (5/8)^{p/q} $. Wait, let me check:

Wait, for example, if x = p/q, then $ (-5/8)^{p/q} = [(-5/8)^{1/q}]^p $. Since q is odd, (-5/8)^{1/q} is real and negative, so equal to - (5/8)^{1/q}. Therefore, raising that to the power p: [ - (5/8)^{1/q} ]^p = (-1)^p \times (5/8)^{p/q}.

Therefore, $ (-5/8)^x = (-1)^p \times (5/8)^{p/q} $.

So, depending on whether p is even or odd, the sign is positive or negative.

Therefore, for each such x = p/q, the value of $ (-5/8)^x $ is either positive or negative, depending on p's parity.

But how does this affect the floor function? Let me think. For example, if x is a positive rational number with odd denominator, then:

- If the exponent p is even, then $ (-5/8)^x $ is positive.
- If the exponent p is odd, then $ (-5/8)^x $ is negative.

Therefore, similar to the integer case. So, the floor function would depend on whether the value is positive or negative.

Additionally, the magnitude of $ (-5/8)^x $ is $ (5/8)^{x} $, which is a decreasing function of x, since 5/8 < 1.

Therefore, for x approaching infinity, $ (5/8)^x $ approaches 0. Therefore, the magnitude is approaching 0 from the positive side if x is even, or negative if x is odd.

Therefore, for positive x:

If x is a positive rational number with odd denominator:

- If x is such that the exponent p is even, then the value is positive, between 0 and 1 (since (5/8)^x is between 0 and 1 for x > 0). Therefore, floor(positive number between 0 and 1) is 0.
- If x is such that the exponent p is odd, then the value is negative, between -1 and 0 (since magnitude is between 0 and 1). Therefore, floor(negative number between -1 and 0) is -1.

However, when x = 0, the value is 1, so floor(1) = 1.

Therefore, similar to the integer case, except now x can be non-integer rational numbers with odd denominators. However, for these x, the value of $ (-5/8)^x $ is either between 0 and 1 (positive) or between -1 and 0 (negative), so the floor function is either 0 or -1. Except for x = 0, which gives floor(1) = 1.

Therefore, the range of f(x) would be { -1, 0, 1 }, same as before. Therefore, even if we consider all x with $ (-5/8)^x $ real (i.e., x rational with odd denominator), the floor function only takes on the values -1, 0, and 1. Therefore, the number of distinct values is 3.

But wait, is there a case where the value of $ (-5/8)^x $ is exactly an integer? For example, when is $ (-5/8)^x $ equal to 1, 0, or -1?

For the value to be 1: $ (-5/8)^x = 1 $. This occurs when x = 0 (as we saw), and also when (-5/8)^x = 1. For real x, this only occurs at x = 0 because any non-zero x would make it less than 1 in magnitude. Similarly, for the value to be -1: $ (-5/8)^x = -1 $. This requires that the magnitude is 1 and the sign is negative. The magnitude is 1 only when x = 0. But $ (-5/8)^0 = 1 $, not -1. So there's no x ≠ 0 for which $ (-5/8)^x = -1 $. Therefore, the floor function never gives -1 when the value is exactly -1, but only when it's between -1 and 0.

Therefore, the floor function can take values -1, 0, 1. Thus, three distinct values.

But wait, for x approaching 0 from the right? If x approaches 0+, then $ (-5/8)^x $ approaches 1, because any number to the power 0 is 1. Therefore, floor(1) = 1. However, if x is positive and approaching 0, then depending on whether x is rational with odd denominator or not, but if x approaches 0, then for small x, the value is near 1. For example, take x approaching 0+, then $ (-5/8)^x = e^{x \ln(-5/8)} $. But in real numbers, this requires x to be rational with odd denominator. For example, x = 1/n with odd n. Then $ (-5/8)^{1/n} $ is real. For x approaching 0+, these values approach 1. So floor of numbers approaching 1 from below (if x is such that the value is slightly less than 1) would be 0? Wait, no. Let me think.

Wait, if x is approaching 0 from the right, what is the value of $ (-5/8)^x $?

If x approaches 0+, then for even x (like x = 2/n where n is a positive integer), the value is $ (5/8)^{x} $, which approaches 1 from below. Therefore, floor( (something just less than 1) ) is 0. However, for x = 0, it's exactly 1, floor is 1. So, when x approaches 0 from the right, floor of the value is 0. However, at x=0, it's 1. Therefore, 1 is included in the range.

But for x approaching infinity, the value approaches 0 from the positive or negative side, so floor is 0 or -1.

Therefore, in the range of f(x), we have 1 (at x=0), 0 (for x > 0 with even exponents, or non-integer x where the value is between 0 and 1), and -1 (for x > 0 with odd exponents, or non-integer x where the value is between -1 and 0). Therefore, the range is { -1, 0, 1 }, so three distinct values.

But let me check if there are cases where the floor function could be -2? For example, suppose $ (-5/8)^x $ is less than -1? Is that possible? Let's see.

Since $ |(-5/8)^x| = (5/8)^x $, which is always between 0 and 1 for x > 0, because 5/8 < 1. Therefore, the magnitude is less than 1, so $ (-5/8)^x $ is between -1 and 1. Hence, the floor function cannot be less than -1, because the value is greater than -1. Similarly, for positive values, it's less than 1, so floor is 0. Except at x=0, where it is exactly 1. Therefore, floor can be 1, 0, or -1. Therefore, three distinct values.

Therefore, regardless of whether x is integer or non-integer, as long as the expression is real, the range of f(x) is { -1, 0, 1 }, so there are 3 distinct values.

Hence, the answer should be 3. But let me check once more with specific examples.

For example, take x = 1/3. Then $ (-5/8)^{1/3} = - (5/8)^{1/3} ≈ -0.855 $. Then floor(-0.855) = -1.

Take x = 1/5: $ (-5/8)^{1/5} ≈ -0.926 $, floor is -1.

Take x approaching 0 from the right, say x=0.1. If we could define it, but x=0.1 is not rational with odd denominator. Wait, 0.1 is 1/10, which has denominator 10, even. Therefore, it's not in the domain. However, for x = 1/3, which is allowed, gives -≈0.855. Similarly, for x approaching 0, rational numbers with odd denominator, say x=1/(2n+1). For example, x=1/3, 1/5, 1/7,... as n increases, the value $ (-5/8)^{x} $ approaches $ (-5/8)^0 = 1 $. But for x approaching 0 from the right, the value approaches 1. However, for example, take x=1/3: value ≈ -0.855, floor is -1. For x=1/5: ≈ -0.926, floor is -1. For x=1/7: ≈ -0.942, floor -1. As x approaches 0 from the right, for odd denominators, the values approach -1 from above? Or below?

Wait, let's compute $ (-5/8)^{x} $ where x approaches 0 from the right, say x=1/n with odd n. As n increases, x approaches 0. Let's compute for large n:

Let me take n approaching infinity, x = 1/n.

Then $ (-5/8)^{1/n} = (-1)^{1/n} \times (5/8)^{1/n} $. Since n is odd, (-1)^{1/n} = -1. Therefore, the value is - (5/8)^{1/n} ≈ - [1 - (1/n) ln(5/8) ] using the approximation (5/8)^{1/n} ≈ 1 - (ln(5/8))/n. Therefore, for large n, the value ≈ -1 + (ln(5/8))/n. Since ln(5/8) is negative, this becomes -1 + (negative)/n. Therefore, it's slightly less than -1? Wait, no:

Wait, ln(5/8) is negative, so (ln(5/8))/n is negative, so -1 + (negative) is less than -1? Wait, that would imply $ (-5/8)^{1/n} $ approaches -1 from below? But that contradicts because when x approaches 0, $ (-5/8)^x $ approaches 1, not -1.

Wait, there's a mistake here. Let me think again.

Wait, x = 1/n, which is positive, and we have $ (-5/8)^{x} $. For x = 1/n, n odd, this is equal to $ [(-5/8)^{1/n}] $. Which is equal to the n-th root of -5/8. Which is - (5/8)^{1/n} because n is odd. Therefore, the magnitude is (5/8)^{1/n}, which approaches 1 as n increases. Therefore, the value is approaching -1 from above (i.e., from the right side of -1). Therefore, for x approaching 0 from the right, $ (-5/8)^x $ approaches -1 from above? Wait, no. Wait, if n approaches infinity, x approaches 0, then (5/8)^{1/n} approaches 1 from below. Therefore, - (5/8)^{1/n} approaches -1 from above (i.e., -1 + something positive). Wait, but - (something approaching 1 from below) is approaching -1 from above? Wait, if something approaches 1 from below, like 0.9, 0.99, 0.999, etc., then -0.9, -0.99, -0.999 approaches -1 from above (i.e., from the right side of -1). Therefore, floor(-0.999) is -1, floor(-0.99) is -1, etc. Therefore, even as x approaches 0 from the right, the value approaches -1 from above, so floor is -1. Except when x = 0, where it is exactly 1.

Therefore, the function f(x) takes value 1 only at x=0, and -1 and 0 for all other x in the domain where the expression is defined. Hence, three distinct values: -1, 0, 1.

Therefore, the answer should be 3.

However, I need to check if there's any x for which the floor function equals something else. For example, if $ (-5/8)^x $ is exactly an integer. For example, is there an x such that $ (-5/8)^x = 0 $? No, because (5/8)^x is never zero. For it to be equal to 1, x has to be 0. For it to be -1, there's no x ≠ 0 satisfying this equation. Therefore, no.

Therefore, the only integers in the range are 1, 0, -1.

Hence, the number of distinct values in the range of f(x) is 3.

But let me check once more. For example, take x=1/2. Wait, 1/2 is not allowed, because denominator is even, so $ (-5/8)^{1/2} $ is not real. So, x=1/2 is not in the domain. How about x=2/3? Then x is rational with denominator 3 (odd). So 2/3 is allowed. $ (-5/8)^{2/3} = [(-5/8)^{1/3}]^2 = [ - (5/8)^{1/3} ]^2 = (5/8)^{2/3} ≈ (0.625)^{0.666} ≈ 0.625^(2/3). Let me compute that. Since 0.625 is 5/8, so (5/8)^(2/3) is equal to (5^(2/3))/(8^(2/3)) = (5^(2/3))/(4). Since 8^(2/3) is 4. 5^(2/3) is approximately 2.924. So 2.924 / 4 ≈ 0.731. Therefore, floor(0.731) = 0. So that's in the range. Similarly, for other x with even numerator: floor is 0. For odd numerator: floor is -1. So yes, same conclusion.

Therefore, the range has three distinct values: -1, 0, 1. Therefore, the answer is 3.

But let me check once again if there's any chance that for some x, the floor function could be something else. For example, suppose $ (-5/8)^x $ is between 0 and 1, but very close to 1. For example, if x approaches 0 from the right, but for x with odd denominator. Wait, earlier analysis shows that for x approaching 0 from the right (with odd denominator), $ (-5/8)^x $ approaches -1 from above, which is between -1 and 0, so floor is -1. For x approaching 0 from the right with even denominators, but those x are not in the domain. Wait, for even denominators, x is not in the domain. So, the only way to have x near 0 with the expression real is x with odd denominators, which approach -1 from above. Therefore, floor is -1. Only at x=0, it's 1.

Therefore, yes, only three distinct values.

Therefore, I think the answer is 3. But let me check again the initial problem statement to make sure I didn't miss anything.

Original problem: "Let $ f(x)=\left\lfloor\left(-\frac{5}{8}\right)^x\right\rfloor $ be a function that is defined for all values of $ x $ in [0, ∞) such that $ f(x) $ is a real number. How many distinct values exist in the range of $ f(x) $?"

So, the key here is that the function is defined for all x in [0, ∞) such that f(x) is real. Which implies that the expression $ (-5/8)^x $ is real. As discussed, this occurs when x is rational with odd denominator, but for the purpose of the problem, even though it's only defined on a subset of [0, ∞), the question is about the range. As shown, the range is { -1, 0, 1 }, so three distinct values.

Therefore, the answer is 3. Hence, \boxed{3} is the answer.

**Final Answer**
\boxed{3}

To determine how many **distinct values** exist in the range of the function
$$
f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor
$$
for $ x \in [0, \infty) $, we begin by understanding the domain and behavior of the function.

---

### **Understanding the Domain**

The function involves raising the negative rational number $ -\frac{5}{8} $ to a real power $ x $. This expression is not generally well-defined for all real values of $ x $, unless the exponent is rational with an **odd denominator**.

- For **integer** $ x $, the expression $ \left( -\frac{5}{8} \right)^x $ is real.
- For **non-integer rational numbers** $ x = \frac{p}{q} $, where $ q $ is odd, the expression is again real.
- If $ x $ is an **irrational number**, the expression becomes **complex** because $ \left( -\frac{5}{8} \right)^x $ cannot be interpreted in the real number system.

Given the problem's wording — particularly "defined for all values of $ x $ in [0, ∞) such that $ f(x) $ is a real number" — it is implied that **only those values of $ x $ for which the expression is real** are considered. This implies the function is **defined only at rational values of $ x $ with odd denominators**.

However, we may interpret **the function's range** as the **set of all attainable real-values** that the floor function can take, regardless of the countable domain. Therefore, we proceed by analyzing the function for all such valid values of $ x $ — both integer and non-integer rational with odd denominators — and determine the distinct values of $ f(x) $.

---

### **Analyzing the Expression**

Let’s analyze the function in two cases: when $ x $ is an **even** integer and when it is an **odd** integer. We will later consider **non-integer** rational values.

#### **Case 1: Even $ x $ — Positive powers**
- $ \left( -\frac{5}{8} \right)^x = \left( \frac{5}{8} \right)^x $
- Since $ \frac{5}{8} < 1 $, the expression decreases toward 0 as $ x \to \infty $
- For all such $ x $ (positive even integers or rational numbers with odd denominators that yield even exponents), the value $ \left( -\frac{5}{8} \right)^x $ lies strictly **between 0 and 1**, and thus:

$$
f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor = 0
$$

#### **Case 2: Odd $ x $ — Negative powers**
- $ \left( -\frac{5}{8} \right)^x = -\left( \frac{5}{8} \right)^x $
- Again, $ \left( \frac{5}{8} \right)^x \in (0, 1) $, so $ \left( -\frac{5}{8} \right)^x \in (-1, 0) $
- Therefore:

$$
f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor = -1
$$

#### **Special Case: $ x = 0 $**
- $ \left( -\frac{5}{8} \right)^0 = 1 $
- $ f(0) = \lfloor 1 \rfloor = 1 $

---

### **Determining the Range**

From the above, the function $ f(x) $ can attain the following **distinct values**:

- $ \boxed{1} $ (only when $ x = 0 $)
- $ \boxed{0} $ (when $ x $ is any even rational with odd denominator, i.e., nonnegative even powers)
- $ \boxed{-1} $ (when $ x $ is any odd rational with odd denominator, i.e., nonnegative odd powers)

No other integer values can be achieved:
- The value $ \left( -\frac{5}{8} \right)^x $ is **always between -1 and 1**, so the floor function cannot exceed -1 or drop below 0.
- The only integer in the interval $ (-1, 1) $ is 0, which we've already accounted for.

---

### **Conclusion**

Thus, the **range** of $ f(x) $ includes only **three** distinct values:

$$
\boxed{-1}, \boxed{0}, \boxed{1}
$$

Therefore, the number of **distinct values** in the range is:

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, let me try to figure out how to solve this problem. The function given is $ f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor $, and we need to find how many distinct values are in its range for $ x $ in $ [0, \infty) $. Alright, let's start by recalling what the floor function does. The floor of a number is the greatest integer less than or equal to that number. So, for example, $ \lfloor 1.2 \rfloor = 1 $, $ \lfloor -0.5 \rfloor = -1 $, etc.

First, I need to understand the behavior of the function $ \left( -\frac{5}{8} \right)^x $ for $ x \geq 0 $. Since the base is negative, and we're raising it to a real exponent, this might not be straightforward. Let me think about how this expression behaves for different values of $ x $.

Let me note that $ -\frac{5}{8} $ is a negative number between -1 and 0. When we raise a negative number to a real exponent, the result can be complex unless the exponent is an integer. But the problem states that $ f(x) $ is defined for all $ x \in [0, \infty) $ such that $ f(x) $ is a real number. So, perhaps the function is only defined when $ \left( -\frac{5}{8} \right)^x $ is real? That would happen when $ x $ is an integer? Wait, but the problem says it's defined for all $ x $ in [0, ∞). Hmm, maybe there's something I'm missing here.

Wait, actually, for real exponents, $ a^x $ is real only if $ a $ is positive or if $ a $ is negative and $ x $ is rational with an odd denominator. But since $ x $ is a real number here, unless the exponent is an integer, the expression might not be real. But the problem says that $ f(x) $ is defined for all $ x \in [0, \infty) $. So, perhaps there's an assumption that $ x $ is an integer? Wait, but the problem says "for all values of $ x $ in [0, ∞)", but maybe they mean that for each real number $ x $, $ \left( -\frac{5}{8} \right)^x $ is real. But that's not true for all real $ x $, unless $ x $ is an integer. So maybe there's a misunderstanding here.

Alternatively, perhaps the problem is considering $ x $ as a real number, and the expression is interpreted in some other way? Wait, maybe the problem is written incorrectly? Or maybe there's a different interpretation. Wait, maybe the problem is actually meant to have $ x $ as an integer? Because otherwise, the expression $ (-5/8)^x $ would not be real for non-integer real numbers. Let me check the original problem again.

It says: "Let $ f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor $ be a function that is defined for all values of $ x $ in $ [0, \infty) $ such that $ f(x) $ is a real number. How many distinct values exist in the range of $ f(x) $?"

So, the function is defined for all $ x \in [0, \infty) $, but only when $ \left( -\frac{5}{8} \right)^x $ is real. So, perhaps the expression is real for some values of $ x $, but not all? But how can that be? If $ x $ is allowed to be any real number in [0, ∞), then for non-integer $ x $, we can't define $ (-5/8)^x $ as a real number. Therefore, maybe the problem is actually intended to have $ x $ as integers? Because for integer $ x $, even if negative, the exponentiation is real. Let me check if the problem mentions anything about $ x $ being an integer. It doesn't. So, perhaps there's a mistake in the problem?

Alternatively, maybe the problem is considering complex numbers? But the floor function is only defined for real numbers. So, if the problem states that $ f(x) $ is a real number, then $ \left( -\frac{5}{8} \right)^x $ must be real, so perhaps the domain is only those $ x $ for which $ (-5/8)^x $ is real. So, that would be when $ x $ is an integer? Or when $ x $ is rational with an odd denominator?

But the problem says "defined for all values of $ x $ in [0, ∞)" which is confusing. Maybe it's a translation issue or a typo? Wait, perhaps the original problem is in Chinese? Or maybe the user made a typo? Alternatively, maybe the base is actually positive? Wait, maybe the user meant $ (-\frac{5}{8})^x $ but actually, for real exponents, that's problematic. Alternatively, maybe it's supposed to be $ (-\frac{5}{8})^x $ but only for integer values of $ x $, and the problem says "for all values of x in [0, ∞)" but actually, it's for all integers x in [0, ∞). But that's speculation.

Alternatively, maybe they consider $ x $ as real numbers and take the principal value? Wait, for example, complex exponentiation? Then, we can write $ (-\frac{5}{8})^x = e^{x \ln(-5/8)} $, but since $ \ln(-5/8) $ is complex, so $ (-\frac{5}{8})^x $ would be a complex number. But since the floor function is for real numbers, the problem must have $ \left( -\frac{5}{8} \right)^x $ real. Therefore, the problem must be considering $ x $ as integers? Otherwise, it's not real. So, perhaps the problem is intended for integer $ x $?

Alternatively, maybe $ x $ is a real number, but the expression is taken as the magnitude? Or maybe there is a different interpretation. Wait, perhaps the problem is written incorrectly, and it's supposed to be $ (-\frac{5}{8})^x $, but with absolute value? Let me check the original problem again.

Original problem says: $ f(x)=\left\lfloor\left(-\frac{5}{8}\right)^x\right\rfloor $. So, the floor of $ (-5/8)^x $. Given that, if the problem is supposed to have real outputs, maybe the domain is restricted to x such that $ (-5/8)^x $ is real. So, for real numbers, when is $ (-5/8)^x $ real?

Well, if $ x $ is rational with an odd denominator, then it's possible. However, since x is in [0, ∞), and the problem says "for all values of x in [0, ∞)", which is confusing. Maybe there's a misunderstanding here. Alternatively, perhaps x is allowed to be any real number, and the expression is defined through complex numbers, but then floor function would not be defined. So, I think there must be something wrong here.

Wait, maybe the problem is from a competition or textbook where they consider exponents with negative bases for real exponents? For example, if x is an integer, then it's defined. Otherwise, they might not. So, perhaps the problem is intended for integer x, and the wording is just off? Maybe "defined for all values of x in [0, ∞)" is supposed to mean that x is an integer in [0, ∞). But the original problem didn't specify. Hmm.

Alternatively, maybe the problem is in the context of a discrete math course where x is an integer? But without more context, it's hard to tell. Since the problem says x is in [0, ∞), which usually is a continuous interval, but given the floor function, maybe x is integer?

Alternatively, perhaps the problem is considering $ (-\frac{5}{8})^x $ as a complex number and then taking the floor of the real part or something? But that's not standard, and the problem says "f(x) is a real number", so perhaps not.

Alternatively, maybe the problem is using a different definition where for non-integer x, the expression is defined? For example, if x is a real number, then $ (-\frac{5}{8})^x $ could be defined as follows: For x rational with odd denominator, but for irrationals, it's not. However, since the problem says "for all values of x in [0, ∞)", which would require the function to be defined for all real numbers x ≥ 0, but since it's not defined for many real numbers, I think that can't be.

Therefore, maybe there's an error in the problem statement, and x is supposed to be an integer? Let's assume that, because otherwise, the problem doesn't make sense. So, let me proceed under that assumption. Let me check if that leads me to a reasonable answer. If x is an integer, then $ (-5/8)^x $ is real, and we can take the floor of it. So, maybe that's the intended problem. Let me try solving it with that assumption.

If x is an integer ≥ 0, then let's analyze $ (-5/8)^x $. For integer x:

- When x is even, $ (-5/8)^x = (5/8)^x $, which is positive.
- When x is odd, $ (-5/8)^x = -(5/8)^x $, which is negative.

So, for each integer x ≥ 0, let's compute $ f(x) $:

First, let me note that $ 5/8 $ is between 0 and 1. So, $ (5/8)^x $ decreases as x increases. For positive x, it's a positive number approaching 0 as x approaches infinity. For negative x, but since x is non-negative, we don't have that here.

So, for even x:

Let x = 2k, where k is a non-negative integer (since x ≥ 0). Then $ (-5/8)^x = (5/8)^{2k} $. So, that's positive, and decreasing as k increases. So, for even x, the value is positive, and approaching zero.

For odd x:

x = 2k + 1. Then $ (-5/8)^x = - (5/8)^{2k + 1} $. So, negative numbers approaching zero from below.

Therefore, the floor function of this would be:

For even x: floor(positive number between 0 and 1). Since $ (5/8)^x $ is between 0 and 1 for x ≥ 1, and for x = 0, it's 1.

Wait, let's compute for x = 0:

x = 0: $ (-5/8)^0 = 1 $, so floor(1) = 1.

x = 1: $ (-5/8)^1 = -5/8 ≈ -0.625 $, floor(-0.625) = -1.

x = 2: $ (-5/8)^2 = 25/64 ≈ 0.390625 $, floor(0.390625) = 0.

x = 3: $ (-5/8)^3 = -125/512 ≈ -0.244140625 $, floor(-0.244) = -1.

x = 4: $ (-5/8)^4 = 625/4096 ≈ 0.1525878906 $, floor(0.152) = 0.

x = 5: $ (-5/8)^5 ≈ -0.09765625 $, floor(-0.0976...) = -1.

x = 6: $ (-5/8)^6 ≈ 0.06103515625 $, floor(0.061) = 0.

x = 7: ≈ -0.03814697265, floor(-0.038) = -1.

x = 8: ≈ 0.0244140625, floor(0.024) = 0.

x = 9: ≈ -0.01525878906, floor(-0.015) = -1.

And so on.

So, for even x ≥ 0:

At x = 0: floor(1) = 1

For x even and ≥ 2: floor( (5/8)^{even} ) which is between 0 and 1, so floor is 0.

For odd x ≥ 1: floor( - (5/8)^{odd} ) which is between -1 and 0, so floor is -1.

Therefore, the possible values of f(x) are 1, 0, and -1. Wait, so does that mean the range has three distinct values? But let me check for x=0: yes, 1. For even x ≥2: 0. For odd x ≥1: -1. So, that would be three distinct values: 1, 0, -1. So, the answer would be 3?

Wait, but wait, let me check if there are more possibilities. Wait, for example, when x=0, the value is 1. For x=1, it's -1. For x=2, 0. For x=3, -1. For x=4, 0. So yes, only 1, 0, -1.

But wait, let me check for x=0: $ (-5/8)^0 = 1 $. So floor(1) = 1. Correct.

For x=1: -5/8 is -0.625, floor is -1.

x=2: 25/64 ≈ 0.390, floor is 0.

x=3: -125/512 ≈ -0.244, floor is -1.

x=4: 625/4096 ≈ 0.152, floor is 0.

x=5: -5^5 / 8^5 = -3125 / 32768 ≈ -0.095, floor is -1.

Similarly, as x increases, the value approaches zero from the positive and negative sides, but since the magnitude is decreasing, the floor function remains 0 or -1, as we saw. So, for even x ≥ 2, floor is 0, for odd x ≥ 1, floor is -1. And for x=0, floor is 1. So, there are three distinct values: 1, 0, -1. Therefore, the number of distinct values is 3. But wait, the problem says "for all values of x in [0, ∞)" but if x is only integer, then the answer is 3. However, maybe there's more?

Wait, but let me think again. What if x is not an integer? Suppose, for example, x is a non-integer real number. But then, as per earlier, $ (-5/8)^x $ is not real. However, if we assume that x is allowed to be real, but the problem says that f(x) is real, then maybe x is restricted to integers? But since the problem says x is in [0, ∞), which is a continuous interval, but maybe the question assumes that x is integer? Or perhaps there is a different interpretation.

Alternatively, maybe the problem considers $ (-5/8)^x $ as a complex number and takes the floor of its real part? But that's not standard. Alternatively, maybe it's absolute value? But the original problem didn't specify that. Alternatively, maybe it's misprinted and the base is positive? For example, $ (5/8)^x $, which would be straightforward. But the original problem says -5/8.

Alternatively, maybe the base is positive, and there's a typo. But assuming the problem is correct, I need to figure out. Wait, maybe the problem actually allows x to be any real number, and for the expression to be real, x must be rational with odd denominator? But then the range might have more values. However, the problem states that f(x) is defined for all x in [0, ∞). But I don't think that's feasible.

Alternatively, maybe the problem is in a context where complex numbers are considered, but floor function is applied to the real part? Or something like that? But without more context, it's hard to tell.

Alternatively, perhaps the problem is considering that for real exponents, $ (-5/8)^x $ is defined by taking the modulus and then applying the exponent, so like $ |(-5/8)|^x \times e^{i\pi x} $, but then taking the real part? But that seems too speculative.

Alternatively, maybe the problem is using the definition of real exponents for negative bases only when x is rational with odd denominator, but then for x in [0, ∞), the function is defined only at those rationals. However, since the problem says "defined for all values of x in [0, ∞)", that's conflicting. Unless the problem is considering x as integers. Since otherwise, it's not defined for all real numbers in [0, ∞).

Given that, perhaps the problem has a typo, and it's supposed to be $ (5/8)^x $ instead of $ (-5/8)^x $. If that were the case, then the function would be $ f(x) = \lfloor (5/8)^x \rfloor $, and we can analyze it. But since the problem has a negative sign, maybe there's a different approach. Wait, maybe the problem is expecting to interpret $ (-\frac{5}{8})^x $ as $ (-1)^x \times (\frac{5}{8})^x $, and then considering the floor function for real x? But then, unless x is an integer, (-1)^x is not real.

But if we assume that x is real, then $ (-1)^x $ can be written as $ e^{i\pi x} $, which is complex. Therefore, multiplying by $ (\frac{5}{8})^x $, which is real, the whole expression is complex. Hence, the floor function is not defined. Therefore, unless the problem is only considering x integer, which is the only way for $ (-\frac{5}{8})^x $ to be real.

So, assuming that the problem is for integer x, then we can proceed as before, getting three distinct values: 1, 0, -1.

But let me confirm again. Let's check x=0: floor(1) = 1. For x=1: floor(-5/8) = -1. For x=2: floor(25/64)=0. For x=3: floor(-125/512)= -1. For x=4: floor(625/4096)=0. Similarly, for higher x, it alternates between -1 and 0. So the range is { -1, 0, 1 }, so three distinct values. Hence, the answer is 3? But let me check if maybe for some x, the floor function gives something else.

Wait, for example, suppose x is a non-integer. If we assume x is a non-integer, but the problem says "defined for all values of x in [0, ∞)", which is conflicting. Therefore, if we take x to be integers, then the answer is 3. However, the problem says "for all values of x in [0, ∞)", which might mean that the function is defined for all real numbers x ≥ 0. But unless there's another way to interpret the expression.

Alternatively, maybe the problem is in a context where the exponent is allowed to be any real number, but the base is interpreted in a different way? For example, maybe the expression is defined as $ (-1)^x \times (5/8)^x $, which is complex, but taking the real part? Then, applying floor function to the real part? But that is highly non-standard. Alternatively, if the problem is in a context where the expression is defined as the magnitude, so $ |(-5/8)^x| $, which would be $ (5/8)^x $, then floor of that. Then, the function would be $ \lfloor (5/8)^x \rfloor $. But then the negative sign is irrelevant. However, the original problem has the negative sign. So, given that, I think this is a stretch.

Alternatively, perhaps the floor function is applied to the absolute value? But again, the problem doesn't specify that. Given that, I think the only way for the problem to make sense is to assume that x is an integer, leading to three distinct values.

However, maybe I need to check if there are more values. Wait, for example, let me suppose that x is a non-integer, but the expression is still real. For example, take x=1/2. Then $ (-5/8)^{1/2} $ is sqrt(-5/8), which is imaginary. Similarly, for x=1/3, $ (-5/8)^{1/3} $ is real because cube root of negative number is negative. So, for x=1/3, it's real. Similarly, for x=2/3, it's $ [(-5/8)^{1/3}]^2 $, which is positive. Similarly, for x= p/q in reduced form, if q is odd, then it's real, else complex.

Therefore, if x is a rational number with odd denominator, then $ (-5/8)^x $ is real. If x is irrational, it's not real. Therefore, the domain of f(x) is the set of real numbers x ≥ 0 for which x can be expressed as a rational number with odd denominator. But the problem says "defined for all values of x in [0, ∞)" such that f(x) is real. Therefore, the domain is not all of [0, ∞), but only those x which are rational numbers with odd denominator. So, for example, x=1/3, x=2/3, x=1, x=1/5, etc. However, the problem says "defined for all values of x in [0, ∞)", which is confusing. Maybe it's a mistranslation or misstatement, and it actually means that f(x) is defined for all x in [0, ∞) where the expression is real? But even so, the function would only be defined at countably many points, which is not the whole interval.

Given that, if we consider x as real numbers where $ (-5/8)^x $ is real, which would be x rational with odd denominator, then the function f(x) is defined at those points. However, analyzing the range would be complicated because there are infinitely many such x. But maybe the floor function takes on only finitely many values? Let me try.

First, consider x positive real numbers. Let me think about x > 0. For x rational with odd denominator, say x = p/q, where q is odd. Then:

If x is rational, then $ (-5/8)^x = (-1)^{p/q} \times (5/8)^{p/q} $. Since q is odd, (-1)^{p/q} = -1^{p/q} = -1^{p} \times (-1)^{something}... Wait, no. Actually, (-1)^{p/q} is equal to the q-th root of (-1)^p. Since p/q is a rational number. If p is even, (-1)^p = 1, so the q-th root of 1 is 1. If p is odd, (-1)^p = -1, so the q-th root of -1 is -1 since q is odd. Therefore, $ (-5/8)^x = (-1)^{p/q} \times (5/8)^{p/q} = (-1)^{p} \times (5/8)^{p/q} $. Wait, let me check:

Wait, for example, if x = p/q, then $ (-5/8)^{p/q} = [(-5/8)^{1/q}]^p $. Since q is odd, (-5/8)^{1/q} is real and negative, so equal to - (5/8)^{1/q}. Therefore, raising that to the power p: [ - (5/8)^{1/q} ]^p = (-1)^p \times (5/8)^{p/q}.

Therefore, $ (-5/8)^x = (-1)^p \times (5/8)^{p/q} $.

So, depending on whether p is even or odd, the sign is positive or negative.

Therefore, for each such x = p/q, the value of $ (-5/8)^x $ is either positive or negative, depending on p's parity.

But how does this affect the floor function? Let me think. For example, if x is a positive rational number with odd denominator, then:

- If the exponent p is even, then $ (-5/8)^x $ is positive.
- If the exponent p is odd, then $ (-5/8)^x $ is negative.

Therefore, similar to the integer case. So, the floor function would depend on whether the value is positive or negative.

Additionally, the magnitude of $ (-5/8)^x $ is $ (5/8)^{x} $, which is a decreasing function of x, since 5/8 < 1.

Therefore, for x approaching infinity, $ (5/8)^x $ approaches 0. Therefore, the magnitude is approaching 0 from the positive side if x is even, or negative if x is odd.

Therefore, for positive x:

If x is a positive rational number with odd denominator:

- If x is such that the exponent p is even, then the value is positive, between 0 and 1 (since (5/8)^x is between 0 and 1 for x > 0). Therefore, floor(positive number between 0 and 1) is 0.
- If x is such that the exponent p is odd, then the value is negative, between -1 and 0 (since magnitude is between 0 and 1). Therefore, floor(negative number between -1 and 0) is -1.

However, when x = 0, the value is 1, so floor(1) = 1.

Therefore, similar to the integer case, except now x can be non-integer rational numbers with odd denominators. However, for these x, the value of $ (-5/8)^x $ is either between 0 and 1 (positive) or between -1 and 0 (negative), so the floor function is either 0 or -1. Except for x = 0, which gives floor(1) = 1.

Therefore, the range of f(x) would be { -1, 0, 1 }, same as before. Therefore, even if we consider all x with $ (-5/8)^x $ real (i.e., x rational with odd denominator), the floor function only takes on the values -1, 0, and 1. Therefore, the number of distinct values is 3.

But wait, is there a case where the value of $ (-5/8)^x $ is exactly an integer? For example, when is $ (-5/8)^x $ equal to 1, 0, or -1?

For the value to be 1: $ (-5/8)^x = 1 $. This occurs when x = 0 (as we saw), and also when (-5/8)^x = 1. For real x, this only occurs at x = 0 because any non-zero x would make it less than 1 in magnitude. Similarly, for the value to be -1: $ (-5/8)^x = -1 $. This requires that the magnitude is 1 and the sign is negative. The magnitude is 1 only when x = 0. But $ (-5/8)^0 = 1 $, not -1. So there's no x ≠ 0 for which $ (-5/8)^x = -1 $. Therefore, the floor function never gives -1 when the value is exactly -1, but only when it's between -1 and 0.

Therefore, the floor function can take values -1, 0, 1. Thus, three distinct values.

But wait, for x approaching 0 from the right? If x approaches 0+, then $ (-5/8)^x $ approaches 1, because any number to the power 0 is 1. Therefore, floor(1) = 1. However, if x is positive and approaching 0, then depending on whether x is rational with odd denominator or not, but if x approaches 0, then for small x, the value is near 1. For example, take x approaching 0+, then $ (-5/8)^x = e^{x \ln(-5/8)} $. But in real numbers, this requires x to be rational with odd denominator. For example, x = 1/n with odd n. Then $ (-5/8)^{1/n} $ is real. For x approaching 0+, these values approach 1. So floor of numbers approaching 1 from below (if x is such that the value is slightly less than 1) would be 0? Wait, no. Let me think.

Wait, if x is approaching 0 from the right, what is the value of $ (-5/8)^x $?

If x approaches 0+, then for even x (like x = 2/n where n is a positive integer), the value is $ (5/8)^{x} $, which approaches 1 from below. Therefore, floor( (something just less than 1) ) is 0. However, for x = 0, it's exactly 1, floor is 1. So, when x approaches 0 from the right, floor of the value is 0. However, at x=0, it's 1. Therefore, 1 is included in the range.

But for x approaching infinity, the value approaches 0 from the positive or negative side, so floor is 0 or -1.

Therefore, in the range of f(x), we have 1 (at x=0), 0 (for x > 0 with even exponents, or non-integer x where the value is between 0 and 1), and -1 (for x > 0 with odd exponents, or non-integer x where the value is between -1 and 0). Therefore, the range is { -1, 0, 1 }, so three distinct values.

But let me check if there are cases where the floor function could be -2? For example, suppose $ (-5/8)^x $ is less than -1? Is that possible? Let's see.

Since $ |(-5/8)^x| = (5/8)^x $, which is always between 0 and 1 for x > 0, because 5/8 < 1. Therefore, the magnitude is less than 1, so $ (-5/8)^x $ is between -1 and 1. Hence, the floor function cannot be less than -1, because the value is greater than -1. Similarly, for positive values, it's less than 1, so floor is 0. Except at x=0, where it is exactly 1. Therefore, floor can be 1, 0, or -1. Therefore, three distinct values.

Therefore, regardless of whether x is integer or non-integer, as long as the expression is real, the range of f(x) is { -1, 0, 1 }, so there are 3 distinct values.

Hence, the answer should be 3. But let me check once more with specific examples.

For example, take x = 1/3. Then $ (-5/8)^{1/3} = - (5/8)^{1/3} ≈ -0.855 $. Then floor(-0.855) = -1.

Take x = 1/5: $ (-5/8)^{1/5} ≈ -0.926 $, floor is -1.

Take x approaching 0 from the right, say x=0.1. If we could define it, but x=0.1 is not rational with odd denominator. Wait, 0.1 is 1/10, which has denominator 10, even. Therefore, it's not in the domain. However, for x = 1/3, which is allowed, gives -≈0.855. Similarly, for x approaching 0, rational numbers with odd denominator, say x=1/(2n+1). For example, x=1/3, 1/5, 1/7,... as n increases, the value $ (-5/8)^{x} $ approaches $ (-5/8)^0 = 1 $. But for x approaching 0 from the right, the value approaches 1. However, for example, take x=1/3: value ≈ -0.855, floor is -1. For x=1/5: ≈ -0.926, floor is -1. For x=1/7: ≈ -0.942, floor -1. As x approaches 0 from the right, for odd denominators, the values approach -1 from above? Or below?

Wait, let's compute $ (-5/8)^{x} $ where x approaches 0 from the right, say x=1/n with odd n. As n increases, x approaches 0. Let's compute for large n:

Let me take n approaching infinity, x = 1/n.

Then $ (-5/8)^{1/n} = (-1)^{1/n} \times (5/8)^{1/n} $. Since n is odd, (-1)^{1/n} = -1. Therefore, the value is - (5/8)^{1/n} ≈ - [1 - (1/n) ln(5/8) ] using the approximation (5/8)^{1/n} ≈ 1 - (ln(5/8))/n. Therefore, for large n, the value ≈ -1 + (ln(5/8))/n. Since ln(5/8) is negative, this becomes -1 + (negative)/n. Therefore, it's slightly less than -1? Wait, no:

Wait, ln(5/8) is negative, so (ln(5/8))/n is negative, so -1 + (negative) is less than -1? Wait, that would imply $ (-5/8)^{1/n} $ approaches -1 from below? But that contradicts because when x approaches 0, $ (-5/8)^x $ approaches 1, not -1.

Wait, there's a mistake here. Let me think again.

Wait, x = 1/n, which is positive, and we have $ (-5/8)^{x} $. For x = 1/n, n odd, this is equal to $ [(-5/8)^{1/n}] $. Which is equal to the n-th root of -5/8. Which is - (5/8)^{1/n} because n is odd. Therefore, the magnitude is (5/8)^{1/n}, which approaches 1 as n increases. Therefore, the value is approaching -1 from above (i.e., from the right side of -1). Therefore, for x approaching 0 from the right, $ (-5/8)^x $ approaches -1 from above? Wait, no. Wait, if n approaches infinity, x approaches 0, then (5/8)^{1/n} approaches 1 from below. Therefore, - (5/8)^{1/n} approaches -1 from above (i.e., -1 + something positive). Wait, but - (something approaching 1 from below) is approaching -1 from above? Wait, if something approaches 1 from below, like 0.9, 0.99, 0.999, etc., then -0.9, -0.99, -0.999 approaches -1 from above (i.e., from the right side of -1). Therefore, floor(-0.999) is -1, floor(-0.99) is -1, etc. Therefore, even as x approaches 0 from the right, the value approaches -1 from above, so floor is -1. Except when x = 0, where it is exactly 1.

Therefore, the function f(x) takes value 1 only at x=0, and -1 and 0 for all other x in the domain where the expression is defined. Hence, three distinct values: -1, 0, 1.

Therefore, the answer should be 3.

However, I need to check if there's any x for which the floor function equals something else. For example, if $ (-5/8)^x $ is exactly an integer. For example, is there an x such that $ (-5/8)^x = 0 $? No, because (5/8)^x is never zero. For it to be equal to 1, x has to be 0. For it to be -1, there's no x ≠ 0 satisfying this equation. Therefore, no.

Therefore, the only integers in the range are 1, 0, -1.

Hence, the number of distinct values in the range of f(x) is 3.

But let me check once more. For example, take x=1/2. Wait, 1/2 is not allowed, because denominator is even, so $ (-5/8)^{1/2} $ is not real. So, x=1/2 is not in the domain. How about x=2/3? Then x is rational with denominator 3 (odd). So 2/3 is allowed. $ (-5/8)^{2/3} = [(-5/8)^{1/3}]^2 = [ - (5/8)^{1/3} ]^2 = (5/8)^{2/3} ≈ (0.625)^{0.666} ≈ 0.625^(2/3). Let me compute that. Since 0.625 is 5/8, so (5/8)^(2/3) is equal to (5^(2/3))/(8^(2/3)) = (5^(2/3))/(4). Since 8^(2/3) is 4. 5^(2/3) is approximately 2.924. So 2.924 / 4 ≈ 0.731. Therefore, floor(0.731) = 0. So that's in the range. Similarly, for other x with even numerator: floor is 0. For odd numerator: floor is -1. So yes, same conclusion.

Therefore, the range has three distinct values: -1, 0, 1. Therefore, the answer is 3.

But let me check once again if there's any chance that for some x, the floor function could be something else. For example, suppose $ (-5/8)^x $ is between 0 and 1, but very close to 1. For example, if x approaches 0 from the right, but for x with odd denominator. Wait, earlier analysis shows that for x approaching 0 from the right (with odd denominator), $ (-5/8)^x $ approaches -1 from above, which is between -1 and 0, so floor is -1. For x approaching 0 from the right with even denominators, but those x are not in the domain. Wait, for even denominators, x is not in the domain. So, the only way to have x near 0 with the expression real is x with odd denominators, which approach -1 from above. Therefore, floor is -1. Only at x=0, it's 1.

Therefore, yes, only three distinct values.

Therefore, I think the answer is 3. But let me check again the initial problem statement to make sure I didn't miss anything.

Original problem: "Let $ f(x)=\left\lfloor\left(-\frac{5}{8}\right)^x\right\rfloor $ be a function that is defined for all values of $ x $ in [0, ∞) such that $ f(x) $ is a real number. How many distinct values exist in the range of $ f(x) $?"

So, the key here is that the function is defined for all x in [0, ∞) such that f(x) is real. Which implies that the expression $ (-5/8)^x $ is real. As discussed, this occurs when x is rational with odd denominator, but for the purpose of the problem, even though it's only defined on a subset of [0, ∞), the question is about the range. As shown, the range is { -1, 0, 1 }, so three distinct values.

Therefore, the answer is 3. Hence, \boxed{3} is the answer.

**Final Answer**
\boxed{3}

To determine how many **distinct values** exist in the range of the function
$$
f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor
$$
for $ x \in [0, \infty) $, we begin by understanding the domain and behavior of the function.

---

### **Understanding the Domain**

The function involves raising the negative rational number $ -\frac{5}{8} $ to a real power $ x $. This expression is not generally well-defined for all real values of $ x $, unless the exponent is rational with an **odd denominator**.

- For **integer** $ x $, the expression $ \left( -\frac{5}{8} \right)^x $ is real.
- For **non-integer rational numbers** $ x = \frac{p}{q} $, where $ q $ is odd, the expression is again real.
- If $ x $ is an **irrational number**, the expression becomes **complex** because $ \left( -\frac{5}{8} \right)^x $ cannot be interpreted in the real number system.

Given the problem's wording — particularly "defined for all values of $ x $ in [0, ∞) such that $ f(x) $ is a real number" — it is implied that **only those values of $ x $ for which the expression is real** are considered. This implies the function is **defined only at rational values of $ x $ with odd denominators**.

However, we may interpret **the function's range** as the **set of all attainable real-values** that the floor function can take, regardless of the countable domain. Therefore, we proceed by analyzing the function for all such valid values of $ x $ — both integer and non-integer rational with odd denominators — and determine the distinct values of $ f(x) $.

---

### **Analyzing the Expression**

Let’s analyze the function in two cases: when $ x $ is an **even** integer and when it is an **odd** integer. We will later consider **non-integer** rational values.

#### **Case 1: Even $ x $ — Positive powers**
- $ \left( -\frac{5}{8} \right)^x = \left( \frac{5}{8} \right)^x $
- Since $ \frac{5}{8} < 1 $, the expression decreases toward 0 as $ x \to \infty $
- For all such $ x $ (positive even integers or rational numbers with odd denominators that yield even exponents), the value $ \left( -\frac{5}{8} \right)^x $ lies strictly **between 0 and 1**, and thus:

$$
f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor = 0
$$

#### **Case 2: Odd $ x $ — Negative powers**
- $ \left( -\frac{5}{8} \right)^x = -\left( \frac{5}{8} \right)^x $
- Again, $ \left( \frac{5}{8} \right)^x \in (0, 1) $, so $ \left( -\frac{5}{8} \right)^x \in (-1, 0) $
- Therefore:

$$
f(x) = \left\lfloor \left( -\frac{5}{8} \right)^x \right\rfloor = -1
$$

#### **Special Case: $ x = 0 $**
- $ \left( -\frac{5}{8} \right)^0 = 1 $
- $ f(0) = \lfloor 1 \rfloor = 1 $

---

### **Determining the Range**

From the above, the function $ f(x) $ can attain the following **distinct values**:

- $ \boxed{1} $ (only when $ x = 0 $)
- $ \boxed{0} $ (when $ x $ is any even rational with odd denominator, i.e., nonnegative even powers)
- $ \boxed{-1} $ (when $ x $ is any odd rational with odd denominator, i.e., nonnegative odd powers)

No other integer values can be achieved:
- The value $ \left( -\frac{5}{8} \right)^x $ is **always between -1 and 1**, so the floor function cannot exceed -1 or drop below 0.
- The only integer in the interval $ (-1, 1) $ is 0, which we've already accounted for.

---

### **Conclusion**

Thus, the **range** of $ f(x) $ includes only **three** distinct values:

$$
\boxed{-1}, \boxed{0}, \boxed{1}
$$

Therefore, the number of **distinct values** in the range is:

$$
\boxed{3}
$$

Solution: Since $-\frac58$ is a negative number, $f(x)$ is only defined for integer values of $x$, and will alternate between positive and negative values. Additionally, $\left|-\frac58\right|< 1$, so $|f(x)|$ will continually decrease and approach 0 as $x$ increases in the interval $x\ge0$. Therefore, the largest positive value will occur at $x=0$, giving us the positive upper bound of $\left\lfloor\left(-\frac58\right)^0\right\rfloor=1$. The negative value that is greatest in magnitude then occurs at the next integer value of $x$: $x=1$, giving us the negative lower bound of $\left\lfloor\left(-\frac58\right)^1\right\rfloor=-1$. This tells us that $-1 \le f(x) \le 1$. Since the $f(x)$ must be an integer, the only possible distinct values contained in the range are -1, 0, and 1. This gives us a total of $\boxed{3}$ values of $f(x)$ when $x\ge0$.


Explanation:
Okay, so I need to figure out how many ways we can form a subcommittee of 5 members from a Senate committee that has 8 Republicans and 6 Democrats. The catch is that the subcommittee must have at least one member from each party. Hmm, let me think about how to approach this.

First, I remember that when dealing with combinations and restrictions, sometimes it's helpful to calculate the total number of possible subcommittees without any restrictions and then subtract the ones that don't meet the criteria. In this case, the restriction is having at least one member from each party. So maybe I should start by finding the total number of ways to choose 5 members from the entire committee, and then subtract the number of subcommittees that are all Republicans or all Democrats. That should leave me with the number of subcommittees that have at least one of each party. Let me check if this logic makes sense.

Total members in the committee: 8 Republicans + 6 Democrats = 14 members. So the total number of ways to choose any 5 members is the combination of 14 taken 5 at a time, which is written as C(14,5). Then, the number of subcommittees with all Republicans would be C(8,5), since we need to choose all 5 members from the Republicans. Similarly, the number of subcommittees with all Democrats would be C(6,5), since there are only 6 Democrats. Therefore, the number of subcommittees with at least one from each party should be C(14,5) - C(8,5) - C(6,5). Let me verify if this is correct.

Wait, but I need to make sure that subtracting those two cases doesn't exclude any other possibilities. Since the only subcommittees that don't meet the "at least one from each party" requirement are those that are all Republicans or all Democrats, right? So yes, subtracting those two cases from the total should give the correct number. Okay, that seems right. Let me compute these values.

First, calculate C(14,5). The formula for combinations is C(n,k) = n! / (k! * (n - k)!). So for C(14,5):

C(14,5) = 14! / (5! * 9!) = (14 × 13 × 12 × 11 × 10) / (5 × 4 × 3 × 2 × 1). Let me compute that. Let's see:

14 × 13 = 182

182 × 12 = 2184

2184 × 11 = 24024

24024 × 10 = 240240

Now divide by 5! which is 120.

So 240240 / 120. Let me divide step by step. 240240 divided by 10 is 24024, then divided by 12 is 24024 / 12. Let's see, 12 × 2000 = 24000, so 24024 - 24000 = 24. So 2000 + 24 / 12 = 2000 + 2 = 2002. So C(14,5) is 2002. Wait, is that right? Let me check another way. I know that C(14,5) is a standard combination. Let me recall that C(14,5) is 2002. Yes, that seems correct.

Now, C(8,5). Let me compute that. C(8,5) is equal to C(8,3) because of the combination identity C(n,k) = C(n, n - k). So C(8,3) is 8! / (3! * 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 56. Let me verify: 8 × 7 is 56, 56 × 6 is 336, divided by 6 is 56. Yes, correct. So C(8,5) = 56.

Next, C(6,5). Similarly, that's C(6,1) = 6. Because choosing 5 out of 6 is the same as leaving out 1. So that's 6. Therefore, the total number of subcommittees with all Republicans or all Democrats is 56 + 6 = 62.

Therefore, the number of subcommittees with at least one from each party is 2002 - 62 = 1940. Wait, 2002 - 62. Let me subtract that. 2002 - 60 is 1942, minus 2 is 1940. So the answer would be 1940? Let me think again if that is correct.

Alternatively, maybe I should approach this problem by considering the different possible distributions of Republicans and Democrats in the subcommittee. Since the subcommittee must have at least one from each party, the possible numbers of Republicans and Democrats can be:

- 1 Republican and 4 Democrats

- 2 Republicans and 3 Democrats

- 3 Republicans and 2 Democrats

- 4 Republicans and 1 Democrat

- 5 Republicans and 0 Democrats? Wait, no, because we need at least one from each party. So 5 Republicans would mean zero Democrats, which is invalid. Similarly, 0 Republicans and 5 Democrats is also invalid. Therefore, the valid splits are the ones where the number of Republicans is from 1 to 4, and Democrats accordingly 4 to 1. So we can calculate each case separately and add them up.

Let me try this method and see if I get the same answer. That might help confirm if my first approach was correct.

So, for 1 Republican and 4 Democrats: C(8,1) * C(6,4)

For 2 Republicans and 3 Democrats: C(8,2) * C(6,3)

For 3 Republicans and 2 Democrats: C(8,3) * C(6,2)

For 4 Republicans and 1 Democrat: C(8,4) * C(6,1)

Let me calculate each term:

First term: C(8,1) * C(6,4). C(8,1) is 8. C(6,4) is equal to C(6,2) = 15. So 8 * 15 = 120.

Second term: C(8,2) * C(6,3). C(8,2) is 28, as calculated earlier. Wait, let me verify. C(8,2) is 8*7 / 2 = 28. C(6,3) is 20. Because 6*5*4 / 6 = 20. So 28 * 20 = 560.

Third term: C(8,3) * C(6,2). C(8,3) is 56 (from earlier, since C(8,5) is 56, and C(8,3) is same as C(8,5)). Wait, no, earlier I had C(8,5) as 56, which is C(8,3) because 8-5=3. Wait, but here I need C(8,3). So 8*7*6/(3*2*1) = 56. Yes. C(6,2) is 15. So 56 * 15. Let me compute that. 56*10=560, 56*5=280, total 560+280=840.

Fourth term: C(8,4) * C(6,1). C(8,4) is 70. Because C(8,4) is 70 (I remember that C(8,4)=70). Let me check: 8*7*6*5 / 4*3*2*1 = (1680)/24 = 70. Yes. C(6,1)=6. So 70 * 6 = 420.

Now, adding all four terms: 120 + 560 + 840 + 420. Let me compute:

First, 120 + 560 = 680

680 + 840 = 1520

1520 + 420 = 1940

So that's the same answer as before, 1940. Therefore, that confirms it. So whether I subtract the invalid cases from the total, or I compute the valid cases by adding up the different distributions, I get 1940. Therefore, the answer should be 1940.

But let me just make sure that there's no miscalculation in my steps. Let me double-check each part.

First method:

Total subcommittees: C(14,5) = 2002. Then subtract the all-Republican and all-Democrat subcommittees. All-Republican: C(8,5) = 56. All-Democrat: C(6,5)=6. So 2002 - 56 - 6 = 2002 - 62 = 1940. Correct.

Second method:

Breaking into cases:

1R4D: 8 * 15 = 120. Correct.

2R3D: 28 * 20 = 560. Correct.

3R2D: 56 * 15 = 840. Correct.

4R1D: 70 * 6 = 420. Correct.

Total: 120 + 560 + 840 + 420 = 1940. Correct.

Therefore, both methods lead to the same answer, so I think it's safe to conclude that the number of ways is 1940. Let me think if there's another way someone might approach this problem incorrectly.

Another possible mistake could be miscalculating the combinations. For example, if someone forgot that C(6,4) is 15 instead of 10, they might get a wrong answer. But since I double-checked, I think that's not the case here. Also, ensuring that we don't have overlapping cases. But since the cases are distinct (different numbers of Republicans and Democrats), there's no overlap.

Alternatively, someone might think that the answer is C(8,5) + C(6,5), but that's actually the number of invalid subcommittees, which we subtract, not add.

Another way to think about it: if someone tried to use inclusion-exclusion incorrectly, but since we did it correctly both ways, I think we are okay.

So, yes, I think 1940 is the correct answer. Therefore, the number of ways is 1940.

**Final Answer**
The number of ways is \boxed{1940}.

To determine the number of ways to form a subcommittee of **5 members** from a Senate committee containing **8 Republicans** and **6 Democrats**, with the constraint that **at least one member must come from each party**, we can use combinatorics to solve the problem systematically.

---

### Step 1: Total Number of Subcommittees

The total number of ways to choose 5 members from 14 (since 8 + 6 = 14) is given by:

$$
\binom{14}{5} = \frac{14!}{5!(14-5)!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002
$$

---

### Step 2: Subtract Invalid Subcommittees

We must exclude the cases where **all 5 members are from one party**—either all Republicans or all Democrats.

- **All Republicans**: Since there are 8 Republicans, the number of ways to choose 5 is:

$$
\binom{8}{5} = \frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1} = 56
$$

- **All Democrats**: There are only 6 Democrats, so the number of ways to choose all 5 from them is:

$$
\binom{6}{5} = 6
$$

Adding these gives:

$$
56 + 6 = 62
$$

---

### Step 3: Calculate Valid Subcommittees

Subtract the invalid cases from the total:

$$
2002 - 62 = \boxed{1940}
$$

---

### Verification (Alternative Approach)

To verify, we can consider all the valid combinations of Republicans and Democrats such that **at least one member from each party is present**. The possible distributions are:

- 1R, 4D
- 2R, 3D
- 3R, 2D
- 4R, 1D

We calculate each and add the results:

- $ \binom{8}{1} \binom{6}{4} = 8 \times 15 = 120 $
- $ \binom{8}{2} \binom{6}{3} = 28 \times 20 = 560 $
- $ \binom{8}{3} \binom{6}{2} = 56 \times 15 = 840 $
- $ \binom{8}{4} \binom{6}{1} = 70 \times 6 = 420 $

Adding these gives:

$$
120 + 560 + 840 + 420 = 1940
$$

---

### Final Answer

$$
\boxed{1940}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways we can form a subcommittee of 5 members from a Senate committee that has 8 Republicans and 6 Democrats. The catch is that the subcommittee must have at least one member from each party. Hmm, let me think about how to approach this.

First, I remember that when dealing with combinations and restrictions, sometimes it's helpful to calculate the total number of possible subcommittees without any restrictions and then subtract the ones that don't meet the criteria. In this case, the restriction is having at least one member from each party. So maybe I should start by finding the total number of ways to choose 5 members from the entire committee, and then subtract the number of subcommittees that are all Republicans or all Democrats. That should leave me with the number of subcommittees that have at least one of each party. Let me check if this logic makes sense.

Total members in the committee: 8 Republicans + 6 Democrats = 14 members. So the total number of ways to choose any 5 members is the combination of 14 taken 5 at a time, which is written as C(14,5). Then, the number of subcommittees with all Republicans would be C(8,5), since we need to choose all 5 members from the Republicans. Similarly, the number of subcommittees with all Democrats would be C(6,5), since there are only 6 Democrats. Therefore, the number of subcommittees with at least one from each party should be C(14,5) - C(8,5) - C(6,5). Let me verify if this is correct.

Wait, but I need to make sure that subtracting those two cases doesn't exclude any other possibilities. Since the only subcommittees that don't meet the "at least one from each party" requirement are those that are all Republicans or all Democrats, right? So yes, subtracting those two cases from the total should give the correct number. Okay, that seems right. Let me compute these values.

First, calculate C(14,5). The formula for combinations is C(n,k) = n! / (k! * (n - k)!). So for C(14,5):

C(14,5) = 14! / (5! * 9!) = (14 × 13 × 12 × 11 × 10) / (5 × 4 × 3 × 2 × 1). Let me compute that. Let's see:

14 × 13 = 182

182 × 12 = 2184

2184 × 11 = 24024

24024 × 10 = 240240

Now divide by 5! which is 120.

So 240240 / 120. Let me divide step by step. 240240 divided by 10 is 24024, then divided by 12 is 24024 / 12. Let's see, 12 × 2000 = 24000, so 24024 - 24000 = 24. So 2000 + 24 / 12 = 2000 + 2 = 2002. So C(14,5) is 2002. Wait, is that right? Let me check another way. I know that C(14,5) is a standard combination. Let me recall that C(14,5) is 2002. Yes, that seems correct.

Now, C(8,5). Let me compute that. C(8,5) is equal to C(8,3) because of the combination identity C(n,k) = C(n, n - k). So C(8,3) is 8! / (3! * 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 56. Let me verify: 8 × 7 is 56, 56 × 6 is 336, divided by 6 is 56. Yes, correct. So C(8,5) = 56.

Next, C(6,5). Similarly, that's C(6,1) = 6. Because choosing 5 out of 6 is the same as leaving out 1. So that's 6. Therefore, the total number of subcommittees with all Republicans or all Democrats is 56 + 6 = 62.

Therefore, the number of subcommittees with at least one from each party is 2002 - 62 = 1940. Wait, 2002 - 62. Let me subtract that. 2002 - 60 is 1942, minus 2 is 1940. So the answer would be 1940? Let me think again if that is correct.

Alternatively, maybe I should approach this problem by considering the different possible distributions of Republicans and Democrats in the subcommittee. Since the subcommittee must have at least one from each party, the possible numbers of Republicans and Democrats can be:

- 1 Republican and 4 Democrats

- 2 Republicans and 3 Democrats

- 3 Republicans and 2 Democrats

- 4 Republicans and 1 Democrat

- 5 Republicans and 0 Democrats? Wait, no, because we need at least one from each party. So 5 Republicans would mean zero Democrats, which is invalid. Similarly, 0 Republicans and 5 Democrats is also invalid. Therefore, the valid splits are the ones where the number of Republicans is from 1 to 4, and Democrats accordingly 4 to 1. So we can calculate each case separately and add them up.

Let me try this method and see if I get the same answer. That might help confirm if my first approach was correct.

So, for 1 Republican and 4 Democrats: C(8,1) * C(6,4)

For 2 Republicans and 3 Democrats: C(8,2) * C(6,3)

For 3 Republicans and 2 Democrats: C(8,3) * C(6,2)

For 4 Republicans and 1 Democrat: C(8,4) * C(6,1)

Let me calculate each term:

First term: C(8,1) * C(6,4). C(8,1) is 8. C(6,4) is equal to C(6,2) = 15. So 8 * 15 = 120.

Second term: C(8,2) * C(6,3). C(8,2) is 28, as calculated earlier. Wait, let me verify. C(8,2) is 8*7 / 2 = 28. C(6,3) is 20. Because 6*5*4 / 6 = 20. So 28 * 20 = 560.

Third term: C(8,3) * C(6,2). C(8,3) is 56 (from earlier, since C(8,5) is 56, and C(8,3) is same as C(8,5)). Wait, no, earlier I had C(8,5) as 56, which is C(8,3) because 8-5=3. Wait, but here I need C(8,3). So 8*7*6/(3*2*1) = 56. Yes. C(6,2) is 15. So 56 * 15. Let me compute that. 56*10=560, 56*5=280, total 560+280=840.

Fourth term: C(8,4) * C(6,1). C(8,4) is 70. Because C(8,4) is 70 (I remember that C(8,4)=70). Let me check: 8*7*6*5 / 4*3*2*1 = (1680)/24 = 70. Yes. C(6,1)=6. So 70 * 6 = 420.

Now, adding all four terms: 120 + 560 + 840 + 420. Let me compute:

First, 120 + 560 = 680

680 + 840 = 1520

1520 + 420 = 1940

So that's the same answer as before, 1940. Therefore, that confirms it. So whether I subtract the invalid cases from the total, or I compute the valid cases by adding up the different distributions, I get 1940. Therefore, the answer should be 1940.

But let me just make sure that there's no miscalculation in my steps. Let me double-check each part.

First method:

Total subcommittees: C(14,5) = 2002. Then subtract the all-Republican and all-Democrat subcommittees. All-Republican: C(8,5) = 56. All-Democrat: C(6,5)=6. So 2002 - 56 - 6 = 2002 - 62 = 1940. Correct.

Second method:

Breaking into cases:

1R4D: 8 * 15 = 120. Correct.

2R3D: 28 * 20 = 560. Correct.

3R2D: 56 * 15 = 840. Correct.

4R1D: 70 * 6 = 420. Correct.

Total: 120 + 560 + 840 + 420 = 1940. Correct.

Therefore, both methods lead to the same answer, so I think it's safe to conclude that the number of ways is 1940. Let me think if there's another way someone might approach this problem incorrectly.

Another possible mistake could be miscalculating the combinations. For example, if someone forgot that C(6,4) is 15 instead of 10, they might get a wrong answer. But since I double-checked, I think that's not the case here. Also, ensuring that we don't have overlapping cases. But since the cases are distinct (different numbers of Republicans and Democrats), there's no overlap.

Alternatively, someone might think that the answer is C(8,5) + C(6,5), but that's actually the number of invalid subcommittees, which we subtract, not add.

Another way to think about it: if someone tried to use inclusion-exclusion incorrectly, but since we did it correctly both ways, I think we are okay.

So, yes, I think 1940 is the correct answer. Therefore, the number of ways is 1940.

**Final Answer**
The number of ways is \boxed{1940}.

To determine the number of ways to form a subcommittee of **5 members** from a Senate committee containing **8 Republicans** and **6 Democrats**, with the constraint that **at least one member must come from each party**, we can use combinatorics to solve the problem systematically.

---

### Step 1: Total Number of Subcommittees

The total number of ways to choose 5 members from 14 (since 8 + 6 = 14) is given by:

$$
\binom{14}{5} = \frac{14!}{5!(14-5)!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002
$$

---

### Step 2: Subtract Invalid Subcommittees

We must exclude the cases where **all 5 members are from one party**—either all Republicans or all Democrats.

- **All Republicans**: Since there are 8 Republicans, the number of ways to choose 5 is:

$$
\binom{8}{5} = \frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1} = 56
$$

- **All Democrats**: There are only 6 Democrats, so the number of ways to choose all 5 from them is:

$$
\binom{6}{5} = 6
$$

Adding these gives:

$$
56 + 6 = 62
$$

---

### Step 3: Calculate Valid Subcommittees

Subtract the invalid cases from the total:

$$
2002 - 62 = \boxed{1940}
$$

---

### Verification (Alternative Approach)

To verify, we can consider all the valid combinations of Republicans and Democrats such that **at least one member from each party is present**. The possible distributions are:

- 1R, 4D
- 2R, 3D
- 3R, 2D
- 4R, 1D

We calculate each and add the results:

- $ \binom{8}{1} \binom{6}{4} = 8 \times 15 = 120 $
- $ \binom{8}{2} \binom{6}{3} = 28 \times 20 = 560 $
- $ \binom{8}{3} \binom{6}{2} = 56 \times 15 = 840 $
- $ \binom{8}{4} \binom{6}{1} = 70 \times 6 = 420 $

Adding these gives:

$$
120 + 560 + 840 + 420 = 1940
$$

---

### Final Answer

$$
\boxed{1940}
$$

Solution: There are a total of $\binom{14}{5}=2002$ ways of selecting a subcommittee of 5 with no restrictions on the membership. Of these committees, the only ones that will violate the given condition are the ones that consist entirely of Republicans or entirely of Democrats. There are $\binom{8}{5}=56$ possible subcommittees that have all 5 members selected from among the 8 Republicans and $\binom{6}{5}=6$ possible subcommittees that have all 5 members selected from among the 6 Democrats. Subtracting the number of subcommittees that don't work from the total number of possible subcommittees gives us our answer: $2002-56-6=\boxed{1940}$.