Explanation:
Okay, so I need to convert the binary number 10101001110₂ to its octal (base 8) equivalent. Hmm, let me think about how to approach this. I remember that converting binary to octal might be easier if I group the binary digits into sets of three, since 8 is 2³. Let me recall the exact method.

First, I think the process involves splitting the binary number into groups of three bits starting from the right. If the number of digits isn't a multiple of three, I might need to add leading zeros to the leftmost group. Then, each group of three bits can be converted to an octal digit. Let me try that.

The given binary number is 10101001110. Let me count the digits to see how many there are. Let's see: 1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. Since 11 divided by 3 is 3 with a remainder of 2, I need to add one leading zero to make it 12 digits, which is divisible by 3. Wait, let me check: 3 groups of 3 would be 9 digits, so 11 digits would need 2 more digits? Wait, no. Wait, if I have 11 digits, then starting from the right, I can split into groups of three. Let me write it down:

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me separate them into groups of three starting from the right. Let me count from the right:

Starting from the rightmost digit, which is the 0. Let me write the number with positions:

Positions (from right, starting at 0): Let's see, the number is 10101001110. Let me write it with spaces for grouping:

Let me write the number again: 1 0 1 0 1 0 0 1 1 1 0

Let me count from the right:

Starting from the right, the digits are:

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

Wait, maybe I should write the number as:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups of three starting from the right. Let me add leading zeros if necessary. Since there are 11 digits, which is 3*3 + 2, so we need to add one leading zero to make it 12 digits. Wait, 11 digits: adding one zero to the left would make it 12 digits. Let me check:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Adding a leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Now, grouping into three:

From left to right: 010 101 001 110

Wait, let me check:

Wait, if I add one leading zero, the number becomes 010101001110. Let me split into groups of three:

Starting from the left: 010 101 001 110. Wait, but that would be four groups. Let me check the number of digits: original was 11, adding one zero makes it 12, so 12 digits divided by 3 is 4 groups. So yes, four groups. Let me confirm:

Original number: 10101001110. Let me write it with the added zero: 010101001110. Now, splitting into groups of three:

First group (leftmost): 010

Second group: 101

Third group: 001

Fourth group: 110

Wait, but let me check the positions again. Let me write the number with the added zero:

0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three from the left:

Group 1: 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4: 1 1 0

Wait, but that seems like splitting from the left. However, usually, when converting binary to octal, you split from the right. Wait, maybe I should split from the right. Let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Starting from the right, let's split into groups of three:

Starting from the right, the last three digits are 1 1 0. Then moving left, the next three digits are 0 0 1. Then the next three digits are 0 1 0. Then the remaining digits are 1. Wait, but that leaves one digit. So we need to add leading zeros to make it three digits. Wait, so:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split from the right:

Group 1 (rightmost): 1 1 0

Group 2: 0 0 1

Group 3: 0 1 0

Group 4: 1 (needs two leading zeros to make three digits)

So adding two leading zeros to the leftmost group: 001

Wait, but then the groups would be:

Group 4: 001

Group 3: 010

Group 2: 001

Group 1: 110

Wait, but that seems different from splitting from the left. Which is correct?

I think splitting from the right is the correct approach. Let me confirm. Because when converting to octal, each group of three bits corresponds to an octal digit, starting from the right. So if the number of digits isn't a multiple of three, we add leading zeros to the leftmost group. So in this case, original number is 11 digits. So adding two leading zeros to make it 13 digits? Wait, no. Wait, 11 digits. To make it a multiple of three, we need to add 2 digits (since 11 + 2 = 13? Wait, no. Wait, 3*4=12, so 11 digits need 1 more digit to make 12. Wait, 11 + 1 = 12. So adding one leading zero. Wait, confusion here.

Wait, let me think again. Let me count the number of digits. The original number is 10101001110. Let me count:

1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. So 11 divided by 3 is 3 with a remainder of 2. Therefore, we need to add one leading zero to make it 12 digits. So adding one zero to the left. Therefore, the number becomes 010101001110. Now, splitting into groups of three from the left:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Wait, but if we split from the left, then each group is three bits. However, if we split from the right, we need to split into groups starting from the right. Let me check both ways.

If I split from the right, with the original number 10101001110, which is 11 digits. Let me write it as:

Positions (from right, starting at 0):

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

So grouping into three from the right:

Group 0: positions 0-2: 0,1,1 → bits 0,1,1 → which is 011? Wait, but positions are from right. Wait, maybe I need to think differently.

Alternatively, maybe it's easier to split the binary number into groups of three starting from the right, adding leading zeros if necessary. Let me try that.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate groups:

Starting from the right, every three digits:

Let me count from the right:

Starting at the end: the last three digits are 1 1 0. Then moving left, the next three digits are 0 0 1. Then the next three digits are 0 1 0. Then there is one digit left: 1. So we need to add two leading zeros to make that group three digits: 0 0 1. Wait, but adding two zeros to the left of the remaining digit?

Wait, let me write the number again:

Original: 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups:

Starting from the right:

Group 1: 1 1 0 (positions 10, 9, 8? Wait, maybe I need to index properly. Let me write the number with positions:

Let me index the digits from right to left, starting at 0:

Position 0: 0 (rightmost)

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1 (leftmost)

So grouping into three:

Group 0: positions 0-2: 0,1,1 → bits: 0 (position 0), 1 (position 1), 1 (position 2) → which is 011? Wait, but when grouping from the right, the first group is the rightmost three bits. So the rightmost three bits are positions 0,1,2: 0,1,1. Wait, but the original number is:

From left to right: 1 0 1 0 1 0 0 1 1 1 0

So from right to left, it's 0 1 1 1 0 0 1 0 1 0 1

Wait, maybe I need to reverse the number to make it easier. Let me reverse the binary number:

Original: 1 0 1 0 1 0 0 1 1 1 0

Reversed: 0 1 1 1 0 0 1 0 1 0 1

Now, grouping into three from the left (since reversed, grouping from the left corresponds to grouping from the right in original):

Group 1: 0 1 1

Group 2: 1 0 0

Group 3: 1 0 1

Group 4: 0 1

Wait, the reversed number is 0 1 1 1 0 0 1 0 1 0 1. Let me count the digits: 11 digits. So grouping into three:

First group: 0 1 1

Second group: 1 0 0

Third group: 1 0 1

Fourth group: 0 1

But the last group has only two digits. So we need to add a leading zero to make it three digits: 0 0 1. Wait, but adding a leading zero to the fourth group? Wait, the fourth group is the leftmost part of the reversed number. Wait, maybe I need to add leading zeros to the left of the reversed number? Wait, confusion again.

Alternatively, since the reversed number is 0 1 1 1 0 0 1 0 1 0 1, which is 11 digits. To make groups of three, starting from the left (which is the original right), we need to add one leading zero to make it 12 digits. So adding a zero at the beginning of the reversed number: 0 0 1 1 1 0 0 1 0 1 0 1. Now, grouping into three:

Group 1: 0 0 1

Group 2: 1 1 1

Group 3: 0 0 1

Group 4: 0 1 0

Group 5: 1 ?

Wait, no. Wait, adding one zero to the reversed number makes it 12 digits. Let me write it:

Original reversed: 0 1 1 1 0 0 1 0 1 0 1

Add one leading zero: 0 0 1 1 1 0 0 1 0 1 0 1

Now, grouping into three:

Group 1: 0 0 1

Group 2: 1 1 1

Group 3: 0 0 1

Group 4: 0 1 0

Group 5: 1 ?

Wait, no, 12 digits divided by 3 is 4 groups. Wait, 0 0 1 1 1 0 0 1 0 1 0 1:

Group 1: 0 0 1

Group 2: 1 1 1

Group 3: 0 0 1

Group 4: 0 1 0

Wait, but the last three digits are 0 1 0? Wait, let me check:

Digits:

Position 0: 0

Position 1: 0

Position 2: 1

Position 3: 1

Position 4: 1

Position 5: 0

Position 6: 0

Position 7: 1

Position 8: 0

Position 9: 1

Position 10: 0

Position 11: 1

Wait, no, maybe I need to index properly. Let me write the reversed number with added zero as:

[0][0][1][1][1][0][0][1][0][1][0][1]

So positions 0 to 11:

Group 1: positions 0-2: 0,0,1 → 001

Group 2: positions 3-5: 1,1,0 → 110

Group 3: positions 6-8: 0,1,0 → 010

Group 4: positions 9-11: 1,0,1 → 101

Wait, that seems better. So the groups are:

Group 1: 001

Group 2: 110

Group 3: 010

Group 4: 101

But remember, this is the reversed number. So when we convert these groups to octal digits, we need to reverse the order of the groups because we reversed the original number. Wait, let me think.

Original number was reversed to make grouping easier. So the groups we found in the reversed number are:

Group 1 (original rightmost three bits): 001

Group 2: 110

Group 3: 010

Group 4: 101

But since we reversed the original number, the group order is reversed. So the original groups, from left to right, would be Group 4, Group 3, Group 2, Group 1. Wait, no. Wait, when we reversed the original number, the first group in the reversed number corresponds to the rightmost group in the original number. Therefore, the octal digits, when converted, should be ordered as Group 4, Group 3, Group 2, Group 1. Wait, maybe not. Let me think.

Let me take an example. Suppose the original number is, say, 1010. Reversed is 0101. Grouping into three, adding a leading zero: 00101. Wait, maybe this is getting too confusing. Let me try a different approach.

Alternatively, maybe I should just convert the binary number to decimal first, then to octal. But that might be more work. However, since the question is about binary to octal, the grouping method is more straightforward. Let me try again.

Original binary number: 10101001110. Let me write it as:

1 0 1 0 1 0 0 1 1 1 0

Let me split into groups of three starting from the right. Since there are 11 digits, adding one leading zero to make it 12 digits:

0 1 0 1 0 1 0 0 1 1 1 0

Now, grouping into three:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Wait, but if I split from the left, then the groups are 010, 101, 001, 110. Then converting each group to octal:

010₂ = 2₈

101₂ = 5₈

001₂ = 1₈

110₂ = 6₈

Therefore, combining them: 2 5 1 6, so 2516₈?

Wait, but let me check if this is correct. Let me verify by converting the original binary number to decimal and then to octal.

First, let me convert 10101001110₂ to decimal.

To convert binary to decimal, we can calculate the value of each bit:

Starting from the right, the positions are 0 to 10.

The binary number is:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, let me write it properly:

The binary number is 1 0 1 0 1 0 0 1 1 1 0

So positions from right (position 0) to left (position 10):

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, actually, position numbering is usually from right, starting at 0. So let me write the binary number with positions:

Let me write the binary number as:

From left to right: positions 10 to 0:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, but that seems confusing. Let me write the binary number as:

Digit positions (from right, starting at 0):

Let me reverse the binary number to make it easier. Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Reversed: 0 1 1 1 0 0 1 0 1 0 1

Now, positions 0 to 10:

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

So the decimal value is sum of (bit * 2^position):

So:

Position 10: 1 * 2^10 = 1024

Position 9: 0 * 2^9 = 0

Position 8: 1 * 2^8 = 256

Position 7: 0 * 2^7 = 0

Position 6: 1 * 2^6 = 64

Position 5: 0 * 2^5 = 0

Position 4: 0 * 2^4 = 0

Position 3: 1 * 2^3 = 8

Position 2: 1 * 2^2 = 4

Position 1: 1 * 2^1 = 2

Position 0: 0 * 2^0 = 0

Adding these up:

1024 + 256 + 64 + 8 + 4 + 2 = Let's compute step by step:

1024 + 256 = 1280

1280 + 64 = 1344

1344 + 8 = 1352

1352 + 4 = 1356

1356 + 2 = 1358

So the decimal value is 1358.

Now, converting 1358 decimal to octal.

To convert decimal to octal, divide by 8 and find remainders.

Let me do that:

1358 ÷ 8 = 169.75 → quotient 169, remainder 1358 - 169*8 = 1358 - 1352 = 6

169 ÷ 8 = 21.125 → quotient 21, remainder 169 - 21*8 = 169 - 168 = 1

21 ÷ 8 = 2.625 → quotient 2, remainder 21 - 2*8 = 21 - 16 = 5

2 ÷ 8 = 0.25 → quotient 0, remainder 2

So writing the remainders from last to first: 2, 5, 1, 6 → 2516₈

So that matches the earlier result from the binary grouping method. Therefore, the octal equivalent is 2516₈.

But let me check again the grouping method. Earlier, when I split the binary number into groups of three starting from the left after adding a leading zero, I had groups 010, 101, 001, 110 which converted to 2, 5, 1, 6, so 2516. Which matches the decimal conversion. Therefore, that seems correct.

But let me check again the grouping process. Original binary number: 10101001110. Adding a leading zero: 010101001110. Split into groups of three:

010 101 001 110 → 2 5 1 6 → 2516. Correct.

Alternatively, if I had split from the right without adding leading zeros, but that might have been confusing. However, since both methods gave me the same result, I can be confident that 2516 is the correct octal equivalent.

But just to make sure, let me try another way. Let me convert each group of three bits to octal digits step by step.

Original binary number: 10101001110. Let me split into groups of three from the right, adding leading zeros if necessary.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups:

Starting from the right:

First group (rightmost three bits): 1 1 0 → 6

Next group: 0 0 1 → 1

Next group: 0 1 0 → 2

Next group: 1 (remaining digit) → add two leading zeros: 0 0 1 → 1

Wait, but that would give groups: 001, 010, 001, 110? Wait, no. Wait, let me do it properly.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split from the right:

Starting at the end:

Group 1: last three digits: 1 1 0 → 6

Group 2: previous three digits: 0 0 1 → 1

Group 3: previous three digits: 0 1 0 → 2

Group 4: remaining digits: 1 → add two leading zeros: 0 0 1 → 1

So the octal digits are Group 4, Group 3, Group 2, Group 1: 1, 2, 1, 6 → 1216? Wait, that contradicts the previous result. Wait, what's happening here?

Wait, this is different. Earlier, when I added a leading zero to the left, I got 2516, but splitting from the right and adding leading zeros to the leftmost group gives me 1 2 1 6? Which is 1216? But that conflicts with the decimal conversion which gave 2516. So there must be a mistake here.

Wait, let me check again. If I split from the right:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Let me write the number with spaces to separate groups:

Starting from the right:

Group 1: 1 1 0

Group 2: 0 0 1

Group 3: 0 1 0

Group 4: 1

So Group 4 is just '1', which needs two leading zeros to make it three bits: 0 0 1.

Therefore, the groups are:

Group 4: 001 → 1

Group 3: 010 → 2

Group 2: 001 → 1

Group 1: 110 → 6

Therefore, the octal digits are Group 4, Group 3, Group 2, Group 1: 1 2 1 6 → 1216? But this contradicts the decimal conversion. So which one is correct?

Wait, but earlier when I converted the binary number to decimal, I got 1358, which is 2516 in octal. So there must be an error in this grouping.

Wait, let me check the grouping again. If I split from the right, adding leading zeros to the leftmost group, then the groups should be:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups from the right:

Group 1 (rightmost): 1 1 0 → 6

Group 2: 0 0 1 → 1

Group 3: 0 1 0 → 2

Group 4: 1 → add two leading zeros: 0 0 1 → 1

But when we write the octal number, the groups are ordered from left to right as Group 4, Group 3, Group 2, Group 1. Wait, but Group 4 is the leftmost group. Wait, no. Wait, if we split from the right, the first group (Group 1) is the rightmost, then Group 2 is to the left of Group 1, etc. Therefore, the order of the octal digits should be Group 4, Group 3, Group 2, Group 1. So 1 (Group 4), 2 (Group 3), 1 (Group 2), 6 (Group 1) → 1216. But this conflicts with the decimal conversion. Therefore, there must be a mistake in this grouping.

Alternatively, maybe I messed up the grouping. Let me write the original number again and split it properly.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate groups of three starting from the right:

Starting from the right, every three bits:

Let me count the digits:

Positions (from right, 0 to 10):

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

So grouping into three:

Group 0: positions 0-2: 0,1,1 → bits: 0 (pos0), 1 (pos1), 1 (pos2) → which is 011 → 3? Wait, but when grouping from the right, the first group is the rightmost three bits. However, the bits are:

From right to left: positions 0-10:

0 (pos0), 1 (pos1), 1 (pos2), 1 (pos3), 0 (pos4), 0 (pos5), 1 (pos6), 0 (pos7), 1 (pos8), 0 (pos9), 1 (pos10)

So grouping from the right:

Group 1 (positions 0-2): 0,1,1 → bits: 0,1,1 → which is 011₂ = 3₈

Group 2 (positions 3-5): 1,0,0 → bits: 1,0,0 → 100₂ = 4₈

Group 3 (positions 6-8): 1,0,1 → bits: 1,0,1 → 101₂ = 5₈

Group 4 (positions 9-10): 0,1 → need to add one leading zero: 0,0,1 → bits: 0,0,1 → 001₂ = 1₈

Wait, so the groups are:

Group 1: 011 (3)

Group 2: 100 (4)

Group 3: 101 (5)

Group 4: 001 (1)

But since we grouped from the right, the order of the octal digits is Group 4, Group 3, Group 2, Group 1. Wait, no. Wait, Group 1 is the rightmost group, Group 2 is next, etc. So the octal number is Group 4 Group 3 Group 2 Group 1: 1 5 4 3 → 1543? But that conflicts with previous results. Wait, this is getting more confusing.

Wait, let me think again. When you split the binary number into groups of three starting from the right, each group corresponds to an octal digit, and the order of the groups is from right to left. Wait, no. Wait, if you split from the right, the first group (rightmost) is the least significant digit in octal, and the next group is the next digit, etc. Therefore, the octal number is constructed by writing the groups from left to right, starting with the leftmost group (which is the most significant octal digit). Wait, this is confusing.

Let me take an example. Let's say the binary number is 1101101. Let's convert it to octal.

Split into groups of three from the right: 1 101 101. Wait, adding leading zeros if necessary. Wait, original number is 1101101 (7 digits). Split into groups of three from the right: 1 101 101. Wait, but that's 1, 101, 101. Wait, but the first group has only one digit. So add two leading zeros: 001, 101, 101. Then converting each group: 001=1, 101=5, 101=5. So octal number is 155. Let me check by converting to decimal: 1101101₂ = 1*64 + 1*32 + 0*16 + 1*8 + 1*4 + 0*2 + 1*1 = 64 + 32 + 8 + 4 + 1 = 109. 155₈ = 1*64 + 5*8 + 5*1 = 64 + 40 + 5 = 109. Correct. So in this case, the groups were 001, 101, 101, which correspond to octal digits 1, 5, 5, so the octal number is 155. So the order is left to right, with the leftmost group being the most significant.

Therefore, applying this to our original problem:

Original binary number: 10101001110. Split into groups of three from the right, adding leading zeros if necessary.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups:

Starting from the right:

Group 1 (rightmost three bits): 1 1 0 → 6

Group 2: next three bits: 0 0 1 → 1

Group 3: next three bits: 0 1 0 → 2

Group 4: remaining bits: 1 → add two leading zeros: 0 0 1 → 1

So the groups are Group 4, Group 3, Group 2, Group 1: 1, 2, 1, 6 → 1216. But earlier decimal conversion gave 2516. So there is a contradiction here. What is wrong?

Wait, let me check the decimal value again. Earlier, I converted the binary number 10101001110 to decimal and got 1358. Let me verify that again.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it as:

From left to right: positions 10 to 0:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Calculating decimal value:

1*2^10 + 0*2^9 + 1*2^8 + 0*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 0*2^0

= 1024 + 0 + 256 + 0 + 64 + 0 + 0 + 8 + 4 + 2 + 0

= 1024 + 256 = 1280

1280 + 64 = 1344

1344 + 8 = 1352

1352 + 4 = 1356

1356 + 2 = 1358. Correct.

Now, converting 1358 to octal:

1358 ÷ 8 = 169.75 → quotient 169, remainder 6

169 ÷ 8 = 21.125 → quotient 21, remainder 1

21 ÷ 8 = 2.625 → quotient 2, remainder 5

2 ÷ 8 = 0.25 → quotient 0, remainder 2

Reading remainders from last to first: 2, 5, 1, 6 → 2516. Correct.

But according to the grouping method, splitting from the right gave me 1216. Which is different. So there must be an error in the grouping.

Wait, let me try splitting the binary number into groups of three from the right again, carefully.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate groups:

Starting from the right, every three bits:

Let me count the digits from the right:

Positions (from right, 0 to 10):

0: 0

1: 1

2: 1

3: 1

4: 0

5: 0

6: 1

7: 0

8: 1

9: 0

10: 1

Group 1 (positions 0-2): bits at positions 0,1,2: 0,1,1 → 011₂ = 3₈

Group 2 (positions 3-5): bits at positions 3,4,5: 1,0,0 → 100₂ = 4₈

Group 3 (positions 6-8): bits at positions 6,7,8: 1,0,1 → 101₂ = 5₈

Group 4 (positions 9-10): bits at positions 9,10: 0,1 → need to add one leading zero: 0,0,1 → 001₂ = 1₈

So the groups are Group 4, Group 3, Group 2, Group 1: 1, 5, 4, 3 → 1543₈. But this is different from both previous results. Now I'm really confused.

Wait, but if I convert 1543₈ to decimal, what do I get?

1*8^3 + 5*8^2 + 4*8 + 3 = 512 + 320 + 32 + 3 = 512 + 320 = 832 + 32 = 864 + 3 = 867. Which is not equal to 1358. So clearly, this is wrong.

But earlier grouping from the left after adding a leading zero gave me 2516, which is correct. So why is there a discrepancy?

Ah! I think the problem is in how I'm grouping the bits. When splitting from the right, the groups should be ordered from left to right as Group 1, Group 2, Group 3, Group 4, but I think I mixed up the order.

Wait, let me think again. When you split the binary number into groups of three from the right, the first group (rightmost) is the least significant octal digit, and the next group is the next digit, etc. So the octal number is constructed by writing the groups from left to right, starting with the leftmost group (which is the most significant octal digit). Wait, no. Wait, if you split from the right, the first group (rightmost) is the least significant digit. So the order of the octal digits is Group 1 (rightmost), Group 2, Group 3, Group 4 (leftmost). So to write the octal number, you need to reverse the order of the groups.

For example, in the earlier example of 1101101₂, splitting into groups from the right gave us groups 001, 101, 101. But since the first group is the least significant, the octal number is 1 (from 001) 5 (from 101) 5 (from 101), which is 155. Wait, but the groups were 001 (LSB), 101, 101 (MSB). So the octal number is written as MSB to LSB: 101 101 001 → 155. Wait, no. Wait, the groups are 001 (LSB), 101, 101 (MSB). So the octal number is 101 101 001, but written as 155. So the order is from MSB to LSB: the first group (from the left) is the MSB. Wait, this is confusing.

Let me think of the binary number as being split into groups, and each group corresponds to an octal digit. The rightmost group is the least significant octal digit, and the leftmost group is the most significant. Therefore, when you split from the right, you get the groups in the order of least significant to most significant. Therefore, to write the octal number, you need to reverse the order of the groups.

So in the example of 1101101₂, splitting from the right gives groups: 001, 101, 101. Reversing the order gives 101, 101, 001 → which is 155₈. Correct.

Applying this to our original problem:

Groups from splitting from the right: Group 1 (LSB): 011 (3), Group 2: 100 (4), Group 3: 101 (5), Group 4 (MSB): 001 (1). Reversing the order of the groups gives Group 4, Group 3, Group 2, Group 1: 1, 5, 4, 3 → 1543₈. But this is not matching the decimal conversion. So something is wrong.

Wait, but earlier when I added a leading zero to the left and split into groups from the left, I got 010 101 001 110 → 2 5 1 6 → 2516, which is correct. So why is there a discrepancy?

Ah! I think the error is in how I'm splitting the groups when splitting from the right. Let me try again.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it as:

From left to right: 1 0 1 0 1 0 0 1 1 1 0

Let me split into groups of three from the right, adding leading zeros if necessary.

Number of digits: 11. To make it divisible by 3, add 1 leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Now, split into groups of three from the left: 010 101 001 110 → which is 2 5 1 6 → 2516. This is correct.

But if I split from the right without adding leading zeros, I need to add leading zeros to the leftmost group. Let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups of three from the right:

Start from the end:

Group 1: last three bits: 1 1 0 → 6

Group 2: previous three bits: 0 0 1 → 1

Group 3: previous three bits: 0 1 0 → 2

Group 4: remaining bits: 1 → add two leading zeros: 0 0 1 → 1

Now, the groups are Group 1 (LSB): 6, Group 2: 1, Group 3: 2, Group 4 (MSB): 1. To write the octal number, we need to reverse the order of the groups: Group 4, Group 3, Group 2, Group 1 → 1, 2, 1, 6 → 1216. But this is wrong. Why?

Wait, but if I add leading zeros to the leftmost group, the groups should be:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Add one leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Split into groups of three from the left: 010 101 001 110 → 2 5 1 6 → 2516. Correct.

But if I split from the right without adding leading zeros, I get groups that need to be reversed, but the result is different. Why?

Ah! Because when adding leading zeros to the leftmost group, the groups are from left to right, which corresponds to the correct order. Whereas splitting from the right and adding leading zeros to the leftmost group, then reversing the order, gives the correct result. Wait, let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split from the right into groups:

Group 1 (LSB): 1 1 0 → 6

Group 2: 0 0 1 → 1

Group 3: 0 1 0 → 2

Group 4 (MSB): 1 → add two leading zeros: 0 0 1 → 1

Now, the groups are [Group 4, Group 3, Group 2, Group 1] = [1, 2, 1, 6], but since Group 4 is the MSB, the octal number is 1 2 1 6 → 1216. But this is wrong. But why?

Wait, but if I add leading zeros to the leftmost group, I should have added them to the left of the original number, not to the leftmost group. Wait, no. When splitting from the right, the leftmost group (Group 4) is the most significant, but it has only one bit. So adding leading zeros to make it three bits, which would be adding to the left of that group. Wait, but adding leading zeros to the left of the entire number.

Wait, this is getting too confusing. Let me try to visualize the binary number with added leading zeros.

Original number: 10101001110 (11 bits)

To split into groups of three from the right, we need to add one leading zero to make it 12 bits: 010101001110.

Now, split into groups of three from the left: 010 101 001 110 → 2 5 1 6 → 2516. Correct.

Alternatively, if I split from the right, I need to add leading zeros to the left of the original number, which is the same as adding leading zeros to make the total number of bits divisible by three. So adding one leading zero gives 010101001110, which is the same as adding leading zeros to the left, and then splitting from the left. So this is the correct method.

Therefore, the correct octal number is 2516. The earlier confusion was due to incorrectly splitting from the right without properly adding leading zeros and reversing the groups. Hence, the correct answer is 2516₈.

To confirm, let me convert 2516₈ back to binary:

2 5 1 6

Each octal digit to 3 bits:

2 → 010

5 → 101

1 → 001

6 → 110

Combined: 010 101 001 110 → 010101001110. Removing the leading zero: 10101001110, which matches the original binary number. Perfect!

Therefore, the correct octal equivalent is 2516.

**Final Answer**
The binary number $10101001110_2$ is equal to \boxed{2516} in base eight.

To convert the binary number $10101001110_2$ to its octal (base 8) equivalent, we can use the fact that **each group of three binary digits corresponds to one octal digit**. This is because $8 = 2^3$, so grouping binary digits into sets of three simplifies the conversion process.

---

### Step 1: Count the Number of Digits

The binary number is:

$$
10101001110
$$

This has **11 digits**. Since 11 is not divisible by 3, we need to **add leading zeros** to make the total number of digits a multiple of 3. The smallest multiple of 3 greater than 11 is 12, so we add **one leading zero**:

$$
010101001110
$$

---

### Step 2: Group into Sets of Three

Now, we split the binary number into groups of three bits, starting from the **left**:

$$
010\ 101\ 001\ 110
$$

---

### Step 3: Convert Each Group to Octal

Now, convert each 3-bit binary group to its octal equivalent:

- $010_2 = 2_8$
- $101_2 = 5_8$
- $001_2 = 1_8$
- $110_2 = 6_8$

---

### Step 4: Combine the Octal Digits

Putting the octal digits together in the same order:

$$
2\ 5\ 1\ 6
$$

So, the octal representation is:

$$
\boxed{2516}
$$

---

### Final Answer

$$
\boxed{2516}
$$

[Assistant]
Explanation:
Okay, so I need to convert the binary number 10101001110₂ to its octal (base 8) equivalent. Hmm, let me think about how to approach this. I remember that converting binary to octal might be easier if I group the binary digits into sets of three, since 8 is 2³. Let me recall the exact method.

First, I think the process involves splitting the binary number into groups of three bits starting from the right. If the number of digits isn't a multiple of three, I might need to add leading zeros to the leftmost group. Then, each group of three bits can be converted to an octal digit. Let me try that.

The given binary number is 10101001110. Let me count the digits to see how many there are. Let's see: 1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. Since 11 divided by 3 is 3 with a remainder of 2, I need to add one leading zero to make it 12 digits, which is divisible by 3. Wait, let me check: 3 groups of 3 would be 9 digits, so 11 digits would need 2 more digits? Wait, no. Wait, if I have 11 digits, then starting from the right, I can split into groups of three. Let me write it down:

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me separate them into groups of three starting from the right. Let me count from the right:

Starting from the rightmost digit, which is the 0. Let me write the number with positions:

Positions (from right, starting at 0): Let's see, the number is 10101001110. Let me write it with spaces for grouping:

Let me write the number again: 1 0 1 0 1 0 0 1 1 1 0

Let me count from the right:

Starting from the right, the digits are:

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

Wait, maybe I should write the number as:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups of three starting from the right. Let me add leading zeros if necessary. Since there are 11 digits, which is 3*3 + 2, so we need to add one leading zero to make it 12 digits. Wait, 11 digits: adding one zero to the left would make it 12 digits. Let me check:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Adding a leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Now, grouping into three:

From left to right: 010 101 001 110

Wait, let me check:

Wait, if I add one leading zero, the number becomes 010101001110. Let me split into groups of three:

Starting from the left: 010 101 001 110. Wait, but that would be four groups. Let me check the number of digits: original was 11, adding one zero makes it 12, so 12 digits divided by 3 is 4 groups. So yes, four groups. Let me confirm:

Original number: 10101001110. Let me write it with the added zero: 010101001110. Now, splitting into groups of three:

First group (leftmost): 010

Second group: 101

Third group: 001

Fourth group: 110

Wait, but let me check the positions again. Let me write the number with the added zero:

0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three from the left:

Group 1: 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4: 1 1 0

Wait, but that seems like splitting from the left. However, usually, when converting binary to octal, you split from the right. Wait, maybe I should split from the right. Let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Starting from the right, let's split into groups of three:

Starting from the right, the last three digits are 1 1 0. Then moving left, the next three digits are 0 0 1. Then the next three digits are 0 1 0. Then the remaining digits are 1. Wait, but that leaves one digit. So we need to add leading zeros to make it three digits. Wait, so:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split from the right:

Group 1 (rightmost): 1 1 0

Group 2: 0 0 1

Group 3: 0 1 0

Group 4: 1 (needs two leading zeros to make three digits)

So adding two leading zeros to the leftmost group: 001

Wait, but then the groups would be:

Group 4: 001

Group 3: 010

Group 2: 001

Group 1: 110

Wait, but that seems different from splitting from the left. Which is correct?

I think splitting from the right is the correct approach. Let me confirm. Because when converting to octal, each group of three bits corresponds to an octal digit, starting from the right. So if the number of digits isn't a multiple of three, we add leading zeros to the leftmost group. So in this case, original number is 11 digits. So adding two leading zeros to make it 13 digits? Wait, no. Wait, 11 digits. To make it a multiple of three, we need to add 2 digits (since 11 + 2 = 13? Wait, no. Wait, 3*4=12, so 11 digits need 1 more digit to make 12. Wait, 11 + 1 = 12. So adding one leading zero. Wait, confusion here.

Wait, let me think again. Let me count the number of digits. The original number is 10101001110. Let me count:

1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. So 11 divided by 3 is 3 with a remainder of 2. Therefore, we need to add one leading zero to make it 12 digits. So adding one zero to the left. Therefore, the number becomes 010101001110. Now, splitting into groups of three from the left:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Wait, but if we split from the left, then each group is three bits. However, if we split from the right, we need to split into groups starting from the right. Let me check both ways.

If I split from the right, with the original number 10101001110, which is 11 digits. Let me write it as:

Positions (from right, starting at 0):

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

So grouping into three from the right:

Group 0: positions 0-2: 0,1,1 → bits 0,1,1 → which is 011? Wait, but positions are from right. Wait, maybe I need to think differently.

Alternatively, maybe it's easier to split the binary number into groups of three starting from the right, adding leading zeros if necessary. Let me try that.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate groups:

Starting from the right, every three digits:

Let me count from the right:

Starting at the end: the last three digits are 1 1 0. Then moving left, the next three digits are 0 0 1. Then the next three digits are 0 1 0. Then there is one digit left: 1. So we need to add two leading zeros to make that group three digits: 0 0 1. Wait, but adding two zeros to the left of the remaining digit?

Wait, let me write the number again:

Original: 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups:

Starting from the right:

Group 1: 1 1 0 (positions 10, 9, 8? Wait, maybe I need to index properly. Let me write the number with positions:

Let me index the digits from right to left, starting at 0:

Position 0: 0 (rightmost)

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1 (leftmost)

So grouping into three:

Group 0: positions 0-2: 0,1,1 → bits: 0 (position 0), 1 (position 1), 1 (position 2) → which is 011? Wait, but when grouping from the right, the first group is the rightmost three bits. So the rightmost three bits are positions 0,1,2: 0,1,1. Wait, but the original number is:

From left to right: 1 0 1 0 1 0 0 1 1 1 0

So from right to left, it's 0 1 1 1 0 0 1 0 1 0 1

Wait, maybe I need to reverse the number to make it easier. Let me reverse the binary number:

Original: 1 0 1 0 1 0 0 1 1 1 0

Reversed: 0 1 1 1 0 0 1 0 1 0 1

Now, grouping into three from the left (since reversed, grouping from the left corresponds to grouping from the right in original):

Group 1: 0 1 1

Group 2: 1 0 0

Group 3: 1 0 1

Group 4: 0 1

Wait, the reversed number is 0 1 1 1 0 0 1 0 1 0 1. Let me count the digits: 11 digits. So grouping into three:

First group: 0 1 1

Second group: 1 0 0

Third group: 1 0 1

Fourth group: 0 1

But the last group has only two digits. So we need to add a leading zero to make it three digits: 0 0 1. Wait, but adding a leading zero to the fourth group? Wait, the fourth group is the leftmost part of the reversed number. Wait, maybe I need to add leading zeros to the left of the reversed number? Wait, confusion again.

Alternatively, since the reversed number is 0 1 1 1 0 0 1 0 1 0 1, which is 11 digits. To make groups of three, starting from the left (which is the original right), we need to add one leading zero to make it 12 digits. So adding a zero at the beginning of the reversed number: 0 0 1 1 1 0 0 1 0 1 0 1. Now, grouping into three:

Group 1: 0 0 1

Group 2: 1 1 1

Group 3: 0 0 1

Group 4: 0 1 0

Group 5: 1 ?

Wait, no. Wait, adding one zero to the reversed number makes it 12 digits. Let me write it:

Original reversed: 0 1 1 1 0 0 1 0 1 0 1

Add one leading zero: 0 0 1 1 1 0 0 1 0 1 0 1

Now, grouping into three:

Group 1: 0 0 1

Group 2: 1 1 1

Group 3: 0 0 1

Group 4: 0 1 0

Group 5: 1 ?

Wait, no, 12 digits divided by 3 is 4 groups. Wait, 0 0 1 1 1 0 0 1 0 1 0 1:

Group 1: 0 0 1

Group 2: 1 1 1

Group 3: 0 0 1

Group 4: 0 1 0

Wait, but the last three digits are 0 1 0? Wait, let me check:

Digits:

Position 0: 0

Position 1: 0

Position 2: 1

Position 3: 1

Position 4: 1

Position 5: 0

Position 6: 0

Position 7: 1

Position 8: 0

Position 9: 1

Position 10: 0

Position 11: 1

Wait, no, maybe I need to index properly. Let me write the reversed number with added zero as:

[0][0][1][1][1][0][0][1][0][1][0][1]

So positions 0 to 11:

Group 1: positions 0-2: 0,0,1 → 001

Group 2: positions 3-5: 1,1,0 → 110

Group 3: positions 6-8: 0,1,0 → 010

Group 4: positions 9-11: 1,0,1 → 101

Wait, that seems better. So the groups are:

Group 1: 001

Group 2: 110

Group 3: 010

Group 4: 101

But remember, this is the reversed number. So when we convert these groups to octal digits, we need to reverse the order of the groups because we reversed the original number. Wait, let me think.

Original number was reversed to make grouping easier. So the groups we found in the reversed number are:

Group 1 (original rightmost three bits): 001

Group 2: 110

Group 3: 010

Group 4: 101

But since we reversed the original number, the group order is reversed. So the original groups, from left to right, would be Group 4, Group 3, Group 2, Group 1. Wait, no. Wait, when we reversed the original number, the first group in the reversed number corresponds to the rightmost group in the original number. Therefore, the octal digits, when converted, should be ordered as Group 4, Group 3, Group 2, Group 1. Wait, maybe not. Let me think.

Let me take an example. Suppose the original number is, say, 1010. Reversed is 0101. Grouping into three, adding a leading zero: 00101. Wait, maybe this is getting too confusing. Let me try a different approach.

Alternatively, maybe I should just convert the binary number to decimal first, then to octal. But that might be more work. However, since the question is about binary to octal, the grouping method is more straightforward. Let me try again.

Original binary number: 10101001110. Let me write it as:

1 0 1 0 1 0 0 1 1 1 0

Let me split into groups of three starting from the right. Since there are 11 digits, adding one leading zero to make it 12 digits:

0 1 0 1 0 1 0 0 1 1 1 0

Now, grouping into three:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Wait, but if I split from the left, then the groups are 010, 101, 001, 110. Then converting each group to octal:

010₂ = 2₈

101₂ = 5₈

001₂ = 1₈

110₂ = 6₈

Therefore, combining them: 2 5 1 6, so 2516₈?

Wait, but let me check if this is correct. Let me verify by converting the original binary number to decimal and then to octal.

First, let me convert 10101001110₂ to decimal.

To convert binary to decimal, we can calculate the value of each bit:

Starting from the right, the positions are 0 to 10.

The binary number is:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, let me write it properly:

The binary number is 1 0 1 0 1 0 0 1 1 1 0

So positions from right (position 0) to left (position 10):

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, actually, position numbering is usually from right, starting at 0. So let me write the binary number with positions:

Let me write the binary number as:

From left to right: positions 10 to 0:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, but that seems confusing. Let me write the binary number as:

Digit positions (from right, starting at 0):

Let me reverse the binary number to make it easier. Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Reversed: 0 1 1 1 0 0 1 0 1 0 1

Now, positions 0 to 10:

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

So the decimal value is sum of (bit * 2^position):

So:

Position 10: 1 * 2^10 = 1024

Position 9: 0 * 2^9 = 0

Position 8: 1 * 2^8 = 256

Position 7: 0 * 2^7 = 0

Position 6: 1 * 2^6 = 64

Position 5: 0 * 2^5 = 0

Position 4: 0 * 2^4 = 0

Position 3: 1 * 2^3 = 8

Position 2: 1 * 2^2 = 4

Position 1: 1 * 2^1 = 2

Position 0: 0 * 2^0 = 0

Adding these up:

1024 + 256 + 64 + 8 + 4 + 2 = Let's compute step by step:

1024 + 256 = 1280

1280 + 64 = 1344

1344 + 8 = 1352

1352 + 4 = 1356

1356 + 2 = 1358

So the decimal value is 1358.

Now, converting 1358 decimal to octal.

To convert decimal to octal, divide by 8 and find remainders.

Let me do that:

1358 ÷ 8 = 169.75 → quotient 169, remainder 1358 - 169*8 = 1358 - 1352 = 6

169 ÷ 8 = 21.125 → quotient 21, remainder 169 - 21*8 = 169 - 168 = 1

21 ÷ 8 = 2.625 → quotient 2, remainder 21 - 2*8 = 21 - 16 = 5

2 ÷ 8 = 0.25 → quotient 0, remainder 2

So writing the remainders from last to first: 2, 5, 1, 6 → 2516₈

So that matches the earlier result from the binary grouping method. Therefore, the octal equivalent is 2516₈.

But let me check again the grouping method. Earlier, when I split the binary number into groups of three starting from the left after adding a leading zero, I had groups 010, 101, 001, 110 which converted to 2, 5, 1, 6, so 2516. Which matches the decimal conversion. Therefore, that seems correct.

But let me check again the grouping process. Original binary number: 10101001110. Adding a leading zero: 010101001110. Split into groups of three:

010 101 001 110 → 2 5 1 6 → 2516. Correct.

Alternatively, if I had split from the right without adding leading zeros, but that might have been confusing. However, since both methods gave me the same result, I can be confident that 2516 is the correct octal equivalent.

But just to make sure, let me try another way. Let me convert each group of three bits to octal digits step by step.

Original binary number: 10101001110. Let me split into groups of three from the right, adding leading zeros if necessary.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups:

Starting from the right:

First group (rightmost three bits): 1 1 0 → 6

Next group: 0 0 1 → 1

Next group: 0 1 0 → 2

Next group: 1 (remaining digit) → add two leading zeros: 0 0 1 → 1

Wait, but that would give groups: 001, 010, 001, 110? Wait, no. Wait, let me do it properly.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split from the right:

Starting at the end:

Group 1: last three digits: 1 1 0 → 6

Group 2: previous three digits: 0 0 1 → 1

Group 3: previous three digits: 0 1 0 → 2

Group 4: remaining digits: 1 → add two leading zeros: 0 0 1 → 1

So the octal digits are Group 4, Group 3, Group 2, Group 1: 1, 2, 1, 6 → 1216? Wait, that contradicts the previous result. Wait, what's happening here?

Wait, this is different. Earlier, when I added a leading zero to the left, I got 2516, but splitting from the right and adding leading zeros to the leftmost group gives me 1 2 1 6? Which is 1216? But that conflicts with the decimal conversion which gave 2516. So there must be a mistake here.

Wait, let me check again. If I split from the right:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Let me write the number with spaces to separate groups:

Starting from the right:

Group 1: 1 1 0

Group 2: 0 0 1

Group 3: 0 1 0

Group 4: 1

So Group 4 is just '1', which needs two leading zeros to make it three bits: 0 0 1.

Therefore, the groups are:

Group 4: 001 → 1

Group 3: 010 → 2

Group 2: 001 → 1

Group 1: 110 → 6

Therefore, the octal digits are Group 4, Group 3, Group 2, Group 1: 1 2 1 6 → 1216? But this contradicts the decimal conversion. So which one is correct?

Wait, but earlier when I converted the binary number to decimal, I got 1358, which is 2516 in octal. So there must be an error in this grouping.

Wait, let me check the grouping again. If I split from the right, adding leading zeros to the leftmost group, then the groups should be:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups from the right:

Group 1 (rightmost): 1 1 0 → 6

Group 2: 0 0 1 → 1

Group 3: 0 1 0 → 2

Group 4: 1 → add two leading zeros: 0 0 1 → 1

But when we write the octal number, the groups are ordered from left to right as Group 4, Group 3, Group 2, Group 1. Wait, but Group 4 is the leftmost group. Wait, no. Wait, if we split from the right, the first group (Group 1) is the rightmost, then Group 2 is to the left of Group 1, etc. Therefore, the order of the octal digits should be Group 4, Group 3, Group 2, Group 1. So 1 (Group 4), 2 (Group 3), 1 (Group 2), 6 (Group 1) → 1216. But this conflicts with the decimal conversion. Therefore, there must be a mistake in this grouping.

Alternatively, maybe I messed up the grouping. Let me write the original number again and split it properly.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate groups of three starting from the right:

Starting from the right, every three bits:

Let me count the digits:

Positions (from right, 0 to 10):

Position 0: 0

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1

So grouping into three:

Group 0: positions 0-2: 0,1,1 → bits: 0 (pos0), 1 (pos1), 1 (pos2) → which is 011 → 3? Wait, but when grouping from the right, the first group is the rightmost three bits. However, the bits are:

From right to left: positions 0-10:

0 (pos0), 1 (pos1), 1 (pos2), 1 (pos3), 0 (pos4), 0 (pos5), 1 (pos6), 0 (pos7), 1 (pos8), 0 (pos9), 1 (pos10)

So grouping from the right:

Group 1 (positions 0-2): 0,1,1 → bits: 0,1,1 → which is 011₂ = 3₈

Group 2 (positions 3-5): 1,0,0 → bits: 1,0,0 → 100₂ = 4₈

Group 3 (positions 6-8): 1,0,1 → bits: 1,0,1 → 101₂ = 5₈

Group 4 (positions 9-10): 0,1 → need to add one leading zero: 0,0,1 → bits: 0,0,1 → 001₂ = 1₈

Wait, so the groups are:

Group 1: 011 (3)

Group 2: 100 (4)

Group 3: 101 (5)

Group 4: 001 (1)

But since we grouped from the right, the order of the octal digits is Group 4, Group 3, Group 2, Group 1. Wait, no. Wait, Group 1 is the rightmost group, Group 2 is next, etc. So the octal number is Group 4 Group 3 Group 2 Group 1: 1 5 4 3 → 1543? But that conflicts with previous results. Wait, this is getting more confusing.

Wait, let me think again. When you split the binary number into groups of three starting from the right, each group corresponds to an octal digit, and the order of the groups is from right to left. Wait, no. Wait, if you split from the right, the first group (rightmost) is the least significant digit in octal, and the next group is the next digit, etc. Therefore, the octal number is constructed by writing the groups from left to right, starting with the leftmost group (which is the most significant octal digit). Wait, this is confusing.

Let me take an example. Let's say the binary number is 1101101. Let's convert it to octal.

Split into groups of three from the right: 1 101 101. Wait, adding leading zeros if necessary. Wait, original number is 1101101 (7 digits). Split into groups of three from the right: 1 101 101. Wait, but that's 1, 101, 101. Wait, but the first group has only one digit. So add two leading zeros: 001, 101, 101. Then converting each group: 001=1, 101=5, 101=5. So octal number is 155. Let me check by converting to decimal: 1101101₂ = 1*64 + 1*32 + 0*16 + 1*8 + 1*4 + 0*2 + 1*1 = 64 + 32 + 8 + 4 + 1 = 109. 155₈ = 1*64 + 5*8 + 5*1 = 64 + 40 + 5 = 109. Correct. So in this case, the groups were 001, 101, 101, which correspond to octal digits 1, 5, 5, so the octal number is 155. So the order is left to right, with the leftmost group being the most significant.

Therefore, applying this to our original problem:

Original binary number: 10101001110. Split into groups of three from the right, adding leading zeros if necessary.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups:

Starting from the right:

Group 1 (rightmost three bits): 1 1 0 → 6

Group 2: next three bits: 0 0 1 → 1

Group 3: next three bits: 0 1 0 → 2

Group 4: remaining bits: 1 → add two leading zeros: 0 0 1 → 1

So the groups are Group 4, Group 3, Group 2, Group 1: 1, 2, 1, 6 → 1216. But earlier decimal conversion gave 2516. So there is a contradiction here. What is wrong?

Wait, let me check the decimal value again. Earlier, I converted the binary number 10101001110 to decimal and got 1358. Let me verify that again.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it as:

From left to right: positions 10 to 0:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Calculating decimal value:

1*2^10 + 0*2^9 + 1*2^8 + 0*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 0*2^0

= 1024 + 0 + 256 + 0 + 64 + 0 + 0 + 8 + 4 + 2 + 0

= 1024 + 256 = 1280

1280 + 64 = 1344

1344 + 8 = 1352

1352 + 4 = 1356

1356 + 2 = 1358. Correct.

Now, converting 1358 to octal:

1358 ÷ 8 = 169.75 → quotient 169, remainder 6

169 ÷ 8 = 21.125 → quotient 21, remainder 1

21 ÷ 8 = 2.625 → quotient 2, remainder 5

2 ÷ 8 = 0.25 → quotient 0, remainder 2

Reading remainders from last to first: 2, 5, 1, 6 → 2516. Correct.

But according to the grouping method, splitting from the right gave me 1216. Which is different. So there must be an error in the grouping.

Wait, let me try splitting the binary number into groups of three from the right again, carefully.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate groups:

Starting from the right, every three bits:

Let me count the digits from the right:

Positions (from right, 0 to 10):

0: 0

1: 1

2: 1

3: 1

4: 0

5: 0

6: 1

7: 0

8: 1

9: 0

10: 1

Group 1 (positions 0-2): bits at positions 0,1,2: 0,1,1 → 011₂ = 3₈

Group 2 (positions 3-5): bits at positions 3,4,5: 1,0,0 → 100₂ = 4₈

Group 3 (positions 6-8): bits at positions 6,7,8: 1,0,1 → 101₂ = 5₈

Group 4 (positions 9-10): bits at positions 9,10: 0,1 → need to add one leading zero: 0,0,1 → 001₂ = 1₈

So the groups are Group 4, Group 3, Group 2, Group 1: 1, 5, 4, 3 → 1543₈. But this is different from both previous results. Now I'm really confused.

Wait, but if I convert 1543₈ to decimal, what do I get?

1*8^3 + 5*8^2 + 4*8 + 3 = 512 + 320 + 32 + 3 = 512 + 320 = 832 + 32 = 864 + 3 = 867. Which is not equal to 1358. So clearly, this is wrong.

But earlier grouping from the left after adding a leading zero gave me 2516, which is correct. So why is there a discrepancy?

Ah! I think the problem is in how I'm grouping the bits. When splitting from the right, the groups should be ordered from left to right as Group 1, Group 2, Group 3, Group 4, but I think I mixed up the order.

Wait, let me think again. When you split the binary number into groups of three from the right, the first group (rightmost) is the least significant octal digit, and the next group is the next digit, etc. So the octal number is constructed by writing the groups from left to right, starting with the leftmost group (which is the most significant octal digit). Wait, no. Wait, if you split from the right, the first group (rightmost) is the least significant digit. So the order of the octal digits is Group 1 (rightmost), Group 2, Group 3, Group 4 (leftmost). So to write the octal number, you need to reverse the order of the groups.

For example, in the earlier example of 1101101₂, splitting into groups from the right gave us groups 001, 101, 101. But since the first group is the least significant, the octal number is 1 (from 001) 5 (from 101) 5 (from 101), which is 155. Wait, but the groups were 001 (LSB), 101, 101 (MSB). So the octal number is written as MSB to LSB: 101 101 001 → 155. Wait, no. Wait, the groups are 001 (LSB), 101, 101 (MSB). So the octal number is 101 101 001, but written as 155. So the order is from MSB to LSB: the first group (from the left) is the MSB. Wait, this is confusing.

Let me think of the binary number as being split into groups, and each group corresponds to an octal digit. The rightmost group is the least significant octal digit, and the leftmost group is the most significant. Therefore, when you split from the right, you get the groups in the order of least significant to most significant. Therefore, to write the octal number, you need to reverse the order of the groups.

So in the example of 1101101₂, splitting from the right gives groups: 001, 101, 101. Reversing the order gives 101, 101, 001 → which is 155₈. Correct.

Applying this to our original problem:

Groups from splitting from the right: Group 1 (LSB): 011 (3), Group 2: 100 (4), Group 3: 101 (5), Group 4 (MSB): 001 (1). Reversing the order of the groups gives Group 4, Group 3, Group 2, Group 1: 1, 5, 4, 3 → 1543₈. But this is not matching the decimal conversion. So something is wrong.

Wait, but earlier when I added a leading zero to the left and split into groups from the left, I got 010 101 001 110 → 2 5 1 6 → 2516, which is correct. So why is there a discrepancy?

Ah! I think the error is in how I'm splitting the groups when splitting from the right. Let me try again.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it as:

From left to right: 1 0 1 0 1 0 0 1 1 1 0

Let me split into groups of three from the right, adding leading zeros if necessary.

Number of digits: 11. To make it divisible by 3, add 1 leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Now, split into groups of three from the left: 010 101 001 110 → which is 2 5 1 6 → 2516. This is correct.

But if I split from the right without adding leading zeros, I need to add leading zeros to the leftmost group. Let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split into groups of three from the right:

Start from the end:

Group 1: last three bits: 1 1 0 → 6

Group 2: previous three bits: 0 0 1 → 1

Group 3: previous three bits: 0 1 0 → 2

Group 4: remaining bits: 1 → add two leading zeros: 0 0 1 → 1

Now, the groups are Group 1 (LSB): 6, Group 2: 1, Group 3: 2, Group 4 (MSB): 1. To write the octal number, we need to reverse the order of the groups: Group 4, Group 3, Group 2, Group 1 → 1, 2, 1, 6 → 1216. But this is wrong. Why?

Wait, but if I add leading zeros to the leftmost group, the groups should be:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Add one leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Split into groups of three from the left: 010 101 001 110 → 2 5 1 6 → 2516. Correct.

But if I split from the right without adding leading zeros, I get groups that need to be reversed, but the result is different. Why?

Ah! Because when adding leading zeros to the leftmost group, the groups are from left to right, which corresponds to the correct order. Whereas splitting from the right and adding leading zeros to the leftmost group, then reversing the order, gives the correct result. Wait, let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Split from the right into groups:

Group 1 (LSB): 1 1 0 → 6

Group 2: 0 0 1 → 1

Group 3: 0 1 0 → 2

Group 4 (MSB): 1 → add two leading zeros: 0 0 1 → 1

Now, the groups are [Group 4, Group 3, Group 2, Group 1] = [1, 2, 1, 6], but since Group 4 is the MSB, the octal number is 1 2 1 6 → 1216. But this is wrong. But why?

Wait, but if I add leading zeros to the leftmost group, I should have added them to the left of the original number, not to the leftmost group. Wait, no. When splitting from the right, the leftmost group (Group 4) is the most significant, but it has only one bit. So adding leading zeros to make it three bits, which would be adding to the left of that group. Wait, but adding leading zeros to the left of the entire number.

Wait, this is getting too confusing. Let me try to visualize the binary number with added leading zeros.

Original number: 10101001110 (11 bits)

To split into groups of three from the right, we need to add one leading zero to make it 12 bits: 010101001110.

Now, split into groups of three from the left: 010 101 001 110 → 2 5 1 6 → 2516. Correct.

Alternatively, if I split from the right, I need to add leading zeros to the left of the original number, which is the same as adding leading zeros to make the total number of bits divisible by three. So adding one leading zero gives 010101001110, which is the same as adding leading zeros to the left, and then splitting from the left. So this is the correct method.

Therefore, the correct octal number is 2516. The earlier confusion was due to incorrectly splitting from the right without properly adding leading zeros and reversing the groups. Hence, the correct answer is 2516₈.

To confirm, let me convert 2516₈ back to binary:

2 5 1 6

Each octal digit to 3 bits:

2 → 010

5 → 101

1 → 001

6 → 110

Combined: 010 101 001 110 → 010101001110. Removing the leading zero: 10101001110, which matches the original binary number. Perfect!

Therefore, the correct octal equivalent is 2516.

**Final Answer**
The binary number $10101001110_2$ is equal to \boxed{2516} in base eight.

To convert the binary number $10101001110_2$ to its octal (base 8) equivalent, we can use the fact that **each group of three binary digits corresponds to one octal digit**. This is because $8 = 2^3$, so grouping binary digits into sets of three simplifies the conversion process.

---

### Step 1: Count the Number of Digits

The binary number is:

$$
10101001110
$$

This has **11 digits**. Since 11 is not divisible by 3, we need to **add leading zeros** to make the total number of digits a multiple of 3. The smallest multiple of 3 greater than 11 is 12, so we add **one leading zero**:

$$
010101001110
$$

---

### Step 2: Group into Sets of Three

Now, we split the binary number into groups of three bits, starting from the **left**:

$$
010\ 101\ 001\ 110
$$

---

### Step 3: Convert Each Group to Octal

Now, convert each 3-bit binary group to its octal equivalent:

- $010_2 = 2_8$
- $101_2 = 5_8$
- $001_2 = 1_8$
- $110_2 = 6_8$

---

### Step 4: Combine the Octal Digits

Putting the octal digits together in the same order:

$$
2\ 5\ 1\ 6
$$

So, the octal representation is:

$$
\boxed{2516}
$$

---

### Final Answer

$$
\boxed{2516}
$$

Solution: Since $2^3=8$, we may convert between base 2 and base 8 representations by replacing each block of three digits in base 2 with its equivalent in base 8. In this case, we begin by noticing that the last three digits are worth $110_2=6_8$. The next block of three digits is $001_2=1_8$. Continuing, we find that the next two digits (moving right-to-left) are $101_2=5_8$ and $010_2=2_8$. Altogether, we find that $10101001110_{2}=\boxed{2516_8}$.


Explanation:
Okay, so I need to figure out the value of 9 & 2 using this operation they defined. Let me start by understanding the problem properly. The operation & is defined for positive integers a and b as a & b equals the square root of (ab + a) divided by the square root of (ab - b). So, mathematically, that's:

a & b = √(ab + a) / √(ab - b)

Alright, so I need to compute 9 & 2. Let me substitute a = 9 and b = 2 into this formula.

First, let me compute the numerator: √(ab + a). Let me calculate ab first. a is 9, b is 2, so ab is 9*2 = 18. Then, adding a to that: 18 + 9 = 27. So the numerator becomes √27.

Now the denominator: √(ab - b). Again, ab is 18, and then subtract b, which is 2. So 18 - 2 = 16. Therefore, the denominator is √16.

So putting it all together, 9 & 2 is √27 divided by √16. Let me write that as:

√27 / √16

Hmm, I can simplify both square roots. Let me recall that √16 is 4, since 4 squared is 16. And √27... Well, 27 is 9*3, so √27 is √(9*3) which is √9 * √3 = 3√3. So substituting back in:

(3√3) / 4

Wait, so is that the answer? Let me check my steps again to make sure I didn't make a mistake.

First, ab is 9*2 = 18. Then numerator is ab + a = 18 + 9 = 27. Denominator is ab - b = 18 - 2 = 16. Then square roots: √27 / √16. Simplify √27 to 3√3 and √16 is 4. So yes, 3√3 over 4. That seems right.

But let me check if there's a way to combine the square roots first before simplifying. Sometimes, when you have a fraction with square roots, you can combine them into a single square root. Let me try that approach to verify.

The original expression is √(ab + a) / √(ab - b). That can be written as √[(ab + a)/(ab - b)]. Let me compute the fraction inside the square root first.

So (ab + a)/(ab - b). Let me factor numerator and denominator. Let's see:

Numerator: ab + a = a(b + 1)

Denominator: ab - b = b(a - 1)

So substituting back, we have [a(b + 1)] / [b(a - 1)]

So then the entire expression becomes √[a(b + 1)/[b(a - 1)]]

So plugging in a = 9 and b = 2:

Numerator inside the square root: 9*(2 + 1) = 9*3 = 27

Denominator inside the square root: 2*(9 - 1) = 2*8 = 16

So again, we get √(27/16) which is √27 / √16 = 3√3 / 4. Same result. So that seems consistent.

Wait, but maybe I can simplify the fraction inside the square root before taking the square root? Let me check.

So, [a(b + 1)] / [b(a - 1)] with a=9, b=2:

[9*(3)] / [2*(8)] = 27 / 16. Yep, same as before. So √(27/16) is indeed 3√3 / 4. So that seems correct.

But let me check if there's any other simplification or if I need to rationalize something? Wait, the problem says to express the answer as a common fraction in simplest radical form. So 3√3 over 4 is already a fraction with a radical in the numerator, which is acceptable. Since √3 can't be simplified further, and 3 and 4 have no common factors, this should be the simplest form.

Wait, but let me check if I did the initial substitution correctly. Let me recompute ab + a and ab - b again.

For a=9, b=2:

ab = 9*2 = 18. Then ab + a = 18 + 9 = 27. Correct. ab - b = 18 - 2 = 16. Correct. So square roots of 27 and 16. So √27 is 3√3, √16 is 4. So 3√3 / 4. Yep.

Alternatively, maybe I can write the entire expression as √( (ab + a)/(ab - b) ) and compute that fraction first. Let me compute 27/16. That's 1.6875, but I don't think that helps. But if I take the square root of 27/16, that's √27 divided by √16, which is the same as before.

Alternatively, maybe there's a mistake in the problem statement? Let me check again. The operation is defined as a & b = sqrt(ab + a)/sqrt(ab - b). So that's exactly what I used. So substituting 9 and 2, I think that's correct.

Wait, but let me check if the problem says "positive integers a and b", which they are, 9 and 2 are positive integers. So the denominators in the original expression: ab - b. Let me check if that's positive. ab - b = b(a - 1). For b=2 and a=9, that's 2*(9 - 1) = 16, which is positive. So the denominator is positive, so the square root is defined. So that's okay.

If I had, say, a=1, then ab - b would be b(1 - 1) = 0, which would be undefined. But since a and b are positive integers, and in this case, a=9, which is greater than 1, so it's okay.

So, I think my answer is correct. 3√3 over 4. Let me check if that's the simplest radical form. Since 3 and 4 are coprime, and √3 is simplified, yes. So I think that's the answer.

But just to be thorough, let me check if there's another way to approach this problem. Maybe by simplifying the expression before plugging in numbers?

Let me try to manipulate the formula algebraically. Let's take the general formula:

a & b = √(ab + a)/√(ab - b) = √[a(b + 1)/b(a - 1)] as I did before. So that's √[a(b + 1)/[b(a - 1)]]. Maybe we can write this as √[a/b * (b + 1)/(a - 1)]? Not sure if that helps, but maybe.

Alternatively, maybe factor something else? Let me see. For specific values, maybe not. But in this case, since we have specific numbers, plugging in is straightforward.

Alternatively, maybe rationalizing something? But since the denominator is already rational (sqrt(16) is 4), and the numerator is 3√3, there's no need to rationalize. So 3√3 / 4 is already simplified.

Alternatively, if I had a radical in the denominator, I would have to rationalize, but here the denominator is rational. So I think that's the answer.

Wait, but let me check if I made a mistake in simplifying √27. √27 is √(9*3) which is √9 * √3 = 3√3. Yes, that's correct. So that's right.

Alternatively, if I had messed up the numerator or denominator, but no, 9*2 + 9 is 18 + 9 = 27. 9*2 - 2 is 18 - 2 = 16. So that's correct.

So, I think I can be confident that 9 & 2 is 3√3 over 4. Therefore, the answer is 3√3⁄4, which as a common fraction in simplest radical form is 3√3/4.

**Final Answer**
The value of $9 \& 2$ is \boxed{\dfrac{3\sqrt{3}}{4}}.

To evaluate the operation $ 9 \& 2 $, we begin by applying the given definition:

$$
a \& b = \frac{\sqrt{ab + a}}{\sqrt{ab - b}}
$$

---

### Step 1: Substitute $ a = 9 $ and $ b = 2 $

We compute the numerator and denominator separately.

#### Numerator:
$$
ab + a = 9 \cdot 2 + 9 = 18 + 9 = 27
$$
$$
\sqrt{ab + a} = \sqrt{27}
$$

#### Denominator:
$$
ab - b = 9 \cdot 2 - 2 = 18 - 2 = 16
$$
$$
\sqrt{ab - b} = \sqrt{16} = 4
$$

---

### Step 2: Combine the results

$$
9 \& 2 = \frac{\sqrt{27}}{\sqrt{16}} = \frac{\sqrt{27}}{4}
$$

Now simplify $ \sqrt{27} $:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

So the expression becomes:

$$
\frac{3\sqrt{3}}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{3\sqrt{3}}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of 9 & 2 using this operation they defined. Let me start by understanding the problem properly. The operation & is defined for positive integers a and b as a & b equals the square root of (ab + a) divided by the square root of (ab - b). So, mathematically, that's:

a & b = √(ab + a) / √(ab - b)

Alright, so I need to compute 9 & 2. Let me substitute a = 9 and b = 2 into this formula.

First, let me compute the numerator: √(ab + a). Let me calculate ab first. a is 9, b is 2, so ab is 9*2 = 18. Then, adding a to that: 18 + 9 = 27. So the numerator becomes √27.

Now the denominator: √(ab - b). Again, ab is 18, and then subtract b, which is 2. So 18 - 2 = 16. Therefore, the denominator is √16.

So putting it all together, 9 & 2 is √27 divided by √16. Let me write that as:

√27 / √16

Hmm, I can simplify both square roots. Let me recall that √16 is 4, since 4 squared is 16. And √27... Well, 27 is 9*3, so √27 is √(9*3) which is √9 * √3 = 3√3. So substituting back in:

(3√3) / 4

Wait, so is that the answer? Let me check my steps again to make sure I didn't make a mistake.

First, ab is 9*2 = 18. Then numerator is ab + a = 18 + 9 = 27. Denominator is ab - b = 18 - 2 = 16. Then square roots: √27 / √16. Simplify √27 to 3√3 and √16 is 4. So yes, 3√3 over 4. That seems right.

But let me check if there's a way to combine the square roots first before simplifying. Sometimes, when you have a fraction with square roots, you can combine them into a single square root. Let me try that approach to verify.

The original expression is √(ab + a) / √(ab - b). That can be written as √[(ab + a)/(ab - b)]. Let me compute the fraction inside the square root first.

So (ab + a)/(ab - b). Let me factor numerator and denominator. Let's see:

Numerator: ab + a = a(b + 1)

Denominator: ab - b = b(a - 1)

So substituting back, we have [a(b + 1)] / [b(a - 1)]

So then the entire expression becomes √[a(b + 1)/[b(a - 1)]]

So plugging in a = 9 and b = 2:

Numerator inside the square root: 9*(2 + 1) = 9*3 = 27

Denominator inside the square root: 2*(9 - 1) = 2*8 = 16

So again, we get √(27/16) which is √27 / √16 = 3√3 / 4. Same result. So that seems consistent.

Wait, but maybe I can simplify the fraction inside the square root before taking the square root? Let me check.

So, [a(b + 1)] / [b(a - 1)] with a=9, b=2:

[9*(3)] / [2*(8)] = 27 / 16. Yep, same as before. So √(27/16) is indeed 3√3 / 4. So that seems correct.

But let me check if there's any other simplification or if I need to rationalize something? Wait, the problem says to express the answer as a common fraction in simplest radical form. So 3√3 over 4 is already a fraction with a radical in the numerator, which is acceptable. Since √3 can't be simplified further, and 3 and 4 have no common factors, this should be the simplest form.

Wait, but let me check if I did the initial substitution correctly. Let me recompute ab + a and ab - b again.

For a=9, b=2:

ab = 9*2 = 18. Then ab + a = 18 + 9 = 27. Correct. ab - b = 18 - 2 = 16. Correct. So square roots of 27 and 16. So √27 is 3√3, √16 is 4. So 3√3 / 4. Yep.

Alternatively, maybe I can write the entire expression as √( (ab + a)/(ab - b) ) and compute that fraction first. Let me compute 27/16. That's 1.6875, but I don't think that helps. But if I take the square root of 27/16, that's √27 divided by √16, which is the same as before.

Alternatively, maybe there's a mistake in the problem statement? Let me check again. The operation is defined as a & b = sqrt(ab + a)/sqrt(ab - b). So that's exactly what I used. So substituting 9 and 2, I think that's correct.

Wait, but let me check if the problem says "positive integers a and b", which they are, 9 and 2 are positive integers. So the denominators in the original expression: ab - b. Let me check if that's positive. ab - b = b(a - 1). For b=2 and a=9, that's 2*(9 - 1) = 16, which is positive. So the denominator is positive, so the square root is defined. So that's okay.

If I had, say, a=1, then ab - b would be b(1 - 1) = 0, which would be undefined. But since a and b are positive integers, and in this case, a=9, which is greater than 1, so it's okay.

So, I think my answer is correct. 3√3 over 4. Let me check if that's the simplest radical form. Since 3 and 4 are coprime, and √3 is simplified, yes. So I think that's the answer.

But just to be thorough, let me check if there's another way to approach this problem. Maybe by simplifying the expression before plugging in numbers?

Let me try to manipulate the formula algebraically. Let's take the general formula:

a & b = √(ab + a)/√(ab - b) = √[a(b + 1)/b(a - 1)] as I did before. So that's √[a(b + 1)/[b(a - 1)]]. Maybe we can write this as √[a/b * (b + 1)/(a - 1)]? Not sure if that helps, but maybe.

Alternatively, maybe factor something else? Let me see. For specific values, maybe not. But in this case, since we have specific numbers, plugging in is straightforward.

Alternatively, maybe rationalizing something? But since the denominator is already rational (sqrt(16) is 4), and the numerator is 3√3, there's no need to rationalize. So 3√3 / 4 is already simplified.

Alternatively, if I had a radical in the denominator, I would have to rationalize, but here the denominator is rational. So I think that's the answer.

Wait, but let me check if I made a mistake in simplifying √27. √27 is √(9*3) which is √9 * √3 = 3√3. Yes, that's correct. So that's right.

Alternatively, if I had messed up the numerator or denominator, but no, 9*2 + 9 is 18 + 9 = 27. 9*2 - 2 is 18 - 2 = 16. So that's correct.

So, I think I can be confident that 9 & 2 is 3√3 over 4. Therefore, the answer is 3√3⁄4, which as a common fraction in simplest radical form is 3√3/4.

**Final Answer**
The value of $9 \& 2$ is \boxed{\dfrac{3\sqrt{3}}{4}}.

To evaluate the operation $ 9 \& 2 $, we begin by applying the given definition:

$$
a \& b = \frac{\sqrt{ab + a}}{\sqrt{ab - b}}
$$

---

### Step 1: Substitute $ a = 9 $ and $ b = 2 $

We compute the numerator and denominator separately.

#### Numerator:
$$
ab + a = 9 \cdot 2 + 9 = 18 + 9 = 27
$$
$$
\sqrt{ab + a} = \sqrt{27}
$$

#### Denominator:
$$
ab - b = 9 \cdot 2 - 2 = 18 - 2 = 16
$$
$$
\sqrt{ab - b} = \sqrt{16} = 4
$$

---

### Step 2: Combine the results

$$
9 \& 2 = \frac{\sqrt{27}}{\sqrt{16}} = \frac{\sqrt{27}}{4}
$$

Now simplify $ \sqrt{27} $:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

So the expression becomes:

$$
\frac{3\sqrt{3}}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{3\sqrt{3}}{4}}
$$

Solution: We have $9\&2 = \frac{\sqrt{(9)(2)+9}}{\sqrt{(9)(2)-2}} = \frac{\sqrt{27}}{\sqrt{16}} = \boxed{\frac{3\sqrt{3}}{4}}.$