When you borrow a number during subtraction, it becomes the number of the base you are subtracting from.

For example, in Base 2 if you borrow a number the number you borrow becomes 2, Likewise in base 5 if you borrow a number it becomes 5.

Now, let’s take some examples.

### Examples

Evaluate the following:**1.** 11000_{two} – 111_{two}**2.** 20204_{six} – 5125_{six}**3.** 7243_{8} – 4536_{8}**4.** 1A35_{16} – F7E_{16}

**Solution** **(1)**

**Explanation**

1st Column (from right to left) = 0 is smaller than 1 so we will have to borrow from the 2nd column. Since it is in base 2 any number borrowed is 2. So borrowing two from the second will give (2 + 0) – 1 = 1

2nd Column = This reduces 0 to -1 in the second column. -1 is smaller than 1 so we have to borrow again from the third column. We now have (2 +( – 1) – 1 = 1 -1 = 0

3rd Column = This also reduces 0 to – 1 in the 3rd column. This too is smaller than 1 so like the second column what we have is 2 -1 – 1 = 0

4th Column = This reduces 1 to 0 in the fourth column. We then have 0 – 0 = 0

5th Column = This column requires no borrowing since 1 is greater than 0. We simply have 1 – 0 = 1

This yields 10001_{two}

**Solution** **(2)**

**Explanation**

1st Column (from right to left) = 4 is smaller than 5 so we will have to borrow from the 2nd column. Since it is in base 6 any number borrowed is 6. So borrowing six from the second will give (6 + 4) – 5 = 5

2nd Column = This reduces 0 to -1 in the second column. -1 is smaller than 2 so we have to borrow again from the third column. We now have 6 – 1 – 2 = 3

3rd Column = This also reduces 2 to 1 in the 3rd column. We can subtract 1 from 1 so we now have 1 – 1 = 0

4th Column = We have 0 – 5. We are going to have to borrow from the 5th column i.e 6 – 5 = 1

5th Column = This reduces 2 in the 5th column to 1. We now have 1 – 0 = 1

This yields 11035_{6}

**Solution** **(3)** **| Octal Subtraction**

**Explanation**

1st Column (from right to left) = 3 is smaller than 6 so we will have to borrow from the 2nd column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the 2nd column will give (8 + 3) – 6 = 5

2nd Column = This reduces 4 to 3 in the second column. We can subtract 3 from 3 so we now have 3 – 3 = 0

3rd Column = 2 is smaller than 5 so we will have to borrow from the 4th column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the 4th column will give (8 + 2) – 5 = 5

4th Column = This reduces 7 in the 4th column to 6. 6 is greater than 4 so we can subtract normally. We now have 6 – 4 = 2

This yields 2505_{8}

** Solution (4)**Â

**| Hex Subtraction**

**Explanation**

Hex subtraction can be computed much the same way as hex addition; by performing the operation while converting between hex and decimal values. The most significant difference between hex and decimal subtraction involves borrowing. When borrowing in hex, the “1” that is borrowed represents **16**

Remember A = 10, B = 11, C= 12, D = 13, E = 14, F = 15

1st Column (from right to left) = 5 is smaller than E (14) so we will have to borrow from the 2nd column. Since it is in base 16 any number borrowed is 16. So borrowing 16 from the second will give (16 + 5) – 14 = 21 – 14 = 7

2nd Column = This reduces 3 to 2 in the second column. 2 is smaller than 7 so we have to borrow again from the third column. We now have (16 + 2) – 7 = 11. B = 11

3rd Column = This also reduces A (10) to 9 in the 3rd column. 9 is smaller than F (15) we are going to have to borrow from the 4th column. We have (16 + 9) – 15 = 10. A = 10

4th Column = This reduces 1 to 0 in the 4th column. We then have 0 – 0 = 0

This yields AB7_{16}

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