Question

Can anyone explain how to check perfect square or not using repeated subtraction?​

Answer 1

Let's say we wanted to check if 16 was a perfect square or not.

What we do is subtract off the list of odd numbers {1,3,5,7,9,...}. If we reach 0 at any point, then the step number is what the square root will be.

So we'll start with 16 and subtract off 1 to get 15. Then from 15, we subtract off 3 to get 12. From 12, we subtract off 5 to get 7. This process is laid out below

1. 16-1 = 15
2. 15-3 = 12
3. 12-5 = 7
4. 7-7 = 0

By step 4, we reach 0. This indicates that
`√(16) = 4`

Note: The numbers to the left of the decimal point indicate the step number. All values mentioned are whole numbers.

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Here's another example. I'll start with 49

1. 49 - 1 = 48
2. 48 - 3 = 45
3. 45 - 5 = 40
4. 40 - 7 = 33
5. 33 - 9 = 24
6. 24 - 11 = 13
7. 13 - 13 = 0

We reach 0 at step 7, therefore
`√(49) = 7`

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If we started with some non-perfect square, say 50, then we'd get this:

1. 50 - 1 = 49
2. 49 - 3 = 46
3. 46 - 5 = 41
4. 41 - 7 = 34
5. 34 - 9 = 25
6. 25 - 11 = 14
7. 14 - 13 = 1
8. 1 - 15 = -14

As you can see, we don't reach 0, which means 50 is not a perfect square. The closest we get is 1 and that happens on the 7th step. This suggests
`√(50)` is closest to 7. It turns out that
`√(50) \approx 7.071` which helps confirm that previous statement.