# Decimal Squares

**URL:** <https://boards.straightdope.com/t/decimal-squares/561539>\
**Category:** Factual Questions\
**Created:** [November 22, 2010, 1:17pm UTC](https://boards.straightdope.com/t/decimal-squares/561539 "2010-11-22T13:17:19Z")\
**Posts on this page:** 2\
**Page:** 2

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [November 24, 2010, 3:16am UTC](https://boards.straightdope.com/t/decimal-squares/561539/21 "2010-11-24T03:16:56Z")

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> [@Dero\_von\_Hesse](#):
>
> Why is the square root of a decimal greater than the decimal value?

Here’s the three-sentence answer that your professor should have given: Start off by assuming that 0 \< x \< 1, and multiply all three numbers by x. Then you see that 0 \< x^2 \< x. That’s why numbers between 0 and 1 get smaller when you square them.

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [November 25, 2010, 7:13am UTC](https://boards.straightdope.com/t/decimal-squares/561539/22 "2010-11-25T07:13:48Z")

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Right on. That is such a clear and simple explanation, I’m ready to prematurely declare it the best.

The one thing I’d take care to explicitly emphasize, though it’s already implicit in that explanation, is that this is just a special case of the fact that multiplying anything by a number between 0 and 1 makes it smaller. That is, I’d note that this doesn’t really have anything to do with squaring in particular; squaring just happens to be one particular case of multiplication.

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