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I understand that the numbers that, when squared, result in 4 are ±2 because both -2 and 2 squared result in four. However, when in a radical, why is it that $\sqrt 4$ is only 2? Why isn't the sign undetermined until it the number is taken out of the square root? Why is the ± needed in front of $\sqrt 4$ to denote that the solution set includes -2?

Blue
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Hello World
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    By definition, if $x$ is a nonnegative real number, then $\sqrt{x}$ denotes the principal (nonnegative) square root of $x$. See this problem. – N. F. Taussig May 16 '18 at 02:01
  • It's a convention. It's good you ask, because there is indeed no mathematical reason for that. – Torsten Schoeneberg May 16 '18 at 02:05
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    @TorstenSchoeneberg I would think that "we want $\sqrt{2}$ to unambiguously refer to a specific number" qualifies as a "mathematical reason" – Gregory J. Puleo May 16 '18 at 02:08
  • see also https://math.stackexchange.com/q/2767829/505767 – user May 16 '18 at 02:27
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    @Gregory: I think what Torsten was getting at is that the choice is mathematically arbitrary -- we could have chosen $\sqrt 2$ to refer to the negative square root and nothing would go wrong. –  May 16 '18 at 03:11

1 Answers1

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Because in many situations we want a function: given a single positive $x$, we want a single $\sqrt x$.

That's the reason why we have the convention that $\sqrt x$, for $x\geq0$, means the positive square root of $x$. It's not a big deal, since you can get the other one as $-\sqrt x$.

The convention does not extend beyond square roots of positive numbers. That's why it is (kind of) poor form to write $i=\sqrt{-1}$. There are two complex numbers whose square is $-1$: namely $i$ and $-i$. When one writes $\sqrt{-1}$, it is not clear which one to choose. So, when working with complex numbers, we use that $i^2=-1$, with no need for writing square roots at all.

Martin Argerami
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