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I can't comprehend this formula. Why does $\sqrt{a^2}=|a|$, why doesn't it equal $a$? Why does it have to be $|a|$?

I'm especially having difficulty when it comes to negative numbers. Take an example, let's say we have $\sqrt{(-2)^2}$. So that means $\sqrt{(-2)^2}=|-2|=2$. I don't understand why can't the answer also be $-2$. $(-2)^2$ still equals $2^2 = 4$.

I found this on mathisfun: "Squaring a makes it positive or zero (for a as a Real Number). Then taking the square root will "undo" the squaring, but leave it positive or zero."

So if we follow that way of reasoning, that means $\sqrt{(-2)^2}=\sqrt{4}$, then the square root "undoes" the squaring to $2$ and the answer is positive. But doesn't every number have $2$ square roots (except $0$)? If so, why not consider $-2$ as the answer as well? Before there was a person who also asked this but the answers didn't help me much.

ntl
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  • By cases if $a\ge 0 \implies \sqrt {a^2}=a=|a|\ge 0$ and if $a<0 \implies \sqrt {a^2}=-a=|a|>0$ – user Jun 09 '23 at 13:40
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    A positive real number $x$ has indeed two real square roots, mutually opposite, and $\sqrt x$ is by definition the positive one. – Anne Bauval Jun 09 '23 at 13:40
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    If your treatment was correct, then since $1^2=(-1)^2,$ $1=\sqrt{1^2}=\sqrt{(-1)^2}=-1.$ There is a way you can treat $\sqrt{\cdot}$ as a multi-values function, so that $\sqrt{a^2}=\pm a,$ but that is tricky, and mathematicians dislike multivalued functions. In particular, you'd want to say $-1\in \sqrt{1}$ rather than $-1=\sqrt1,$ because otherwise you'd have cases where $a=b$ and $b=c$ but $a\neq c.$ – Thomas Andrews Jun 09 '23 at 13:49
  • @ThomasAndrews So that means this formula is just generalizing the answers, and there is still a possibility for it to be a negative? – ntl Jun 09 '23 at 13:52
  • No, we choose the positive square root to make it a single-valued function. That is our definition of square root. We could have chosen the negative square root instead, but the positive square root has advantages, like $\sqrt{ab}=\sqrt{a}\sqrt{b}.$ If we chose the negative we'd have $\sqrt {ab}=-\sqrt a\sqrt b,$ slightly less elegant. @ntl – Thomas Andrews Jun 09 '23 at 13:54
  • @user1176409 Hi, I'm new so I didn't read through the tour, sorry for my mistake and thank you for telling me! – ntl Jun 09 '23 at 13:54
  • @ThomasAndrews ah okay, thank you – ntl Jun 09 '23 at 13:55
  • @ThomasAndrews if mathematicians dislike multivalued functions then why they created complex functions $?$ as for example logarithm of a negative real number to the base $e$ – MathStackexchangeIsNotSoBad Jun 09 '23 at 14:42
  • Disliking them doesn't mean they avoid them completely. @MathStackexchangeIsNotSoBad And even in complex functions, they usually pick branches of the function which keeps it single-valued. For example, to keep $\ln$ meromorphic we usually make a branch cut at the negative reals (leaving it undefined) or define the function on a Reimann surface as a single-valued function. – Thomas Andrews Jun 09 '23 at 14:44
  • @ThomasAndrews ohh okk...I see thanks...im a grade 12 student and those terms are intimidating – MathStackexchangeIsNotSoBad Jun 09 '23 at 14:48
  • Suffice it to say, mathematicians go to great lengths to try to make complex functions single-valued functions. @MathStackexchangeIsNotSoBad – Thomas Andrews Jun 09 '23 at 14:52

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