(Dis)Prove $\forall x \in \mathbb{R}, \sqrt{x^2} = x$

Question

I'm in my discrete math class, and I'm being asked to prove or disprove:

$\forall x \in \mathbb{R}, \sqrt{x^2} = x$

I think that would be false because I know that when you square a negative number, the result is positive. HOWEVER the reason that I'm uncertain is because I learned in algebra that every positive real number has $2$ real square roots, one of those being negative, which would make the above statement true.

It is highly unlikely that $\sqrt{\cdot}$ is meant to be a multivalued function, but it could be in some contexts. Without knowing your context, it is hard to be sure. — Thomas Andrews, Jan 29 '24 at 21:40
It is true algebrically that $(\pm 2)^2 =4$ but by definition the function $f(x)=\sqrt x\ge 0$ for any $x\ge 0$. — user, Jan 29 '24 at 21:44

score 4 · Answer 1 · answered Jan 29 '24 at 21:40

4

Since the square root always takes the positive root, we get that $\sqrt{x}\geq0$ so for negative numbers, the statement doesn't hold.

Since you want that there is only one function value for each input, we "make a choice" and define $\sqrt{x}$ as the positive square root of $x$.

answered Jan 29 '24 at 21:40

Vincent Batens

1,762

1

So that means that I can just go with a basic counterproof to finish that off, and use a negative value for x. Thank you so much. – Hayden Jan 29 '24 at 21:42

score 1 · Answer 2 · edited Jan 29 '24 at 21:48

Every positive real number $y$ has two real numbers $r$ and $-r$, each of which squares to $y$, i.e. the equation $$ x^2 = y $$ has two solutions, namely $x = +r$ and $x = -r$.

However, the square root of $y$, written $\sqrt{y},$ is the output of a function which (by convention) is chosen to be the positive root. In other words, $$ \sqrt{y} = x \quad\iff\quad y = x^2 \text{ and } x \geq 0. $$

(Dis)Prove $\forall x \in \mathbb{R}, \sqrt{x^2} = x$

2 Answers2