$\forall a,b \in {\mathbb R^+}$: if $a=b$ then $\sqrt a =\sqrt b$

Question

Is this statement:

$\forall a,b \in {\mathbb R^+}$: if $a=b$ then $\sqrt a =\sqrt b$

True or False? If False, then how?

I know we could take $a = b = 9$ and then continue with $ 9 = (-3)^2 = 3^2$ and try that. But my professor said in class that the negative sign arises 'externally' or when we use the convention $\pm4$ when square rooting a number, say, $16$ it comes from outside. This has got me pretty confused.

Answer 1

The square root is a function from $[0,+\infty[$ to $[0,+\infty[$. For any function $f$, if $a=b$, then $f(a)=f(b)$ (provided $f$ is defined at $a$).

Your professor is probably referring to the situation where you look for solutions of the equation $x^2=a$ with $a\ge0$, and then there are two solutions, $+\sqrt{a}$ and $-\sqrt{a}$. But the sign is not "part" of the function $x\to\sqrt x$.

Answer 2

Typically, the square root sign means the positive value only, so $\sqrt{9} = 3$. However, whether or not you want both the positive and negative solutions depends on the context in which you are finding the square root. For example, in the quadratic formula you want both, hence $$ ax^{2} + bx + c = 0 \iff x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}. $$ Observe that we specify we want both roots with the $\pm$ sign.

For this reason, if $a, b \in \mathbb{R}^{+}$, then it is true that $\sqrt{a} = \sqrt{b}$. However, the set of solutions to $a^{2} = b^{2}$ is $a = \pm b$. The reason that we are finding square roots tells us whether we need to consider the negative counterpart as well.

Answer 3

$x^2 = 4$ has two solutions

i.e. $x = 2$ and $x = -2$

But $\sqrt 4 = 2$ and $\sqrt 4 \ne - 2$

Which means that $\sqrt x$ is not exactly the inverse operation of $x^2$

We have defined the radical ($\sqrt{}$ symbol) to always refer to the "principal root." This means for one input we get one output. And, $\sqrt x$ is a function.

And if we indeed want to refer to both roots we need the $\pm$ prefix.

Answer 4

By convention the $\sqrt{M}$ refers specifically to the positive square root of $M$. So if $a=b=9$ then $\sqrt{a} = \sqrt{b} = \sqrt{9} = 3\ne -3$ even though $(-3)^2 = 9$. Just because $x^2 = k$ does NOT mean $x = \sqrt {k}$. It means $|x| =\sqrt {k}$ and $x =$ either $\sqrt{k}$ or $-\sqrt{k}$ but not both.

But that's irrelevent.

Even if we were to define something else (and math is only definitions ) as $squareroot(9) =\{3,-3\}$ so $squareroot(k)=\{$ all things $x$ so that $x^2 = k\}$. Then if $a = b$ then $squareroot(a) = squareroot(b)$.

The thing is, if $a = b$ than $a$ and $b$ are the same thing. So whatever you say about $a$ has to be the same as you say about $b$. So $\mu^{\omega}_{gumpty}(a) = \mu^{\omega}_{gumpty}(b)$ even if you have not freaking idea what $\mu^{\omega}_{gumpty}(x)$ means.