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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.

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    $\begingroup$ If $a= b$ then any statement involving $a$ will be true if it is said about $b$. So true. $\endgroup$ – fleablood May 23 '18 at 19:16
  • $\begingroup$ @fleablood +1 That is an incredibly intuitive way to deal with this question. $\endgroup$ – Bill Wallis May 23 '18 at 21:17
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    $\begingroup$ to be honest, I'm perplexed that there is any other way to think of the question. Even if we make some weird assumption/definition that a square root can flip back and forth between positive and negative values we can't say "The square root of 9 will be 3 when we use one variable but it will be 9 when we use a different variable". I wonder if the question was supposed to go the other way. If $\sqrt a = \sqrt b$ then $a = b$, true false? (Hint: true... but not without a reason.) $\endgroup$ – fleablood May 23 '18 at 21:26
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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 $x\to\sqrt x$.

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$x^2 = 4$ 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.

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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.

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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.

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