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True or False

$$a\in \mathbb R\implies \sqrt{a^2} = a$$

a positive or negative, will always be equal to a, so for me it is true, but the teacher says that the expression is false, but I can not understand why?

amWhy
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Brian
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    Indeed, $\sqrt {a^2}=|a|$. Note: I don't think you know what "simplicial" means. – lulu Sep 05 '18 at 17:43
  • Hi @lulu, what do you say by the tag? I wanted to say simplification of expressions, but I did not find something similar – Brian Sep 05 '18 at 17:46
  • Maybe [algebra-precalculus]? – lulu Sep 05 '18 at 17:47
  • @BrianVanegasParra This is precalculus, so the [algebra-precalculus] tag should be the way to go. In fact, when you hover over the tag, you can read a description, and it is rather apparent that [simplicial-stuff] has nothing to do with this. –  Sep 05 '18 at 17:48
  • Suppose $a = -3$. Then is it true that $\sqrt{(-3)^2} = \sqrt 9 = -3 = a$? – amWhy Sep 05 '18 at 18:20
  • for me a is always a real number, reason to prove that it is true – Brian Sep 05 '18 at 18:23

10 Answers10

4

If

$0 > a \in \Bbb R, \tag 1$

then

$a^2 > 0, \tag 2$

whence

$0 < \sqrt{a^2} = \vert a \vert \ne a < 0 \tag 3$

Note: for $r > 0$, it is generally understood that $\sqrt r > 0$.

(See comment below that the function $\sqrt \cdot$ is defined for non-negative arguments.) End of Note.

amWhy
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Robert Lewis
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  • my teacher particularized the value of a = -4 to say that the expression is false, but I think it is wrong: $$ \sqrt{-4}^2 = \sqrt{16} = 4 $$ 4 ≠ a ????? I do not understand your argument, for me to be a or -a $$∈ {R}$$ – Brian Sep 05 '18 at 17:55
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    @BrianVanegasParra: with $a = -4$, $\sqrt {a^2} = \sqrt{(-4)^2} = \sqrt{16} = 4 \ne -4$, so $\sqrt{a^2} \ne a$ in this case. – Robert Lewis Sep 05 '18 at 18:03
  • Yes, I understand, but -4 is a real number, can I prove that it is true because its result is in the real numbers? – Brian Sep 05 '18 at 18:08
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    @BrianVanegasParra: of course $-4 \in \Bbb R$; it's just that not all real numbers have real square roots; the function $\sqrt \cdot$ is only defined for non-negative arguments. – Robert Lewis Sep 05 '18 at 18:11
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    Thank you, your comments were very helpful – Brian Sep 05 '18 at 18:40
  • @BrianVanegasParra: my pleasure sir! – Robert Lewis Sep 05 '18 at 18:42
  • a can perfectly well be such that $a = 0$, then $\sqrt {a^2} = \sqrt{0} = |0| = 0$. $$0 \leq \sqrt{a^2} = \vert a \vert \geq 0$$ – amWhy Sep 05 '18 at 19:11
  • @amWhy: of course we may perfectly well have $a = 0$ and your result clearly and cleanly follows, which is why I stipulated $a \ne 0$ in my answer, which begins with the line "If $0 > a$ . . . – Robert Lewis Sep 05 '18 at 19:14
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    I realize you took care. Note: I added a reference to your comment in comments about the domain of definition for the square root "function" to be a function, because this point was missed in other answers here, and is an important response to "why" |a|.... You can remove it, or expand it. I think you made the point well. In any case +1 – amWhy Sep 05 '18 at 19:23
  • @amWhy: A'all good. I don't usually leave edits to my post alone, but I will in this case. Cheers! – Robert Lewis Sep 05 '18 at 19:25
1

Here we are using a convention, which is that by $\sqrt{x}$ (for $x\in\mathbb{R}^+$) we intend the positive number $\alpha\in\mathbb{R}$ such that $\alpha^2=x.$

It is a convention, we might take the negative one (which is $-\alpha$) but it would be pointless.

Back to your question, if you take $a<0,$ square it and take the square root you obtain a positive number, that thus can't be $a,$ which is negative.

In general you have that $\sqrt{a^2}=|a|.$

Riccardo Ceccon
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1

By definition, the square root $\sqrt{x}$ of a real number $x\ge 0$ is the unique number $y\ge 0$ such that $y^2=x$.

Quite obviously there exist numbers $a\in\mathbb R$ with $a<0$. Those are not the square root of any number, because the square root is, by definition, not negative. However, the number $a^2$ does have a square root, as the square of a negative number is positive, and furthermore $(-a)^2 = a^2$. And when $a<0$ then $-a>0$, and therefore we have $$\sqrt{a^2} = \begin{cases} a & \text{for } a\ge 0\\ -a & \text{for } a<0 \end{cases}$$ Now there is a function that is defined exactly that way, and that is the absolute value function $|x|$. Therefore the above equation can be written as $$\sqrt{a^2} = |a|.$$

amWhy
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celtschk
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  • Thanks, but the in between part is just another definition for $|a|$. Hence I'd argue $\sqrt {a^2} = |a|$, and then elaborate, that means, $|a| = \begin{cases} a & \text{for } a\ge 0\ -a & \text{for } a<0 \end{cases}$ – amWhy Sep 05 '18 at 18:27
  • @amWhy: The case definition is more elementary: It works for any totally ordered abelian group. – celtschk Sep 05 '18 at 19:57
  • Next time an asker asks about |a| in a post tagged "order theory" and "abelian groups", I'll link them to your answer. Here, I think either you or me is nitpicking; in any case, I found the original answers to the duplicated posts, and others here, far more helpful than yours. – amWhy Sep 05 '18 at 20:18
1

Think about this ...

$2 = \sqrt{4} \neq \sqrt{-2}\sqrt{-2} = \sqrt{2}i\sqrt{2}i = -\sqrt{2}\sqrt{2} = -2$.

amWhy
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Anik Bhowmick
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1

Since for $x\ge 0$ we define

$$y=\sqrt {x} \iff y^2=x \quad y\ge 0,$$

we have that

$$\forall a\in \mathbb{R} \implies \sqrt {a^2}=|a|.$$

Indeed, for example, for $a=\pm 2$

  • $\sqrt {2^2}=\sqrt 4=2$
  • $\sqrt {(-2)^2}=\sqrt 4=2$
amWhy
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user
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  • The second statement doesn't follow from the first. We have, rather, for real $x, y$, $ y= \sqrt{x^2} = |x|$. We don't need to know that $y^2 = x$ – amWhy Sep 05 '18 at 18:48
  • @amWhy I mean that by definition for $x\ge 0$ we define $y=:\sqrt x$ the real number $y\ge 0$ such that $y^2=x$. From that definition $\sqrt {a^2}=|a|$ follows. Where am I wrong? – user Sep 05 '18 at 18:54
  • You are not "wrong", just made things a tad more complicated than need be. But I have not said, nor voted, to suggest you are wrong. It was merely the seventh or eight time it had already been said. This is just me, I'm sure, but I prefer not to repeat an answer that's already been repeated, and said in just about every possible permutations of words in the definition as possible to express the same definition. – amWhy Sep 05 '18 at 19:17
  • @amWhy Without the first definition it is not so clear why $\sqrt {(-a)^2}=-a$ is not an acceptable conclusion, for that I gave that explanation. For the duplicates I think that it is unavoidable that after many years some topic is presented many times as a question and as related answers. For expert users that can be bad thing but for many contributors those topics are discussed for the first time. I think that the system for the closure of duplicates works very well and allow us to behave as we prefer. Thanks, Bye. – user Sep 05 '18 at 19:29
0

For $a \in \mathbb R$, $\sqrt{a^2} = |a|$ by definition, so $\sqrt{(-5)^2} = |-5| =5.$

amWhy
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maveric
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    Well, not quite the definition. There is a very specific reason why this identity holds (which of course follows from the definition of $\sqrt\bullet$). –  Sep 05 '18 at 17:46
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    for me a is always a real number, reason to prove that it is true – Brian Sep 05 '18 at 18:23
  • actually, it can be be always visualluzed as distance of point from (0,0). – maveric Sep 22 '18 at 20:20
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For $b>0$, the symbol $\sqrt b$ means, by definition, the nonnegative solution of $x^2=b$.

The solutions of $x^2=b$ are $\pm\sqrt b$.

lhf
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  • But the problem isn't $x^2 = b$: what is $b$?. We are given $a \in \mathbb R$ (positive, negative, not complex). Then the solution, for any real $a$ in \mathbb R$, $\sqrt{a^2} = |a|$. – amWhy Sep 05 '18 at 18:23
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Well it really should be

$$a \in \mathbb{R} \implies \sqrt{a^2} \in \{-a,a\}$$

as $(-a)^2=a^2$.

[I think your teacher was asking you to recognize the ambiguity, that both $-a$ and $a$--instead of only $a$--square to $a^2$.]

Mike
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    I think $\sqrt a$ is general taken to be positive for positive $a$. So it's not really correct to say $\sqrt a \in {-a, a }$. What is correct is $\pm \sqrt a \in {-a, a}$. Cheers! – Robert Lewis Sep 05 '18 at 17:52
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    I am not positive that everyone follows that convention though: some may take $\sqrt{r}$ as any value that squares to $r$. I suspect the teacher really was asking the OP to recognize that $-a$ also squares to $a^2$. If I were writing a paper and it wasn't clear I would be explicit that by $\sqrt{r}$ for real positive $r$, I mean the positive real number that squares to $r$. – Mike Sep 05 '18 at 17:54
  • yes, it can be, but that does not exclude it from the real numbers – Brian Sep 05 '18 at 18:01
  • @Mike I fullt understand you point but according to the usual definition, it is not correct claim your answer. Indeed we have that $\sqrt{a^2}=|a|$ and of course as $|a|^2=a$ and $(-|a|)^2=a$. That is the square root $\sqrt x$ of a non negative value $x\ge 0$ is the unique non negative value $y\ge 0$ such that $y^2=x$ andm of course $(-y)^2=x$. – user Sep 05 '18 at 19:35
  • These "easy" questions always seem to get the most debate. I think we can all agree that both $-a$ and $a$ square to the same thing, I think that was what the teacher was asking the student to recognize that, and that is reflected in my answer. [Even though I have a PhD I always forget the convention of the sign of $\sqrt{r}$ and I don't think I am the only one. In practice typically the author makes it explicit whether $\sqrt{r}$ is always nonnegative if it wasn't clear already.] – Mike Sep 05 '18 at 19:40
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The words we often say aren't quite precise. For example, we see $\sqrt{9}$, and say "the square root of $9$." But actually $9$ has two square roots, namely $3$ and $-3$. What we should say when we see $\sqrt{9}$ is "the nonnegative square root of $9$" (also known as the principal square root of $9$).

In short, $\sqrt{x}$ for nonnegative $x$ means (always and only) the nonnegative square root of $x$.

paw88789
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0

suppose a = -4

$$ a = -4 ⟹ \sqrt{-4}^2 = \sqrt{16} = 4 ≠ -4 $$

It is false

Brian
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