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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?

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10 Answers

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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.

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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 $\sqrt{a^2}=|a|.$

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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 $$\sqrt{a^2} = |a|.$$

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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 $a=\pm 2$

• $\sqrt {2^2}=\sqrt 4=2$
• $\sqrt {(-2)^2}=\sqrt 4=2$

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Think about this ...

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

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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 $\pm\sqrt b$.

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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 $a^2$.]

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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 $x$.

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

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suppose a = -4

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

It is false

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