Is $\sqrt{x^2} = x$?

Question

Does the $\sqrt{x^2}$ always equal $x$? I am trying to prove that $i^2 = -1$, but to do that I need to know that $\sqrt{(-1)^2} = -1$. If that is true then all real numbers are imaginary, because an imaginary number is any number that can be written in terms of $i$. For example, 2 can be written as $i^2 + 3$. Does this work or did I make an error?

Imaginary numbers are of the form $ib$ where $b$ is real. It is not right to say that 'in terms of $i$'. — Extremal, Feb 27 '15 at 01:33
@EpsilonDelta, it's not an imaginary number since $i^2 + 3 = -1 + 3 = 2$ so it's in the form $a$. Mostly because $a,b$ in $a+bi$ are restricted to real numbers — jameselmore, Feb 27 '15 at 01:34
${\sqrt x}^2$ always equals $x$. $\sqrt{x^2}$ equals $\lvert x\rvert$ for real values of $x$. (By the way: $2=i^2+3$, yes.) — Akiva Weinberger, Feb 27 '15 at 01:35
I think you are talking about complex numbers, $\mathbb{R} \subset \mathbb{C}$ in the same way that $\mathbb{N} \subset \mathbb{R}$ — Gonate, Feb 27 '15 at 01:35
possible duplicate of When is square root the inverse of the square? — MJD, Feb 27 '15 at 01:36
You shouldn't be trying to prove that $i^2 = -1$, since it's true by definition. I can't think of a case when you can prove that a definition is true; it's defined to be so. — pjs36, Feb 27 '15 at 02:01

score 2 · Answer 1 · answered Feb 27 '15 at 01:52

2

Not always. $\sqrt{(-1)^2}=\sqrt{1}=1\neq -1$. In general $\sqrt{x^2}=|x|$

answered Feb 27 '15 at 01:52

Ryan

102

5

Equation $\sqrt{x^2} = |x|$ is only for real $x$. – GEdgar Feb 27 '15 at 02:05
In general $\sqrt{x^2}=\pm x$. – Milo Brandt Feb 27 '15 at 04:00
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@Meelo, the symbol $\sqrt{}$ by definition denotes the nonnegative square root. – Santiago Canez Feb 27 '15 at 04:30
@SantigoCanez Oh, I was intending my equation to be read as "$\sqrt{x^2}$ is either $x$ or $-x$" not "$\sqrt{x^2}$ is both $x$ and $-x$." This might have been unclear. (I used $\pm$ instead of $|\cdot |$ to emphasize that, for complex input, the principal value can't be taken to be the non-negative square root since no such thing exists) – Milo Brandt Feb 27 '15 at 04:36

score 0 · Answer 2 · edited Jun 12 '20 at 10:38

It is not true that $\sqrt{x^2} = x$. As a very simple example, with $x=-2$, we obtain $$ \sqrt{(-2)^2} = \sqrt{4} = 2 \ne -2. $$ In general, if $x \in \mathbb{R}$, then $\sqrt{x^2} = |x|$. Things get more complicated when you start working with complex numbers, but I think that a discussion of "branches of the square root function" is quite a bit beyond the scope of this question.
There is a serious problem of definitions in the question. The question asserts "...all real numbers are imaginary, because an imaginary number is any number that can be written in terms of $i$." However, this is not the definition of an imaginary number. An imaginary number is a number $z$ such that there is some real number $y$ such that $z = iy$, where $i$ is the imaginary unit. A number such as $i^2 + 3$ is not an imaginary number, since there is no real number $y$ such that $2 = i^2 + 3 = iy$.

On the other hand, it is reasonable to say that every real number is a complex number. A complex number is a number $z$ such that there are $x,y\in\mathbb{R}$ such that $z = x + iy$. In the case of the example give, we have $$ i^2 + 3 = (-1) + 3 = 2 = 2 + i0. $$

Is $\sqrt{x^2} = x$?

2 Answers2