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Q1 - Why is $\sqrt{-x}\times\sqrt{-x}=-x?$

Q2 - I was thinking it would be: $\sqrt{-x}\times\sqrt{-x}=\sqrt{-x\times-x}=\sqrt{x^2}$ but apparently not (why not?)

Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}\times\sqrt{-x}=i\sqrt{x}\times i\sqrt{x}=-\sqrt{x^2}=-x$.

GambitSquared
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    The square root function cannot be properly defined over the complex numbers so that the rule $\sqrt x \sqrt y = \sqrt{xy}$ holds. – Stephen Montgomery-Smith Jul 15 '15 at 22:26
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    Even more confusing: $i=\sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}=-i$. Thinking as for real numbers can easily give you nonsense. – A.Γ. Jul 15 '15 at 22:29
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    A square-root of a number is something whose square is that number. So, for any $z$, if there is such a thing as $\sqrt{z}$, then its square is $z$. – GEdgar Jul 15 '15 at 22:31
  • Can I arrive at $\sqrt{-x}*\sqrt{-x}=-x$ without using complex numbers? By the way, in the problem I try to solve I know that x is negative. – GambitSquared Jul 15 '15 at 22:45
  • @ImreVégh: If you know $x$ is negative, then $-x$ is positive, and $\sqrt{-x}\sqrt{-x}=-x$ is then just the familiar $\sqrt a\sqrt a=a$ for positive $a$ in the case that $a=-x$. Under this assumption you also do have that $\sqrt{x^2}=-x$, so the answers in your Q1 and Q2 are the same. – hmakholm left over Monica Jul 15 '15 at 22:48
  • I see, but then why doesn't $\sqrt{-x*-x}=\sqrt{x^2}=x$ work? – GambitSquared Jul 15 '15 at 22:50
  • @ImreVégh: Because $\sqrt{x^2}$ is by definition the positive number whose square is $x^2$. Both $x$ and $-x$ square to $x^2$, and $\sqrt{x^2}$ is the one of $x$ and $-x$ that is not negative. And if $x$ is negative then that means it is $-x$. – hmakholm left over Monica Jul 15 '15 at 22:52
  • @Imre: See my answer below. Let me know if you have any questions. – Cameron Buie Jul 15 '15 at 22:53

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By definition, a square root of $u$ is something that, when squared, gives $u$. Hence, if $\sqrt{-x}$ exists, then $\sqrt{-x}\cdot\sqrt{-x}=-x$ by definition.

Now, the principal square root of a nonnegative real number $u$ is the nonnegative real number whose square is $u$. We denote this by $\sqrt u.$ What this means is that, for $\sqrt{-x}$ to be defined, we need $-x$ to be a nonnegative real number, which means that $x$ is nonpositive real. Now, if $x=0,$ this is no problem, and you can say that $$\sqrt{-x}\cdot\sqrt{-x}=x,\tag{$\star$}$$ since $-0=0.$ If $x$ is positive, then the left hand side of $(\star)$ isn't even defined, so $(\star)$ is false. If $x$ is negative, then the right hand side of $\star$ is a negative number, while the left hand side is the square of a positive number, so is positive, and so $(\star)$ is again false.

However, we can conclude that, if $\sqrt{-x}$ is defined (that is, if $x$ is nonpositive real), then $$\sqrt{-x}\cdot\sqrt{-x}=\sqrt{x^2}.$$ How can this be? Well, remember that a principal square root has to be nonnegative real, so for any real $u,$ we have in general that $$\sqrt{u^2}=\lvert u\rvert.$$ In particular, then, since $x$ is nonpositive real, then $$\sqrt{-x}\cdot\sqrt{-x}=\sqrt{x^2}=\lvert x\rvert=-x.$$

Cameron Buie
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$\sqrt{-x}*\sqrt{-x}=(\sqrt{-x})^2=-x$ (the square root and square cancels out)

Chuks
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The square root function is not uniquely defined, but is rather multivalued. Therefore by declaring that $\sqrt{1}=1$ you are making a branch cut and eliminating the -1. By changing the choice of branch you can obtain the correct answer and use the property that @StephenMontgomery-Smith mentioned (which is not valid for the entire domain of complex numbers but can be restricted to a certain domain).

MBP
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  • We don't need to care about the branch cut here; any branch satisfies $(\sqrt{x})^2=x$. (The problematic equation is rather $\sqrt{(x^2)}=x$ which cannot hold everywhere) – Milo Brandt Jul 15 '15 at 23:57
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The square root function is defined for $\left\{ x \in \mathbb{R}: x \geq 0\right\}$. When you have $-x$ under the square root, the implication is that $x$ is negative. The square root function doesn't work the same way for complex numbers as it does for reals.

Race Bannon
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Here is a "proof" that $1=-1$:

$$1=\sqrt{1}=\sqrt{1^2}=\sqrt{(-1)^2}=\sqrt{(-1)(-1)}=\underline{\sqrt{-1}\sqrt{-1}}=i^2=-1$$

The problem is in the underlined step. The rule of $\sqrt{xy}=\sqrt{x}\sqrt{y}$ only works when $x,y>0$.

Moya
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