Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$
Q2 - I was thinking it would be:
$\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)
Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.
Answer
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 $$\sqrt{-x}\cdot\sqrt{-x}=\sqrt{x^2}=\lvert x\rvert=-x.$$
No comments:
Post a Comment