Is this form of proof circular reasoning? This is a general question about trivial proofs. Recently in a textbook I was reading there was simple exercise in the form:

Prove that $\frac{1}{1 - \sqrt2} = -(1 + \sqrt2)$

I can prove this simply by multiplying like this:
$\frac{1 * (1 + \sqrt2)}{(1 - \sqrt2) * (1 + \sqrt2)} = \frac{1 + \sqrt2}{-1} = -(1 + \sqrt2)$
However, my question is this, can I also prove questions like this by manipulating both sides of the supposed equality, or does that in some way assume what I'm trying to prove? I'll give an example of what I mean with the above exercise.
So instead of just multiplying the numerator and denominator to reach my answer, I multiply both sides of the equation(?) by $1 -\sqrt2$ so that:
$\frac{1}{1 - \sqrt2} * (1 - \sqrt2) = -(1 + \sqrt2) * (1 - \sqrt2) = $
$1 = -(1 - 2) = -(-1) = 1$
I.e. we've arrived at a truth (that 1 = 1) from our supposed truth. Would this also constitute a proof? I feel like it doesn't, but I can't quite explain why. It seems circular, but maybe it's not.
If anyone could clarify that would be great!
 A: You are right, but it's not circular reasoning. It's another type of fallacious argument called affirming the consequent. Just because something implies a true statement doesn't mean it is true. For instance, the statement "for all $a$ and $b,$ $a+b = a\cdot b$" implies the true statement $2+2=2\cdot 2,$ but it's obviously false.
However, any statement that implies a false statement must be false, so if you'd derived $1=0$ you'd be justified in concluding the original equation was false.
A: Doing this in general is not valid. Trivially, we could multiply both sides by $0$ and get $0=0$ no matter what.  However, if $f$ is any function that is bijective, that is it has an inverse $f^{-1}$ such that $f(f^{-1}(y)) = y$ and $f^{-1}(f(x))=x$, then we generally have $x = y \iff f(x) = f(y)$.  The domain of $f$ matters for whether $f$ is a bijection or not. $f(x) = x^2$ is a bijection if we restrict the domain to either numbers $\geq 0$ or numbers $\leq 0$. So it's valid to say $-1 = -2/2$ because $(-1)^2 = (-2/2)^2$, but it's not valid to say $-1 = 1$ because $(-1)^2 = 1^2$ as the latter requires using $f(x)=x^2$ at both positive and negative numbers. Similarly, for real numbers $f(x)=x^3$ is bijective, but it isn't for complex numbers.  Multiplying both sides by an invertible value or adding a value are always bijective operations. So the logical structure of a proof that "does the same thing to both sides" is $$a = b \impliedby f(a) = f(b) \impliedby g(f(a)) = g(f(b)) \impliedby h(g(f(a))) = h(g(f(b)))$$ where $f$, $g$, and $h$ are bijections, and $h(g(f(a)))=h(g(f(b)))$ is obviously true, e.g. $1=1$. Each bijection corresponds to one "step" of the proof.
So your second proof is valid because $f(x)=(1-\sqrt{2})x$ is a bijection.
