Logic and proof I had an assignment from class,
to proof 
for all real numbers $R$, $x$ is subsets of $R$, if $x^2 - 2x\ne -1$,
then $x\ne 1$.
in contrapositive proof and contradiction.
So far with my knowledge, I should construct a direct proof, and show that it is not possible. I have no idea how to start with.
Ironically, I could construct a proof by contraposition which assume $x=1$, and 
'$x^2 - 2x = -1$'.
As we apply $x=1$ in to '$x^2 - 2x$', we get $-1$ which is a negation of the hypothesis of the original theorem - we proved it with proof by contraposition.
I have no idea about proof by contradiction by the way.
p.s.
Sorry I don't know how to write words into mathematical symbols, but I think this should still work.
 A: To prove statment T by contradiction, assume not T is a true statement, show that it then leads to a false statement, ie a statment that contradicts some other statement already shown to be true or generally accepted as true or an axiom.
To prove a statement "by contraposition" that is the statement T implies R, prove not T implies not R. By law of contraposition the statement T implies R; is true when ever not T implies not R or simply it's contrapositive. 
It seems you are to do the proof a certain way, as there are multiple ways to prove such a statement. It is not exactly clear but it seems you should prove the contrapositive of the negation of the statement leads to a contradiction. 
Assuming your statement you wish to prove is
$$  (\forall x: x \subseteq \Bbb R) \Rightarrow ((x^2-2x \neq-1 )\Rightarrow (x \neq 1)) $$
if $ R = \forall x: x \subseteq \Bbb R$
and $T = (x^2-2x \neq-1 )\Rightarrow (x \neq 1))$
What you want to show is, possibly $ R \Rightarrow \lnot T$ proof by contradiction (it will lead to contradiction).
Or $ \lnot R \Rightarrow \lnot T$ proof of the contrapositive, implying original statement's truth.
Or $ \lnot R \Rightarrow T$ leads to a contradiction (proof by showing contradiction showing the truth of the contrapositive, implying the original statement). I think this is the method you are to use, however it is hard to be certain due to the informal nature of the way the question was asked.
A: required to prove$$x^{2}-2x\neq-1\:\:implies\:\:x\neq1$$
so if hypothesis is false, then:$$x^{2}-2x=-1\:\:\: for\: some\:\:\:x\neq1$$ but then we have $$x^{2}-2x+1=0$$ $$(x-1)^2=0$$ $$x=1$$ but this is a contradiction because we supposed $$x\neq1$$
