# 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 $x\ne 1$.

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.

p.s.

Sorry I don't know how to write words into mathematical symbols, but I think this should still work.

• To prove by contradiction: assume the hypothesis is not true, and manage to show that this assumption leads to a contradiction. Commented Jun 8, 2015 at 4:31
• Should I start with assuming x^2−2x=-1 ? Commented Jun 8, 2015 at 4:46
• And then the next thing would be leading to the contradiction where x2−2x=−1, then x=1. Than, is the proof succeeded? Commented Jun 8, 2015 at 4:47
• Yes, exactly :) Commented Jun 8, 2015 at 4:47
• I think I'm almost there :) thanks for editing btw Commented Jun 8, 2015 at 5:06

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.

• Greatly helped setup the big picture of proof methods :) Commented Jun 8, 2015 at 5:14
• Sorry I miss interpreted your original statement earlier. But the outline provided still stands (switch R and T to the equalities). Assume $\lnot R \Rightarrow T$, then proceed to show it leads to a contradiction by solving the quadratic equation who's truth implies T under assumption. That quadratic also implies the negation of T which is the contradiction that resulted from the assumption of $\lnot R \Rightarrow T$. Therefore that statement as a whole is false, it's negation must be true. That is $\lnot R \Rightarrow \lnot T$. by contrapositive law this implies $R \Rightarrow T$. Commented Jun 8, 2015 at 7:17
• Still it is interesting to consider the extension of the theorem to set's not contained in the reals. Commented Jun 8, 2015 at 7:18

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 $$x\neq1$$

• I already had x=1 but I had no idea how to finish the proof. Commented Jun 8, 2015 at 5:05
• I don't understand where we did 'supposed x≠1'. Would you explain? Thought we only assumed x2−2x=−1 Commented Jun 8, 2015 at 5:05
• this seems to be a special case where the proof by contrapositive is very similar to contradiction. somehow we have to use that x=1 causes a contradiction Commented Jun 8, 2015 at 5:09
• This writing style is hard to follow. Where does one thought end and the next begin? Commented Jun 8, 2015 at 5:12
• Okay. It seemed to me like I was almost there leading contradiction, but I wasn't so sure about the x=1. Thanks Commented Jun 8, 2015 at 5:13