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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.

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

2 Answers 2

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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.

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  • $\begingroup$ Greatly helped setup the big picture of proof methods :) $\endgroup$
    – Minjae
    Commented Jun 8, 2015 at 5:14
  • $\begingroup$ 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 $R \Rightarrow T$. $\endgroup$ Commented Jun 8, 2015 at 7:17
  • $\begingroup$ Still it is interesting to consider the extension of the theorem to set's not contained in the reals. $\endgroup$ Commented Jun 8, 2015 at 7:18
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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 $$x\neq1$$

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  • $\begingroup$ I already had x=1 but I had no idea how to finish the proof. $\endgroup$
    – Minjae
    Commented Jun 8, 2015 at 5:05
  • $\begingroup$ I don't understand where we did 'supposed x≠1'. Would you explain? Thought we only assumed x2−2x=−1 $\endgroup$
    – Minjae
    Commented Jun 8, 2015 at 5:05
  • $\begingroup$ 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 $\endgroup$ Commented Jun 8, 2015 at 5:09
  • $\begingroup$ This writing style is hard to follow. Where does one thought end and the next begin? $\endgroup$
    – Théophile
    Commented Jun 8, 2015 at 5:12
  • $\begingroup$ Okay. It seemed to me like I was almost there leading contradiction, but I wasn't so sure about the x=1. Thanks $\endgroup$
    – Minjae
    Commented Jun 8, 2015 at 5:13

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