Prove the assertion: if $x<-1$, then $x^2>1$
I am trying to prove this by contrapositive and so far I have
$x^2\ge1$ then $x\ge-1$
$x^2-1\ge0$
$(x-1)(x+1)\ge0$
$-1\le x \le 1$
which does not prove what I am trying to prove.
Error in proof by contrapositive.
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$\begingroup$
logic
proof-verification
foundations
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0The implication from second last to last line is incorrect. – 2017-01-25
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0Alternately, you can see this http://math.stackexchange.com/questions/2112657/supply-a-proof-for-the-assertion/2112676#2112676 – 2017-01-25
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4You didn't do contrapositive, which would be: if $\;x^2\rlap{\,\,/}>1\;$ , then $\;x\rlap{\;\,/}<-1\;$ , or what is the same : $\;x^2\le1\implies x\ge-1\;$ . – 2017-01-25
1 Answers
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Why not prove it directly?
$$x<-1\implies \text{ (since}\;x\;\text{ is negative)} \;\; x^2>-x>1$$
the last inequality being the very first times $\;-1\;$ .