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let $x>1$
Obviously $x^2>x$
Then $x^2>x>1$
Taking $x^2>1$, we can assert that this holds true for all the values for $x>1$ and $x<-1$
But if I take $-5$ such that $x<-1$, then $x^2>x$ holds but $x>1$ doesn't. Why is it so? Doesn't $x^2>x>1$ mean that all of the three must be true?

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    $\begingroup$ Yes, if you say $x^2 > x > 1$ then both inequalities should hold. Also, you're assuming in the very beginning that $x>1$ so this inequality can't be applied to $x < -1$. $\endgroup$
    – M. Wang
    Jun 2 '20 at 10:01
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When you deduced that $𝑥^2>𝑥>1$, you specifically had the constraint that $x > 1$. Of course, this inequality is not then applicable since now you have $x > -5$ which is a larger domain than $x > -1$.

More specifically, $x > 1 \Rightarrow x^2 > x > 1$ is a true statement, but $x^2 > x \iff x> 1$ is clearly not a true statement as you have demonstrated.

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Yes, $x^2>x>1$ is really shorthand for "$x^2>x$ and at the same time $x>1$". So it doesn't hold for $x = -5$.

You start with "let $x>1$". This is not true for $x = -5$. You do get $x^2 > 1$ in this case, but your argument as written won't get you there. You need a different argument.

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