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?

• 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 $x < -1$. Jun 2 '20 at 10:01

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.

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.