I need to prove that if $a > 1$ then $a^2 > a$

At first glance it seems that this is true, if letting $a = 2$ for example than,

$$2 > 1$$ and $$2^2 > 2$$

What I tried to do is work with $a^2 > a$ to someone show that a > 1:

$$a^2 > a$$ $$\frac{1}aa^2 > \frac{1}aa$$ $$a > 1$$

But to me this doesn't seem like a proof at all so I'm wondering what is the correct approach to proving this statement?

  • $\begingroup$ Just do the inverse of your last three steps. $\endgroup$ – user122283 Feb 7 '14 at 1:15
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    $\begingroup$ How does $\tfrac{1}{a}a^2>\tfrac{1}{a}a$ imply $a^2>a$? It seems like it's using the result "$x>y$ and $b>0$ implies $bx>by$" which is a generalization of the result you want to prove. Seems dangerously circular. $\endgroup$ – Rebecca J. Stones Feb 7 '14 at 1:26

If $a>1$ then $a>0$ so you can just multiply both sides by $a$ without changing the sign so that $a\cdot a=a^2>1\cdot a=a$ and $a^2>a$


The proof you found is rigorous, because since $a>1$, $\frac{1}{a}$ is a positive real number, preserving the inequality.


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