Is this proof about $a^3>a \rightarrow a^5>a$ correct?

Basically what I need is to know if this proof is correct

what I need to prove is:
if $a^3 > a$ then $a^5>a$ so, what i did was this:
$a^3 a^2 > a a^2$
$a^5 > a^3$
because $a^5>a^3$ I can say that $a^5>a$

EDIT: $a,b \in \mathbb{R}$, sorry I totally forgot to write it

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It would be good to specify the context: Do you want a proof for the case that $a$ is a real number? Do you want it for the case of arbitrary ordered fields? If it's not the latter, tagging abstract-algebra is perhaps not the most suitable tag. I guess algebra-precalculus or inequality would be better. \\ This in fact doesn't really matter to the way it's proved, but I think it's good to state your assumptions whenever you post a question. –  Martin Sleziak Jan 11 '12 at 9:42

Looks ok to me. Since $a^2$ is positive no matter what, it doesn't change the sense of the inequality. And then you simply deduced from the fact $a^5 > a^3$ and $a^3 > a$ then $a^5 > a$. Looks ok to me. And it also seems kind of obvious that the answer is yes.

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It may be true but I would say it is not totally obvious for negative $a$. –  Henry Jan 11 '12 at 8:47
It doesn't even hold for negative a. If we take -2, we get $-8 > -2$ which is obviously false. That's because you're dealing with odd powers (where the sign counts). So, his proof is correct and the fact that $a^3 > a^1$ sort of implies you're dealing with $a \in \mathbb{R}_+$. –  andreas.vitikan Jan 11 '12 at 8:58
It does hold in $(-1,0)$ even if this is not obvious. For example $\left(-\frac{1}{2}\right)^3 \gt \left(-\frac{1}{2}\right)$ and $\left(-\frac{1}{2}\right)^5 \gt \left(-\frac{1}{2}\right)$ –  Henry Jan 11 '12 at 9:56
@andreas.vitikan: $a^{3}>a^{1}$ does not imply $a\in \mathbb{R}_{+}$. Consider e.g. $a=-\frac{1}{2}$. –  Thomas E. Jan 11 '12 at 10:10
Yes, you are right. Now that I think of it, it implies that a is different than 0 and $a > -1$. –  andreas.vitikan Jan 11 '12 at 11:16

To clarify this proof further, I would additionally say, given $a^3 > a$, that $a \ne 0$ and also that $a^2 \gt 0$. This justifies the algebraic step of multiplying both sides of the inequality by $a^2$.

EDIT: This answer is completely different from my original one which was all wrong as pointed out by Henry.

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As a counterexample $\left(-\frac{1}{2}\right)^3 \gt \left(-\frac{1}{2}\right)$ but $\left(-\frac{1}{2}\right)^2 \lt 1$. –  Henry Jan 11 '12 at 8:45
@Henry, indeed. The other answer nailed it, actually. I'll delete this so as not to sow confusion once I recover from my embarrassment... –  Dan Brumleve Jan 11 '12 at 8:52

How do you deduce $a^3a^2>aa^2$? You have to assume that $a^2>0$. You should either:

• separately consider the case $a=0$ (where of course $0^3\not>0$), or
• just deduce $a^3a^3 \geq aa^2$ (since $a^2\geq 0$) so $a^5\geq a^3 > a$.
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Well, when $a = 0$ then simply everything is 0. 0 at any power except 0 is 0. But I don't think that's what they had in mind when they made the exercise and put it in the textbook. –  andreas.vitikan Jan 11 '12 at 9:24
But your solution should note that case anyway, that is how to make a complete proof. –  GEdgar Jan 11 '12 at 14:03