Prove If $a^3>a$ then $a^5>a$

Question

Here was my go at it: Assume $a^3>a$. Then $$a^3>a\Rightarrow a^3-a>0\Rightarrow a(a+1)(a-1)>0$$ Solving this inequality gives the truth set $\{x\in\mathbb{R}:-1<x<0\lor x>1\}$. Then solving the inequality $a^5>a$ I get the same truth set, since $x\in\mathbb{R}$. Does this prove that $$a^3>a\Rightarrow a^5>a$$ Does this also mean $$a^5>a\Rightarrow a^3>a$$ I'm just learning how to write proofs and any help would be appreciated. Thanks!

The set ${x\in\mathbb{R}:-1<x<0,x>1}$ is empty: that is probably not what you meant. — Mariano Suárez-Álvarez, Dec 15 '13 at 19:39

score 8 · Accepted Answer · answered Dec 15 '13 at 19:35

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Suppose $a<a^3$. We must have $a\neq0$ and then $a^2>0$. It follows that we can multiply the inequality by $a^2$ without breaking it, and we get $a^3<a^5$.

answered Dec 15 '13 at 19:35

Mariano Suárez-Álvarez

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It took me so long to figure out $a<a^3$ is the same as $a^3>a$. For some reason I was reading it as $a^3<a$. Time for more coffee I guess :). So I think I understand. Just so I'm not misunderstanding, that also shows that $a^3>a\Rightarrow a^5>a$ right? Edit. Just read Newb's answer so now this makes sense. Thank you so much! – dserver Dec 15 '13 at 20:04
@dserver I didn't realize this was the same as my answer either. I also found it unintuitive that the inequality should proceed < --- I'm used to them being written > – Newb Dec 15 '13 at 20:44

score 3 · Answer 2 · answered Dec 15 '13 at 20:00

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All the other approaches are way too complicated. You can prove this straight from the field axioms, and the rules of order. You know $a^3 > a$. I assume we know $a \neq 0 \in \mathbb{R}$.

Now take $a^2$. We know $a^2 > 0$. So $a^2$ is definitely positive. (Just in case: if you don't know this, I'll write the proof in a comment.)

By the rules of order: multiplication by positive elements, we know that $a^3 \cdot a^2 > a \cdot a^2$. That is, $a^5 > a^3$. So $a^5 > a^3 > a$. By transitivity, $a^5 > a$.

answered Dec 15 '13 at 20:00

Newb

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Proof that $\forall a \neq 0 \in \mathbb{R}, a^2 > 0$:
By trichotomy, there are only two possibilities: $a > 0$ or $a < 0$. If $a >0$, then $a \cdot a > a \cdot 0$, which we know from the rules of order: multiplication by positive elements. If $a < 0$, then $-a > 0$, and $-a \ cdot -a > -a \cdot 0 \Longrightarrow (-1)(-1)a^2 > 0 \Longrightarrow a^2 > 0$.
– Newb Dec 15 '13 at 20:00
Ok, this makes more sense to me. Thanks! – dserver Dec 15 '13 at 20:06

Christopher K · Answer 3 · 2013-12-15T19:50:30.757

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The easiest way to look at it is: $a^2 + 1 > 0$. Why?

$a^5 - a = a(a^2-1)(a^2+1) = (a^2+1)(a^3-a)$ and $a^2+1\geq1>0$ and $a^3-a>0$ by hypothesis. Hence, $a^5-a>0$.

As for truth sets, they are overkill and definitely not the best way to proceed.

edited Dec 15 '13 at 19:50

answered Dec 15 '13 at 19:39

Christopher K

3,383

I don't understand what that inequality does here... – Mariano Suárez-Álvarez Dec 15 '13 at 19:40
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Well... $a^5-a = a(a^4-1) = (a^2+1)\cdot (a^3-a)$, where $a^2+1>0$ and $a^3-a>0$. So, $a^5-a>0$. – Christopher K Dec 15 '13 at 19:44
Why should $a^2-1$ be greater than $0$? Try for instance $a=-\frac12$. – TonyK Dec 15 '13 at 19:44
@TonyK: thank you for pointing that out, I corrected it. – Christopher K Dec 15 '13 at 19:44
It would be beest if instead of a «the proof follows» you wrote down a complete argument, really. The connection from $a^2+1$ to the wanted inequality is quite non-obvious! – Mariano Suárez-Álvarez Dec 15 '13 at 19:46
Hence the downvote? However, I stand by not giving a complete argument here; it was meant to guide the OP not hold their hands. – Christopher K Dec 15 '13 at 19:47
Please, do not assume that it was me who downvoted —you'd be wrong. I do not think your answer is a good one, though, as it stands. – Mariano Suárez-Álvarez Dec 15 '13 at 19:47
I don't disagree. Your criticism was justified. – Christopher K Dec 15 '13 at 19:48
I can understand why truth sets would be overkill for someone who can see solutions to easy inequalities at a glance, but can you elaborate on why they aren't a good way to proceed? Is there a case when using them will lead me in the wrong direction? At this point I'm only working on proofs of $P(x)\Rightarrow Q(x)$ – dserver Dec 15 '13 at 20:26

Prove If $a^3>a$ then $a^5>a$

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