Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

3 Answers 3

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.

share|improve this answer
    
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.

share|improve this answer
    
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 $a^5\geq a^3 > a$.
share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.