Verification if a>b>0 then cube root of a is bigger than cube root of b

Question

I am trying to do this proof. Please tell me if it is correct. Please help me also with the mathematical symbols.

If a>b>0 then cube root of a is bigger than cube root of a.

Case 1:

Suppose for the search of contradiction that cube root of b is

bigger than cube root of a.

Then (cube root of b)(cube root of b)>(cube root of a)(cube root of a)

By extension, (cube root of b)(cube root of b)(cube root of b) > (cube root of a)(cube root of a)(cube root of a)

Therefore, b>a, a contradiction of the starting condition.

Case 2:

Suppose that cube root of b is equal to cube root of a.

By the same method:

(cube root of b)(cube root of b)(cube root of b) = (cube root of a)(cube root of a)(cube root of a)

Therefore, b=a, also a contradiction of the starting condition.

Then, if a>b>0 then cube root of a is bigger than cube root of b. Q.E.D.

@LinAlg if negative, multiplication reverses the inequality. But then again the second multiplication reverses that back. Just needs some care in handling the negative cases. — Macavity, Oct 20 '16 at 18:51
@LinAlg Thank you for your fast answer. As a>b>0, the cube roots are positive, Then the case that you mention does not exist. Correct? — Beginner, Oct 20 '16 at 18:54
@Macavity As a and b are positive, the cube root of them are positive too. Correct? — Beginner, Oct 20 '16 at 18:55
@copper.hat Thank you for your comment. I believe that I fixed the mistake. Correct? — Beginner, Oct 20 '16 at 18:57
@LinAlg I used a>b>0 in the contradictions, to assume that the cube roots are positive, and to multiply both sides without changing the inequality. — Beginner, Oct 20 '16 at 18:59
I edited one formula to show you how to format using MathJax (essentially TeX/LateX). Either edit the answer to see or right click on the formula and select 'Show Math As'->'TeX Commands'. — copper.hat, Oct 20 '16 at 19:03
@copper.hat I am sorry. I tried to follow what you did, but I was not successful. Even, it seems that I erased what you did, that was very nice. — Beginner, Oct 20 '16 at 19:13
NP. There are a lot of little details. If you look at the edit history (click on edited $n$ mins ago) you will see what I did. — copper.hat, Oct 20 '16 at 19:18

score 3 · Accepted Answer · answered Oct 20 '16 at 19:07

The proof is correct, with the addition of what you state as obvious: if $\sqrt[3]{b}>\sqrt[3]{a}$, then $$ \sqrt[3]{b}\sqrt[3]{b}>\sqrt[3]{a}\sqrt[3]{b}>\sqrt[3]{a}\sqrt[3]{a} $$ and similarly for the next step.

However, the proof can be simplified: you don't need to do separate cases for $>$ and $=$, for instance.

On the other hand, you can recall $x^3-y^3=(x-y)(x^2+xy+y^2)$, so $$ 0<a-b=(\sqrt[3]{a}-\sqrt[3]{b}) \underbrace{((\sqrt[3]{a})^2+\sqrt[3]{a}\sqrt[3]{b}+(\sqrt[3]{b})^2)}_{>0} $$ and therefore $\sqrt[3]{a}-\sqrt[3]{b}>0$.

Verification if a>b>0 then cube root of a is bigger than cube root of b

1 Answers1