# Thread: Help with True/False algebra inequality

1. ## Help with True/False algebra inequality

Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!

True or False:
If a<b then a3 <b3

2. Originally Posted by sleigh
Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!

True or False:
If a<b then a3 <b3
0 < b3 - a3

true if a and b are positive

3. Originally Posted by sleigh
Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!
True or False:
If a<b then a3 <b3
Originally Posted by 11rdc11
0 < b3 - a3 true if a and b are positive
It is true period. The function $f(x)=x^3$ is increasing for all x.

4. Originally Posted by sleigh
Hi. I had the problem below for homework. I think it is true, but I don't know why. Can anyone give me a reason why this is true, because I need to provide an explination. Thanks!

True or False:
If a<b then a3 <b3
I assume you mean: if $a < b$ then $a^3 < b^3$.

It is actually true, but needs a fair amount of work to demonstrate.

First, you can show it's true when both $a$ and $b$ are positive, as follows.

You know that if $a, b, c < 0$ then if $a < b$ then $a c < b c$.

Okay, so $a < b$ implies $a^2 < a b$ (putting $c = a$ in the above).

Then $a < b$ implies $ab < b^2$ (putting $c = b$ in the above).

So you've shown $a < b$ implies $a^2 < b^2$.

Do the same (similar) thing to prove for $a^3 < b^3$.

You go through a similar process when $a$ and $b$ are both negative, and again when $a$ is negative and $b$ positive.

An alternative approach makes use of the fact that the cube function is monotonically increasing - but that's really on a calculus course.

5. Originally Posted by Plato
It is true period. The function $f(x)=x^3$ is increasing for all x.
Thanks I didn't realise he meant cubed

6. $a < b = (a

7. Originally Posted by Mozart
$a < b = (a
You can't just put an inequality into a cube like that, surely?

8. I trust this works?

9. How sure are we that $a^2 + ab + b^2 > 0$?

10. $a^{2}+ab+b^{2}=\frac{4a^{2}+4ab+4b^{2}}{4}=\frac{( 2a+b)^{2}+3b^{2}}{4}>0.$

Though you can't manipulate what you want to prove.