# Help with True/False algebra inequality

• Sep 9th 2009, 01:57 PM
sleigh
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

• Sep 9th 2009, 02:07 PM
11rdc11
Quote:

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
• Sep 9th 2009, 02:16 PM
Plato
Quote:

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

Quote:

Originally Posted by 11rdc11
0 < b3 - a3 true if a and b are positive

It is true period. The function $\displaystyle f(x)=x^3$ is increasing for all x.
• Sep 9th 2009, 02:17 PM
Matt Westwood
Quote:

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 $\displaystyle a < b$ then $\displaystyle a^3 < b^3$.

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

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

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

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

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

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

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

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

An alternative approach makes use of the fact that the cube function is monotonically increasing - but that's really on a calculus course.
• Sep 9th 2009, 03:57 PM
11rdc11
Quote:

Originally Posted by Plato
It is true period. The function $\displaystyle f(x)=x^3$ is increasing for all x.

Thanks I didn't realise he meant cubed
• Sep 9th 2009, 07:56 PM
Mozart
$\displaystyle a < b = (a<b)^3 = \boxed{a^3 < b^3}$
• Sep 9th 2009, 10:09 PM
Matt Westwood
Quote:

Originally Posted by Mozart
$\displaystyle a < b = (a<b)^3 = \boxed{a^3 < b^3}$

You can't just put an inequality into a cube like that, surely?
• Sep 9th 2009, 11:48 PM
Logic
I trust this works?
• Sep 10th 2009, 01:00 PM
Matt Westwood
How sure are we that $\displaystyle a^2 + ab + b^2 > 0$?
• Sep 10th 2009, 04:11 PM
Krizalid
$\displaystyle a^{2}+ab+b^{2}=\frac{4a^{2}+4ab+4b^{2}}{4}=\frac{( 2a+b)^{2}+3b^{2}}{4}>0.$

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