Help with True/False algebra inequality

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September 9th 2009, 02:57 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
September 9th 2009, 03:07 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
September 9th 2009, 03:16 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 $f(x)=x^3$ is increasing for all x.
September 9th 2009, 03:17 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 $a < b$ 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.
September 9th 2009, 04:57 PM
11rdc11

Quote:

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
September 9th 2009, 08:56 PM
Mozart

$a < b = (a<b)^3 = \boxed{a^3 < b^3}$
September 9th 2009, 11:09 PM
Matt Westwood

Quote:

Originally Posted by Mozart

$a < b = (a<b)^3 = \boxed{a^3 < b^3}$

You can't just put an inequality into a cube like that, surely?
September 10th 2009, 12:48 AM
Logic

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I trust this works?
September 10th 2009, 02:00 PM
Matt Westwood

How sure are we that $a^2 + ab + b^2 > 0$ ?
September 10th 2009, 05:11 PM
Krizalid

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

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