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
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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 - a3Quote:
Originally Posted by sleigh
true if a and b are positive
Quote:
Originally Posted by sleigh
It is true period. The function $\displaystyle f(x)=x^3$ is increasing for all x.Quote:
Originally Posted by 11rdc11
I assume you mean: if $\displaystyle a < b$ then $\displaystyle a^3 < b^3$.Quote:
Originally Posted by sleigh
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.
Thanks I didn't realise he meant cubedQuote:
Originally Posted by Plato
$\displaystyle a < b = (a<b)^3 = \boxed{a^3 < b^3}$
You can't just put an inequality into a cube like that, surely?Quote:
Originally Posted by Mozart
I trust this works?
How sure are we that $\displaystyle a^2 + ab + b^2 > 0$?
$\displaystyle a^{2}+ab+b^{2}=\frac{4a^{2}+4ab+4b^{2}}{4}=\frac{( 2a+b)^{2}+3b^{2}}{4}>0.$
Though you can't manipulate what you want to prove.
