One to One proof

• Sep 8th 2011, 04:02 PM
JSB1917
One to One proof
Let f: R+ --> R+ where R+ is the positive reals, and f(x)=x^4. Prove or disprove that it's one to one.

Ok, I understand the whole f(a)=f(b) thing down to a=b. But the book says it's not one to one. But wouldn't it be one to one since it's R+ --> R+? Or am I missing something?
• Sep 8th 2011, 04:08 PM
alexmahone
Re: One to One proof
Quote:

Originally Posted by JSB1917
Let f: R+ --> R+ where R+ is the positive reals, and f(x)=x^4. Prove or disprove that it's one to one.

Ok, I understand the whole f(a)=f(b) thing down to a=b. But the book says it's not one to one. But wouldn't it be one to one since it's R+ --> R+? Or am I missing something?

$\displaystyle f(a)=f(b)$

$\displaystyle \implies a^4=b^4$

$\displaystyle \implies (a^2+b^2)(a+b)(a-b)=0$

$\displaystyle \implies a=b$ or $\displaystyle a=-b$

So, f is not one-to-one.
• Sep 8th 2011, 04:25 PM
Plato
Re: One to One proof
Quote:

Originally Posted by JSB1917
Let f: R+ --> R+ where R+ is the positive reals, and f(x)=x^4. Prove or disprove that it's one to one.

Quote:

Originally Posted by alexmahone
$\displaystyle f(a)=f(b)$
$\displaystyle \implies a^4=b^4$
So, f is not one-to-one.

This function is one-to-one on its domain $\displaystyle R^+$
$\displaystyle a^4=b^4$ implies $\displaystyle a=b$ because $\displaystyle a\in R^+~\&~b\in R^+$.

Note that $\displaystyle a^4=b^4\text{ if and only if }|a|=|b|$ but they are both posiive.