# One to One proof

• September 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?
• September 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?

$f(a)=f(b)$

$\implies a^4=b^4$

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

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

So, f is not one-to-one.
• September 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
$f(a)=f(b)$
$\implies a^4=b^4$
So, f is not one-to-one.

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

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