Proving function is injective

**m112358** · February 29th 2012, 11:10 AM

Hey guys.

I've got the following problem:

Let $U = (0, \infty ) \times R \subset R^2$ og let $f : U \rightarrow R^2$ be defined by : $f(x,y) = (x, y^3+xy)$

(a) Show that $f(U) = U$ Done that already.
(b) Show that $f$ is injective
(Hint: When is $g: R \rightarrow R$ given by $g(y)= y^3 + ay + b$ monotonic?)
(c)... some more questions.

I've already found the solution to (a), but I cant quite figure out how to solve (b). This is pretty much what I've tried so far:
For $f : U \rightarrow R$ to be injective, the following has to be true : $q \neq p \Rightarrow f(q) \neq f(p) \forall q, p \in U$ .
So, for proof by contradiction, we assume that for a pair $q \neq p \in U$ we have that $f(p) = f(q)$ .
Since $f(q) = f(p)$ we have that $x_q = x_p \wedge y_q^3+x_q y_q = y_p^3+x_p y_p$ (1).
But since $q \neq p$ we have that $x_q \neq x_p \vee y_q \neq y_p$ .
According to (1) we know that $x_q = x_p$ so $y_q \neq y_p$ must be the case.

From here I dont know exacly where to go. I can see that I'm not really using the hint.
I know that all monotonic functions are injective, but how can I prove that f is monotonic?
I can see by visualization that f will be monotonic, because x is defined to be positive, but how do I prove that rigorously?

Thanks for your help guys.

Morten

**arlingtonbassett** · February 29th 2012, 01:26 PM

the function is clearly continuous, since each component function is continuous. there is a theorem that says that a continuous real-valued function is injective if and only if it is strictly monotonic. but im not sure if there is an analogous theorem for $\mathbb{R}^2$

**m112358** · February 29th 2012, 01:37 PM

Yes I understand that, but how do I prove that it is monotonic?

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