Question

See attached.

Answer 1

Do you know what the definition of one-one is?

Can you tell me how far you've gotten in the proof?

Answer 2

Eh, since you're offline, I'll outline it for you in case you come back:

We want to show if x = y, then f(x) = f(y). That will show f is one-one. To do this, assume x = y. We want to get x + e*g(x) = y + e*g(y). For that to happen, we need to show g(x) = g(y). We know g' is bounded, so, by definition:

|g(x) - g(y)|/(x-y) <= B, for some B in the reals. Figure out what B needs to equal (in terms of epsilon) to show that g(x) = g(y). If you come back and I'm still around, just tag me if you're still confused.

Fun problem.

Answer 3

@DanielxAK So I got to here and again got stuck:
|g(x) - g(y)| < B |x-y| (by the MVT) because g' is bounded by B

Answer 4

Okay, here's another hint ( a big one):

Let B = 1/epsilon

So, now you have: |g(x) - g(y)| <= (x-y)/epsilon

Shuffle the equation, and you have epsilon*|g(x) - g(y)| <= x - y

This is true for all x and y. But, we're assuming x = y, so:

epsilon*|g(x) - g(y)| <= x - y = 0

Epsilon is positive by definition, so g(x) = g(y). From there, the rest of the proof follows.