# Intermediate Value Theorem question

1. Jul 5, 2017

### Portuga

Consider a continuous function $f$ in $[a,b]$ and $f(a) < f(b)$. Suppose that $\forall s \neq t$ in $[a,b]$, $f(s) \neq f(t)$. Proof that $f$ is strictly increasing function in $[a,b]$.

2. Relevant equations

I.V.T: If $f$ is continuous in $[a,b]$ and $\gamma$ is a real in $[f(a),f(b)]$, then there'll be at least one $c$ in $[a,b]$ such that $f(c) = \gamma$.

3. The attempt at a solution

This exercise is very strange to me. Besides I can apply the I.V.T to show that for any sub interval in [a,b] there will be an intermediate value in $f(a), f(b)$, I can easily draw and counter example of what it pretends:
https://www.dropbox.com/s/dtj28xo4ilaai4z/pf.eps?dl=0

I am missing something important?

2. Jul 5, 2017

### Ray Vickson

Your example violates the hypotheses of the claim: your function $f(x)$ has $f(s) = f(t)$ for several pairs $(s,t)$ with $s \neq t$.

3. Jul 5, 2017

### BvU

Picture doesn't make it. Blank screen. Can you copy/paste it in the post ?

4. Jul 5, 2017

5. Jul 5, 2017

### Portuga

I did the first one in libreoffice, so I made the mistake pointed by Ray.

6. Jul 5, 2017

### Staff: Mentor

Your drawing violates the assumption that $f(s) \neq f(t)$
Minor point. The verb is "to prove". The noun is "proof".

7. Jul 5, 2017

### Portuga

Thank you! Now I got the point! Sorry for my poor English!! Thank you very much. Now it's clear for me!

8. Jul 5, 2017

### Staff: Mentor

No need for an apology. Lots of native speakers of English also get this wrong (prove vs. proof), sometimes spelling "prove" as "proove."

9. Jul 8, 2017

### SammyS

Staff Emeritus