# Is the proof complete?

1. Sep 19, 2014

### baconeater

Hello, think i have proved it but is the proof complete, is there any more i should do?

1. The problem statement, all variables and given/known data
Let f:[0,1] --> [0,1], f be continuous, f(0)=0, f(1)=1
and let f(f(x)) = x, for all x in [0,1]

prove that f(x) = x.

2. Relevant equations

3. The attempt at a solution
(*) f(f(x)) = x, for all x in [0,1]
(**) f(0) = 0, f(1) = 1

Assume that f(a) = b>a, then it follows by (*) that f(f(a)) = f(b) = a<b.
Since f is continuous it follows by "Intermediate"- theorem that there exist
a point c where a< c <b such that f(c) = c.

Now we have that f(b) = a < c = f(c), since then there must exist a point d
where b< d <1 such that f(d) = c.
(if b=1 then f(b) = a <1, witch contradicts (**))
But then f(f(d)) = c < b witch is an contradiction!

Assume that f(a) = b < a, then it follows by (*) that f(f(a)) = f(b) = a>b.
Since f is continuous it follows by "Intermediate"- theorem that there exist
a point c where b< c <a such that f(c) = c.

Now we have that f(c) = c < a = f(b), since then there must exist a point d
where 0< d <b such that f(d) = c.

(if b=0 then f(b) = a > b, witch contradicts (**))

But then f(f(d)) = c < a witch is an contradiction!

2. Sep 20, 2014

### Fredrik

Staff Emeritus
There are some issues with the presentation at the start. First you left $a$ undeclared, and at the end of the line (the one that starts with "assume that"), you wrote things down in an order that makes it unnecessarily hard to understand you. These aren't big issues, but I assume that you want feedback on the presentation too, so here's a pedantic version of the start of your proof:

Let $a\in(0,1)$ be arbitrary. We will prove that f(a)=a by deriving a false statement from the assumption that this is not true. So suppose that $f(a)\neq a$. Then either f(a)>a or f(a)<a. We will only discuss the former case in detail. The other can be treated similarly. So suppose that f(a)>a. Define b=f(a). These statements and (*) imply that b=f(a)>a=f(f(a))=f(b). In particular, we have a<b and f(b)<b.​

After this, you said that the intermediate value theorem implies that there's a real number c such that a<c<b and f(c)=c. You may be right (I don't know, I didn't think it through), but you didn't make a convincing argument. I'd say that what the intermediate value theorem says is that for each $y\in(a,b)$, there's a $c\in(a,b)$ such that $f(c)=y$.

Also, note the spelling of the word "which".

Last edited: Sep 20, 2014