# Thread: Proof of intermediate value theorem for a particular case

1. ## Proof of intermediate value theorem for a particular case

Let
f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that
f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists
a x' such that f(x')=0

S is non empty as x=0 is a member and bounded above by 1, so by the completness axiom Sup(S) exists. Sup(S)_> x E S and Sup(s) _< 1. Not sure where to go from here. Thanks in advance.

2. Denote c = sup S and suppose f ( c ) < 0 or f ( c ) > 0 . In both cases you'll get a contradiction. So, necessarily f ( c ) = 0 .

3. Originally Posted by poirot
[LEFT]Let f be a continuous function on the closed interval [0; 1], and suppose that f(0) < 0
and that
f(1) > 0. By considering the set S= {x E [0,1] :f(x)_<0) prove that there exists a x' such that f(x')=0
This proof relies upon the following theorem.
If $f$ is continuous and then there is a neighbored of $c$ on which $f$ has the same sign as $f(c)$.
If you take and suppose that then you can find a contradiction.

4. let c=Sup(S)
Assume F(c) < 0. I don't see a contradiction

5. Originally Posted by poirot
I don't see it
If then .
Can't you find a contradiction there???

6. I don't know if c is a member of the set. If it is then surely sigma will be 1-c.

7. ok I'm in a better maths mood today. I got frustated by this yesterday.
assume F(c)<0 which implies the theorem.
c<_1 so F(c)<0 implies c is a member of the set
So for all m E [0,1] with m > c implies that F(m) > 0 but get stuck here. Its sort of intuitive that it can't be because of the density of reals but I don't know how to prove it convincingly.

8. Originally Posted by Plato
If then .
Can't you find a contradiction there???
It follows that that means .
That is a contradiction because c is the greatest number with that property

9. this is only true if c+ sigma/2 <1

10. Originally Posted by poirot
this is only true if c+ sigma/2 <1
Well of course. But you have to pick delta to make that true.
Look, I am not doing this question for you.
If you do not understand this approach, please do not use it.

11. sorry but you have put 'there exists a delta >0 '. So you have to show such a delta does not exist, not picking a counter example. Thats just bad logic. I am close to understanding.

12. poirot: I don't understand which is the problem. According to Plato's answer (#5):

$f(c)<0\Rightarrow \exists \delta >0: f(x)<0 \;,\;\forall x\in (c,c+\delta)$

and this, evidently contradicts the choice of c.