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This proof looks sound to me, but it seems too simple to work.

Proof of intermediate value theorem:

Suppose $t \in [m,M]$, with $m$ and $M$ the minimum and maximum values of $f$ on $[a,b]$, respectively, and $f(c) \neq t$ for any $c\in [a,b]$. By the extreme value theorem, $ m \leq f(c) \leq M$ $ \forall c \in [a,b]$. Therefore $f(c) \in [m,M] $ $\forall c \in [a,b].$ Since $t \in [m,M]$, there is a $c \in [a,b]$ such that $f(c) = t$.

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    $\begingroup$ it is completely wrong. What justifies the last sentence? $\endgroup$ – mookid Nov 2 '14 at 21:46
  • $\begingroup$ I was with you until the last sentence. $\endgroup$ – vadim123 Nov 2 '14 at 21:46
  • $\begingroup$ No, this is not a proof. The same argument would give us that there is an integer between $2$ and $3$. You still have not said anything that would not apply to that (false) case. For instance, nowhere have you mentioned (or used) continuity of $f$, which I assume is one of your assumptions. $\endgroup$ – Andrés E. Caicedo Nov 2 '14 at 21:47
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The proof is not sound. From "$f(c)\in[m,M]$ for all $c\in[a,b]$" and $t\in[m,M]$ you cannot infer "There exists $c\in[a,b]$ such that $f(c)=t$". For example

  • $n^2 \in \mathbb N_0$ for all $n\in\mathbb N_0$
  • $2\in \mathbb N_0$
  • therefore(?) $n^2=2$ for some $n\in\mathbb N_0$
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