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I have been looking at various proofs of the IVT, and, perhaps the simplest I have encountered makes use of the Completeness Axiom for real numbers and Bolzano’s Theorem, which, honestly, I find a bit of an overkill. For an informal proof, we could write something like this:

“If $f$ is continuous on $[a,b]$ then for every value L between the end points $f(a)$ and $f(b)$ there exists $f(c)=L$, for, if this were to be untrue then it would mean $f$ is not continuous as it may be not well-defined for a particular $L$”.

Now, my question does not concern opinions, rather, I am asking if this informal proof ,which perhaps mostly use intuition and definitions, is mathematically incorrect since I do know it is not rigorous.

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The proof you gave, even with the allowance of informality, is flawed:

To see this, apply your reasoning to $f(x) = 0$ using an interval of $[0,2]$. Your statement is that for every value $L$ between the interval (in particular for $L = 1$ which is surely “between the interval” $[0,2]$) there exists a $c$ such that $f(c) = L = 1$. That is obviously false.

Now say you modify your statements to say “for every $L$ between the endpoint values $f(a)$ and $f(b)$”. Then the proof is closer but still not “correct,” because we don’t care, in the definition of “continuous,” whether $f$ is defined for a particular $L$ (which you are stating is a particular value of $f(x)$).

The “proof” you are trying to say is that if for some $L$ between $f(a)$ and $f(b)$ there is no $c$ such that $f(c) = L$, then $f$ is not continuous because at the values of $c$ where $f(c)$ is closest to $L$, there has to be a gap in $f(x)$ on one side of $c$. That informal proof is almost correct, but then you have to worry about what if there is some $L$ such that $f(x)$ is never $L$, but $f(x)$ comes arbitrarily close to $L$.

The value of the rigorous proofs is that they deal with such thorny issues, so that you can trust the IVT without asking anything about $f$ other than that it is continouous.

In essence, the Intermediate Value Theorem bridges the gap between a very informal understanding of continuity as a sort of large-scale connectedness, and our formal definition of continuity as a property of a function at an individual point (with a function being continuous over an interval if it is continuous at each point in the interval): it proves that the formal definition does more-or-less correspond to our very informal understanding, enough to justify its being called “continuity”.

As a result, there’s really no way to prove it informally: if we replace our formal definition of continuity with our informal understanding, then we simply get the Intermediate Value Theorem directly, as an informal axiom. There’s nothing to prove it *from*. This is why (as David C. Ullrich notes above) your proof has to resort to saying that the theorem is true because it’s true.

For a quick reference, Ch $7$ of Spivak’s Calculus, (I’m using the Third Edition) is called “Three Hard Theorems”, which he proves in Chapter $8$. However in chapter $7$ he proves some of their consequences namely, the intermediate value theorem, which is the conclusion of theorems $4$ and $5$ of this chapter $(7)$. It follows from the validity of the first three:

**Theorem $1$** If $f$ is continuous on $[a,b]$ and $f(a)<0<f(b)$, then there is some $x \in [a,b]$ such that $f(x)=0$.

**Theorem $2$** If $f$ is continuous on $[a,b]$, then $f$ is bounded above on $[a,b]$, that is, there is some number $N$ such that $f(x)\leq N$ for all $x \in [a,b]$.

**Theorem $3$** If $f$ is continuous on $[a,b]$, there is some number $y$ in $[a,b]$ such that $f(y)\geq f(x)$ for all $x \in [a,b]$.

Proving these three theorems relies heavily on the completeness of the real numbers.

Formalism and rigor are two aspects which are normally present in a modern mathematical proof. Out of these formalism is about style/fashion and rigor is the substance so that you may have an informal proof, but “a non-rigorous proof” is self-contradictory. Your proof unfortunately lacks both formalism and rigor.

An informal proof of IVT is as follows:

It is sufficient to prove that *if a continuous function $f$ has opposite signs at end-points of a closed interval then it vanishes somewhere in the interior of that interval*.

Let’s assume that $f$ does not vanish anywhere in the given interval. If we divide the interval into two equal sub-intervals then one of these two sub-intervals is such that $f$ has opposite signs at its end-points. Repeat the procedure of dividing this interval into two sub-intervals of equal length and get another small interval such that $f$ has opposite signs at its end-points. This way we get a sequence of nested intervals whose length tends to $0$ and $f$ has opposite signs at end-points of each interval in the sequence.

By Nested Interval Principle there is a unique point $c$ which lies in all the intervals of the sequence. Since $f(c)$ is non-zero, by continuity $f$ maintains its sign in a small neighborhood of $c$ and this neighborhood contains one of the intervals of the sequence described earlier (because length of the intervals in the sequence tends to $0$ and each interval of the sequence contains $c$) and hence $f$ changes its sign in this neighborhood of $c$. This contradiction shows that our assumption about non-vanishing of $f$ in the original interval is false. This completes the informal proof.

A formal version of the above proof only adds more symbols to make things strictly unambiguous and should be considered something like “legalese” of mathematics.

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