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.

  • 38
    $\begingroup$ That's not an informal proof, or a non-rigorous proof. That's simply no proof at all! Really. You're supposed to prove that a continuous function has a certain property, and you "prove" this by simply saying that otherwise the function would not be continuous. That's simply proving the theorem by saying the theorem is true. $\endgroup$ Aug 4, 2015 at 19:33
  • 6
    $\begingroup$ Any proof (formal or informal) would need to deal with somebody thinking $f(x)=x^2$ might be a continuous function on the rational numbers as there are no discrete jumps, with $f(0)=0$ and $f(1)=1$ but no rational $x$ with $f(x)=\frac12$. $\endgroup$
    – Henry
    Aug 4, 2015 at 19:37
  • 7
    $\begingroup$ You're making no sense at all. Sorry, you asked. $\endgroup$ Aug 4, 2015 at 19:44
  • 9
    $\begingroup$ By the way, one can construct functions which have the Intermediate Value Property and are not continuous. $\endgroup$ Aug 4, 2015 at 19:47
  • 6
    $\begingroup$ Using the completeness axiom is not overkill, it's simply necessary, because if we work in a number system (such as the rationals) where the completeness axiom is not satisfied, then the IVT may be false. A proof of the IVT must use the completeness axiom (in some form). $\endgroup$
    – littleO
    Aug 4, 2015 at 20:09

4 Answers 4


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.

  • 4
    $\begingroup$ If I could upvote this 10 times I would. You took the trouble to give a serious and helpful reply to the OP. Which is precisely the purpose of the site. Bravo sir. $\endgroup$ Aug 4, 2015 at 19:52
  • $\begingroup$ @MarkFischer Sorry, in my "proof" I was actually thinking about an $L$ between $f(a)$ and $f(b)$ which explains all of the comments. $\endgroup$ Aug 4, 2015 at 19:55

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.


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.


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.

  • $\begingroup$ Well just theorem one actually is needed for your purpose but I believe that's the bolzano property you mention. $\endgroup$
    – snulty
    Aug 4, 2015 at 20:07
  • $\begingroup$ Why are theorems #2 and #3 needed? It is easy to prove (using "epsilon-delta"-style proofs) that if $g$ is continuous then $f(x) = g(x) - c$ is also. Combining this with theorem #1, we can immediately conclude that if $g$ is continuous on $[a,b]$ and $g(a) < c < g(b)$ then there is some $x \in [a,b]$ such that $g(x) = c$. What am I missing? (Is Spivak using a definition of continuity that doesn't rely on "epsilon"-delta-style limits?) $\endgroup$
    – ruakh
    Aug 4, 2015 at 22:45
  • $\begingroup$ @ruakh No that's why I tacked in that comment above, I had to leave earlier, and I meant to update this. So yeah only theorem 1 is necessary for the op, but I was more making reference to spivaks chapter singling out these non-trivial theorems which relied on completeness, just to make the op aware that it was an important property of $\Bbb{R}$. Apologies $\endgroup$
    – snulty
    Aug 4, 2015 at 22:48
  • $\begingroup$ Isn't Theorem 2 trivially implied by Theorem 3?  Just set $N=f(y)$. $\endgroup$ Aug 5, 2015 at 3:07
  • 1
    $\begingroup$ @scott it would be only that you use theorem 2 to prove theorem 3 in Spivaks chapter 8 :) $\endgroup$
    – snulty
    Aug 5, 2015 at 11:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .