7

I would like to know why the intermediate value theorem is so important. So my questions are:

  1. Which important theorems do we prove using the intermediate value theorem?
  2. Are there direct applications of the intermediate value theorem outside mathematics?
  3. Does the intermediate value theorem have a historically importance?
user42912
  • 23,582
  • 1
    This may be a good place to start https://en.wikipedia.org/wiki/Intermediate_value_theorem – Harry Alli Dec 10 '17 at 06:59
  • Extremely short version: The IVT is the reason people care about continuous functions as a category (in much the same way as the Fundamental Theorem of Arithmetic is the reason people care about prime numbers as a category). – Kevin Dec 10 '17 at 07:11
  • 1
    @Kevin: that's not entirely true. Intermediate value property is not exclusive privilege of continuous functions as many popular expositions try to present. There are discontinuous functions which possess this property. The importance of the theorem is mainly due to the fact that it is an obvious result which no one bothered to prove. When people started to prove it, they were simply unable. This led to many ideas regarding completeness of reals. – Paramanand Singh Dec 10 '17 at 09:33
  • @Paramanand: I never said that. – Kevin Dec 10 '17 at 17:07
  • @Kevin: you state that IVT is the only thing due to which people care about continuous functions. I just wanted to highlight that these are different properties. And IVT is not the only nice feature of continuous functions. There are other aspects like being integrable, or being bounded and attaining min max values on a closed interval which are also noteworthy properties of continuous functions. I never really understood how IVT became so popular and overshadowed other features of continuous functions so much so that most people often believe it to be synonymous with continuity. – Paramanand Singh Dec 10 '17 at 17:15
  • @ParamanandSingh: I never said "only" either. I'm beginning to think you're arguing with someone entirely different from me by proxy. – Kevin Dec 10 '17 at 17:42

3 Answers3

4

We can use the intermediate value theorem to compute equations like, for example, $\cos(x)=x$.

consider the function $f:[0,\pi/2]\mapsto\Bbb R, x\to\cos x-x$.

this function is continuous and $f(0)=1,f(\pi/2)=-\pi/2$. by the intermediate value theorem we know that $f(c)=0,c\in[0,\pi/2]$. this doesn't help us too much, so let's divide $[0,\pi/2]$ into $[0,\pi/4],[\pi/4,\pi/2]$.

now we have $f(0)=1,f(\pi/4)\approx -0.0782913822,f(\pi/2)=-\pi/2$.

using the intermediate value theorem we know that $c\in[0,\pi/4]$, great! now let's divide our interval into 2 smaller intervals again:$[0,\pi/4]$ into $[0,\pi/8],[\pi/8,\pi/4]$

now we have $f(0)=1,f(\pi/8)\approx0.531180451,f(\pi/4)\approx -0.0782913822$.

hence $c\in[\pi/8,\pi/4]$. by doing this over and over again we can get to pretty nice approximation

this theorem is important in physics where you need to construct functions using results of equations that we know only how to approximate the answer, and not the exact value, a simple example is 2 bodies collide in $\mathbb R^2$. in this case you will have system of 2 equations in similar form to the example of the first part.

I know about only one historically importance, before there was a definition of continuity people use this theorem, become if this is true for all sub-intervals in the function $f$ then $f$ is continuous

ℋolo
  • 10,006
  • 1
    The past paragraph is not correct. Continuity implies intermediate value property, but intermediate value property does not imply continuity. This is the typical wrong belief propagated by some texts and instructors alike. You may wish to edit your last paragraph. – Paramanand Singh Dec 10 '17 at 09:37
  • this was true before there was a definition for continuity, this is not true for the current definition but this definition didn't exist when this theorem first appear – ℋolo Dec 10 '17 at 09:43
  • 1
    I am saying about "this is true only for continuous function..." and that's wrong. It is true for discontinuous functions also (for examples derivatives possess this property without being necessarily continuous). – Paramanand Singh Dec 10 '17 at 10:48
  • you are right, i'll delete this part – ℋolo Dec 10 '17 at 10:50
2

A bit too long for a comment ...

In general, the IVT states that continuous maps preserve connectedness.

For point (1):

Let $f$ be a real-valued continuous function defined on an interval $J$ of $\mathbb R$. Claim: If $f$ is injective, then it is strictly monotone.

Define $g\colon D:=\{(x,y)\in J^2|x>y\}\to\mathbb R$, $g(x,y):=f(x)-f(y)$. As $f$ is continuous so is $g$. Now $D$ is connected, hence $g(D)$ is connected, that is, an interval of the reals. Since $f$ is injective, $g(D)$ doesn’t contain $0$.

Michael Hoppe
  • 18,103
  • 3
  • 32
  • 49
1

there exist two antipodal points on the equator that have the same temperature.

Can you prove that at all times there is a pair of opposite points on the equator where the temperature at one is equal to the temperature of the other?

To see how we can prove this, pick two arbitrary points. Label the points A and B with the temperature of A greater than B. Find the difference in temperature of the two points. If the difference is zero, then you've found a pair of points. If not, move point A to point B and point B to point A. Now the difference in temperature is negative. Because the difference in temperature is a continuous function, there exists a point on the equator where the difference is zero. Those two points have the same temperature.

Eric Zhang
  • 141