What does the intermediate value theorem mean?

2 Answers
Jan 5, 2016

It means that a if a continuous function (on an interval AA) takes 2 distincts values f(a)f(a) and f(b)f(b) (a,b in Aa,bA of course), then it will take all the values between f(a)f(a) and f(b)f(b).

Explanation:

In order to remember or understand it better, please know that the math vocabulary uses a lot of images. For instance, you can perfectly imagine an increasing function! It's the same here, with intermediate you can imagine something between 2 other things if you know what I mean. Don't hesitate to ask any questions if it's not clear!

Jan 5, 2016

You could say that it basically says the Real numbers have no gaps.

Explanation:

The intermediate value theorem states that if f(x)f(x) is a Real valued function that is continuous on an interval [a, b][a,b] and yy is a value between f(a)f(a) and f(b)f(b) then there is some x in [a,b]x[a,b] such that f(x) = yf(x)=y.

In particular Bolzano's theorem says that if f(x)f(x) is a Real valued function which is continuous on the interval [a, b][a,b] and f(a)f(a) and f(b)f(b) are of different signs, then there is some x in [a,b]x[a,b] such that f(x) = 0f(x)=0.

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Consider the function f(x) = x^2-2f(x)=x22 and the interval [0, 2][0,2].

This is a Real valued function which is continuous on the interval (in fact continuous everywhere).

We find that f(0) = -2f(0)=2 and f(2) = 2f(2)=2, so by the intermediate value theorem (or the more specific Bolzano's Theorem), there is some value of x in [0, 2]x[0,2] such that f(x) = 0f(x)=0.

This value of xx is sqrt(2)2.

So if we were considering f(x)f(x) as a rational valued function of rational numbers then the intermediate value theorem would not hold, since sqrt(2)2 is not rational, so is not in the rational interval [0, 2] nn QQ. To put it another way, the rational numbers QQ have a gap at sqrt(2).

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The big thing is that the intermediate value theorem holds for any continuous Real valued function. That is there are no gaps in the Real numbers.