Balzano's Theorem
A particular case of the intermediate value theorem is Bolzano's theorem. This theorem proposes that if a function is continuous on a closed interval, 
and 
and 
have opposite sign then there is at least one value of x for which 
. Usually, that value of x is denoted with "c" in many textbooks.
Bolzano's theorem does not indicate the value or values of c, it only confirms their existence. Let's prove this theorem.
Bolzano's Theorem Proof
Imagine a continuous function where point a is the minimum point and point b is the maximum point, 
and 
. Now we will divide the interval into two equal parts and call the mid point c, therefore, there will be two intervals, 
. There are three possibilities in this scenario and they are:
If the first condition occurs then you already proved this theorem's existence. On the other hand, the last two possibilities have some contradictions. Let's say that 
, since 
, we will consider the 
interval. If 
then we will consider the 
interval where . In both conditions, there is one term a negative entity while the other term is a positive entity.
Now we will select a sub-interval such that 
at its negative end points and 
at its positive end points. We will denote this sub-interval by 
where 
.
The next step is to bisect the close intervals into two equal parts as shown above. This can either result in or new sub-interval which we will name 
. Where 
, now we will again bisect the sub-interval.
Since, , therefore:
This is following a sequence which is 
. This process can go till infinity. Thus we get a sequence of nested intervals.
Example
Verify that the equation 
has at least one real solution in the interval 
.
We consider the function 
, which is continuous on because it is polynomial. We study the sign in the extremes of the interval:
First, consider the function , which is continuous in because it is polynomial. Then, study the sign in the extremes of the interval:
As the signs are different, Bolzano's theorem can be applied which determines that there is a 
such that 
. This process demonstrates that there is a solution in this interval.
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