Bolzano’s Theorem
What is Bolzano’s theorem?
The Bolzano theorem states that if a function is continuous at all points of a closed interval [a, b] and holds the image of “a” and “b” (low function) have opposite signs, then exist by at least one point “c” in the open interval (a, b), in such a way that the function evaluated in “c” will be equal to 0.
This theorem was enunciated by the philosopher, theologian and mathematician Bernard Bolzano in 1850. This scientist, born in what is now the Czech Republic, was one of the first mathematicians in history to make a formal proof of the properties of continuous functions.
Explanation of the theorem
Bolzano’s theorem is also known as the intermediate values theorem, which helps in determining specific values, particularly zeros, of certain real functions of a real variable.
In a given function f (x) continues —that is, f (a) and f (b) are connected by a curve—, where f (a) is below the xaxis (it is negative), and f (b) by above the xaxis (it is positive), or vice versa, graphically there will be a cutoff point on the xaxis that will represent an intermediate value “c”, which will be between “a” and “b”, and the value of f (c) will be equal to 0.
By graphically analyzing Bolzano’s theorem, it can be seen that for every continuous function f defined on an interval [a, b], where f (a) _{*} f (b) is less than 0, there will be at least one root “c ”Of that function within the interval (a, b).
This theorem does not establish the number of points in that open interval, it only states that there is at least 1 point.
Proof of Bolzano’s theorem
To prove Bolzano’s theorem, it is assumed without loss of generality that f (a) <0 and f (b)> 0; thus, there can be many values between “a” and “b” for which f (x) = 0, but only one needs to be shown.
We begin by evaluating f at the midpoint (a + b) / 2. If f ((a + b) / 2) = 0 then the proof ends here; otherwise, then f ((a + b) / 2) is positive or negative.
One of the halves of the interval [a, b] is chosen, such that the signs of the function evaluated at the extremes are different. This new interval will be [a1, b1].
Now, if f evaluated at the midpoint of [a1, b1] is not zero, then the same operation as before is performed; that is, one half of this interval is chosen that fulfills the condition of the signs. Let this new interval be [a2, b2].
If you continue with this process, then you will have two sequences {an} and {bn}, such that:
{an} is increasing and {bn} is decreasing:
a ≤ a1 ≤ a2 ≤… ≤ an ≤…. ≤…. ≤ bn ≤…. ≤ b2 ≤ b1 ≤ b.
If the length of each interval [ai, bi] is calculated, we will have:
b1a1 = (ba) / 2.
b2a2 = (ba) / 2².
….
bnan = (ba) / 2 ^ n.
Therefore, the limit as n approaches infinity of (bnan) is equal to 0.
Using that {an} is increasing and bounded and {bn} is decreasing and bounded, we have that there exists a value “c” such that:
a ≤ a1 ≤ a2 ≤… ≤ an ≤… .≤ c ≤…. ≤ bn ≤…. ≤ b2 ≤ b1 ≤ b.
The limit of an is “c” and the limit of {bn} is also “c”. Therefore, given any δ> 0, there is always an “n” such that the interval [an, bn] is contained within the interval (cδ, c + δ).
Now, it must be shown that f (c) = 0.
If f (c)> 0, then since f is continuous, there exists an ε> 0 such that f is positive over the entire interval (c – ε, c + ε). However, as mentioned above, there is a value “n” such that f changes sign in [an, bn] and, furthermore, [an, bn] is contained within (c – ε, c + ε), which which is a contradiction.
If f (c) <0, then since f is continuous, there exists an ε> 0 such that f is negative throughout the interval (c – ε, c + ε); but there exists a value “n” such that f changes sign in [an, bn]. It turns out that [an, bn] is contained within (c – ε, c + ε), which is also a contradiction.
Therefore, f (c) = 0 and this is what we wanted to prove.
What is Bolzano’s theorem for?
From its graphical interpretation, Bolzano’s theorem is used to find roots or zeros in a continuous function, through bisection (approximation), which is an incremental search method that always divides the intervals by 2.
Then an interval [a, c] or [c, b] is taken where the sign change occurs, and the process is repeated until the interval is smaller and smaller, in order to be able to approach the desired value; that is, to the value that the function makes 0.
In summary, to apply Bolzano’s theorem and thus find the roots, limit the zeros of a function or give a solution to an equation, the following steps are carried out:

It is verified if f is a continuous function on the interval [a, b].

If the interval is not given, one must be found where the function is continuous.

It is verified if the extremes of the interval give opposite signs when evaluated in f.

If no opposite signs are obtained, the interval must be divided into two subintervals using the midpoint.

Evaluate the function at the midpoint and verify that the Bolzano hypothesis is satisfied, where f (a) _{*} f (b) <0.

Depending on the sign (positive or negative) of the value found, the process is repeated with a new subinterval until the aforementioned hypothesis is fulfilled.
Solved exercises
Exercise 1
Determine if the function f (x) = x ^{2} – 2, has at least one real solution in the interval [1,2].
Solution
We have the function f (x) = x ^{2} – 2. Since it is polynomial, it means that it is continuous in any interval.
It is asked to determine if it has a real solution in the interval [1, 2], so now it is only necessary to substitute the extremes of the interval in the function to know the sign of these and to know if they fulfill the condition of being different:
f (x) = x ^{2} – 2
f (1) = 1 ^{2} – 2 = 1 (negative)
f (2) = 2 ^{2} – 2 = 2 (positive)
Therefore, sign of f (1) ≠ sign f (2).
This ensures that there is at least one point “c” that belongs to the interval [1,2], in which f (c) = 0.
In this case, the value of “c” can be easily calculated as follows:
x ^{2} – 2 = 0
x = ± √2.
Thus, √2 ≈ 1,4 belongs to the interval [1,2] and fulfills that f (√2) = 0.
Exercise 2
Show that the equation x ^{5} + x + 1 = 0 has at least one real solution.
Solution
Let’s first notice that f (x) = x ^{5} + x + 1 is a polynomial function, which means that it is continuous on all real numbers.
In this case, no interval is given, so values must be intuitively chosen, preferably close to 0, to evaluate the function and find the sign changes:
If you use the interval [0, 1] you have to:
f (x) = x ^{5} + x + 1.
f (0) = 0 ^{5} + 0 + 1 = 1> 0.
f (1) = 1 ^{5} + 1 + 1 = 3> 0.
As there is no sign change, the process is repeated with another interval.
If you use the interval [1, 0] you have to:
f (x) = x ^{5} + x + 1.
f (1) = (1) ^{5} + (1) + 1 = 1 <0.
f (0) = 0 ^{5} + 0 + 1 = 1> 0.
In this interval there is a sign change: sign of f (1) ≠ sign of f (0), which means that the function f (x) = x ^{5} + x + 1 has at least one real root “c” in the interval [1, 0], such that f (c) = 0. In other words, it is true that x ^{5} + x + 1 = 0 has a real solution in the interval [1,0].