# Irational Number

Irrational numbers are number with infinite and non-repeat decimal representation. Probably the most common example of an irrational number is π, the Ludolphian number. We can estimate the value of π to 3.1415 but is has much more digits, infinite amount of non-repeating digits. We know at least 12 trillion digits of π by now. It’s probably not very useful to know all these digits but you now, everyone has its own hobby.

## How do we Know Irational Numbers Even Exist?

Maybe it’s some non-sense crazy mathematicians invented to confuse ordinary people. Well, irational numbers really exist in real world and we can simply prove it. Let’s suppose isosceles triangle

with given lengths: |AB| = 1 and |BC| = 1. What’s the length of the side AC? We can compute it using the Pythagorean theorem:

|AC| = √{12 + 12} = √{1 + 1} = √2

So the length of the hypotenuse AC is equal to √2. What number is √2? Is it an integer? Clearly not. Is it a rational number? Well, maybe. If it is a rational number, it can be written in the form of fraction, i. e. there must be two integers a, b for which the equation

holds. Now, each integer is either odd or even. That’s simple.

• We can prove that square of even integer is again an even integer, i. e. if n is an even integer then n2 is again even. E. g. 42 = 16. Both 4 and 16 are even numbers. And the converse holds too: if n2 is even, then n is even.
• How we can prove it? Every even integer n can be written in the form n = 2k for some integer k. E. g. the number 32 can be written as 32 = 2 · 16. Thus k = 16. Then the square of even number n2 is equal to (2k)2. We can simplify it to 4k2 and then we can adjust it to 2 · 2 · k2. We can see by now that the expression 2 · 2 · k2 is an even number.
• Then, if n is an odd integer, then n2 is odd too. E. g. 72 = 49. Both 7 and 49 and odd integers. The converse holds too, if n2 is odd, then n is odd.
• The prove is similar. Each odd integer n can be written as n = 2k + 1. The square of n is then equal to n2 = (2k + 1)2. We can simplify i to (2k + 1)2 = 4k2 + 4k + 1 and them modify it to the final form: 2(2k2 + 2k) + 1.We can see that this is an odd number.

We can use this knowledge to prove that √2 is an irrational number. We’re trying to find integers a, b such that

We can safely assume that the numbers a and b have no common divisor. Meaning that the fraction cannot be simplified. E. g. the fraction 12/8 can be simplified to 3/2 and it’s still the same number. The numbers 12 and 8 had the common divisor 4. This innocent precondition will be crucial.

We can compute squares of both sides of equation:

We can simplify it little bit:

a2 = 2b2

Now:

• 2b2 is clearly an even number since it’s in the form 2k where k = b2.
• Since the right side of the equation is an even number, the left side must be also even. I. e. a2 is an even number and moreover the number a itself is an even number. Thus there must be integer k such that a = 2k and from this we have

a2 = 4k2.

We know that a2 = 2b2 thus we can write

a2 = 4k2 = 2b2

Now we have the equation 4k2 = 2b2 which we can simplify to the final form

2k2 = b2.

From the last equation we can derive that the number b2 is even and thus the number b itself is even. As a corollary we now have both numbers a and b even.

But… but wait. We assumed that the number a and b have no common divisor. Our first assumption was the fraction cannot be simplified because the number a and b have no common divisor. And if both numbers a and b are even then it means they have a common divisor, the number 2. This violates our precondition and thus there are no integers a and b such that

which proves that the √2 cannot be a rational number since it cannot be expressed as a fraction. So there are numbers in this world that cannot be expressed as a fraction and you find them anywhere around you. Just keep looking :-).