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

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

\frac{a}{b}=\sqrt{2}

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

\frac{a}{b}=\sqrt{2}.

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:

\left(\frac{a}{b}\right)^2=\sqrt2^2

We can simplify it little bit:

\frac{a^2}{b^2}=2

And adjust it:

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

\frac{a}{b}=\sqrt{2}

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 :-).