Euclid's Proof that √2 is Irrational

square root of 2

Euclid proved that √2 (the square root of 2) is an irrational number.

He used a proof by contradiction.

First Euclid assumed √2 was a rational number.

A rational number  is a number that can be in the form p/q where p and q are integers and q is not zero.

He then went on to show that in the form p/q it can always be simplified.

But we can't go on simplifying an integer ratio forever, so there is a contradiction.

So √2 must be an irrational number.

We will go into the details of his proof, but first let's take a look at some useful facts:

Rational Numbers and Even Numbers

First, let's look at some interesting facts about even numbers and rational numbers:

Any integer multiplied by 2 gives an even number.

Examples:
  • 2×3 = 6, 6 is an even number.
  • 2×16 = 32, 32 is an even number.
  • etc

The square of an even number is always an even number (multiplying two even numbers gives an even number).

Likewise if a number is even and is a square of an integer, then its square root must be even.

Example: the square of 14 is 196. 14 is an even number, and so is 196.
Example: 256 is even, and is the square of an integer, so its square root must be even.

In fact the square root of 256 is 16.

Rational numbers or fractions must have a simplest form.

Example: 16/64 is the same as 1/4 (by dividing top and bottom by 16). That is as simple as we can get.
Example: 28/100 can be simplified to 14/50 (by dividing top and bottom by 2).

We can go further and simplify it to 7/25.

But it cannot be simplified further since 7 and 25 have no common factors.

The Proof

Euclid's proof starts with the assumption that √2 is equal to a rational number p/q.

Assume: √2 = p/q
Square both sides: 2 = p2/q2
Rewrite it as: 2q2 = p2

p2 must be even (since it is 2 multiplied by some number).

Since p2 is even, then p is also even (square root of a perfect square is even).

Since p is even, it can be written as 2m where m is some other whole number (because an even number can be written as 2 multiplied by a whole number).

Substituting p=2m in the above equation:

Start with: 2q2 = p2
Substitute p=2m: 2q2 = (2m)2
Simplify: 2q2 = 4m2
Divide both sides by 2: q2 = 2m2

q2 is an even number (since it is written as 2 multiplied by some number).

So q is an even number.

Since q is even, it can be written as 2n where n is some other whole number.

Now we have p = 2m and q = 2n and remember we assumed that √2 =  p/q:

Assume: √2 = p/q
Substitute p=2m and q=2n: √2 = 2m/2n
Simplify: √2 = m/n
We now have a fraction m/n that is simpler than p/q.


But now we can repeat the whole process again using m/n and simplify it to something else (say g/h).

We can then do that again ... and again ... and again ... !

But a rational number cannot be simplified forever. There must eventually be a simplest rational number, but in our case there is not: we have a contradiction!

So something is definitely wrong here. √2 cannot be written as p/q or it can be simplified forever.

So √2 cannot be rational and so must be irrational.


Note: The method used is known as Infinite Descent because it uses the fact that there is no infinite sequence of decreasing positive integers, and is a special case of Proof by Contradiction.