# The Square Root of Two is a Real Irrational Number

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## The Pythagorean Desire for Rational Numbers

Pythagoras and his students had a religious belief in the power of mathematics as derived from geometry. Part of this belief was that rational numbers could handle any situation.

Pythagoras is most famous for the “Pythagorean Theorem,” which states that the square of the hypotenuse of a right triangle equals the sum of the squares of the other two sides.

## What is the Hypotenuse of a Right Triangle?

A right triangle has one “right angle” of exactly 90°. The “hypotenuse” is the side opposite that angle. In other words: the only side of a right triangle that does not touch the right angle, is called the “hypotenuse.”

Obviously the hypotenuse of a right triangle is the longest side.

## The Proof that the Square Root of Two is Irrational

This ancient Greek proof is a “proof by contradiction”. It makes an assumption, begins to make conclusions from that assumption, and finds a contradiction. This demonstrates that the assumption was false. Georg Cantor used a “proof by contradiction” many centuries later in last week’s article.

It is not clear whether the credit for this proof should go to Hippasus, a student of Pythagoras who may have been killed for his efforts, or to Euclid.

## Clarifying the Language for this Proof

To say that “the square root of two is irrational” means the following. Let ‘i’ and ‘j’ be any two integers, with ‘j’ not zero. The claim that 2^(1/2) is irrational states there is no pair of integers ‘i’, ‘j’ such that “i/j = 2^(1/2).”

Next, let’s put the Pythagorean theorem into a more modern framework. Let ABC be a right triangle, with angle <ACB as the right angle. So side AB is the hypotenuse opposite angle <ACB. Let’s say that the length of side AB is ‘h’, and the lengths of the other sides are ‘x’ and ‘y.’

Isosceles Right Triangle: Image by Mike DeHaan

## Proof by Contradiction

We have the right triangle ABC, and the sides have lengths ‘h’ (the long hypotenuse), ‘x’ and ‘y’ (the two shorter sides). So “h^2 = x^2 + y^2” by the Pythagorean theorem.

Let’s make this an isosceles right triangle, as shown in the image. Therefore both of the short, non-hypotenuse sides have the same length. Therefore “x^2 = y^2”, and therefore “x^2 + y^2 = 2*(x^2)”. Therefore “h^2 = 2*(x^2).”

Taking the (positive) square root of both sides of “h^2 = 2*(x^2)” results in “h = ( 2^(1/2) )*x.”

Since ‘x’ is a positive, non-zero length, we divide both sides by ‘x’ to find “h/x = 2^(1/2)”. In the image, the hypotenuse has the length “root two” and each of the other sides of the triangle has a length of “one.”

Therefore the square root of two is the ratio of the hypotenuse to one side of an isosceles right triangle.

#### The Square Root of Two is not an Integer

We take note that “1^2 = 1”, “2^2 = 4” and “3^2 = 9”. Without belabouring this issue with a definitive proof, it seems clear that the square root of two is a non-integer with a value somewhere between one and two.

#### The Square of Odd Numbers and Even Numbers

We also take note that, by definition, even numbers are divisible by 2. In other words, ‘2’ is a factor of an even number. By a similar definition, odd numbers do not have ‘2’ as a factor.

We also notice that any integer ‘n’ is the product of a set of prime numbers {p[z]}. If n = p[1] * p[2] *…*p[z], then n^2 = ( p[1] * p[2] *…*p[z] )*( p[1] * p[2] *…*p[z] ). This only introduces a new factor of ‘2’ for each p[z] that happened to have the value ‘2.’

Therefore, the square of an odd number remains an odd number; and the square of an even number becomes divisible by 2*2=4.

#### Making the Invalid Assumption

State the belief that this ratio, “h/x”, is indeed a rational number. Then “h/x = i/j” for some integers ‘i’ and ‘j’. Without loss of generality, the integers ‘i’ and ‘j’ have no common factors; the fraction is in its lowest terms.

In particular, this means that ‘i’ and ‘j’ cannot both be even numbers. If both were even, then there would be a common factor, ‘2’. We could just keep dividing both ‘i’ and ‘j’ by the common factor of ‘2’ until either (or both) is an odd number.

#### Drawing the Contradictory Result

Since we assumed that “i/j = 2^(1/2)”, we can square both sides of the equation.

Now we have “(i/j)^2 = (i^2)/(j^2) = 2”. Let’s multiply both sides by “j^2.”

Now we have “(i^2) = 2*(j^2).”

Since the right-hand side is a multiple of two, that implies that “i^2” is an even number. Let’s say that “i^2 = 2*k” where ‘k’ is an integer. (This implies “k = j^2”.)

However, we had noticed that “the square of an odd number remains an odd number; and the square of an even number becomes divisible by 2*2=4”. If “i^2 is even”, then it must be divisible by 4.

But if “i^2 is divisible by 4”, then let’s divide both sides of “(i^2) = 2*(j^2)” by 2.

Now “( (i^2) )/2 = j^2”. We still could divide “( (i^2) )/2” by 2, since the square of an even number is divisible by 4. But that means that both “( (i^2) )/2”, and also “j^2”, have a factor of ‘2’ remaining. Therefore “j^2” is even; so ‘j’ must be even.

But the assumption was that ‘i’ and ‘j’ did not have any common factors! We’ve just demonstrated that they both must have a common factor of ‘2’. This is the fatal contradiction for the claim that the square root of two could be a rational number.

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