In a sense, the difference between rational and irrational numbers is a simple one: a rational number can be expressed as the quotient of two integers, and an irrational number cannot. For example, 10 3/16 can be written as 163/16. What is not immediately clear is that there are numbers which cannot be so expressed.

To show this, we will use the demonstration developed in the 5th century BC by Hippasus of Metapontum. Hippasus was a follower of Pythagoras, who taught that all numbers were rational. One story relates that the Pythagoreans were so incensed at this proof that one of Pythagoras’ fundamental ideas was wrong that they drowned Hippasus. It can be dangerous to be right!

We are going to show that the square root of 2 is not a rational number. The process, known as reductio ad absurdum, is to assume that √2 is rational, and then use that assumption to obtain a logical impossibility.

If we have a right triangle with sides 1 unit in length, the Pythagorean theorem tells us that the length of the hypotenuse is √(1²+1²) = √2. Supposing that √2 is rational means that there exist integers m and n such that √2 = m/n. If m/n has been reduced to its lowest common form (e.g., 13/6 is the lowest common form of 26/12), then one of the following three relations is true:

m and n are both odd

m is even and n is odd, or

m is odd and n is even.

Now square both sides of √2 = m/n to obtain 2 = m²/n². Multiplying both sides of this equation tells us that 2n² = m².

Because m² is an integral multiple of 2 (equal to 2 times an integer), it is an even number, and hence m is also an even number.

[How do we know this? If m² is even, then it is equal to 2 times some other integer, say j, so that m² = 2j. If m is not an even number, then m=2h+1, where h is an integer. Then m² = (2h+1)² = 2j. Thus 4h²+4h+1 = 2j, or equivalently j = 2h²+2h+½. This means that j cannot be an integer even though we assumed it was. The contradiction implies that when m² is an even if and only if m is also even. ]

Looking at the three relations above tells us that if m is even, n must be odd.

But if m is even, then m=2k, where k is another integer. Squaring this relationship gives us m² = 4k² = 2n². It follows that n² = 2k², meaning that n², and hence n, are even numbers.

We have now shown that n must be odd, and also that n must be even. The route out of this contradiction is that √2 cannot be expressed as the ratio of two integers, and hence is not a rational number. Mathematicians call these numbers irrational.