Quicker Root 2 Irrationality Proof.

Firstly, a rational number is any number expressible as the quotient of 2 integers, hence the group of rational numbers is labelled by the letter `Q`. E.g., 5 = 10/2, among many more possible integer fractions, hence 5 is rational. Any irrational number is a number inexpressible as an `integer-only` fraction.

To prove (square) root 2 is irrational, let`s assume it is rational, and show that this leads to a contradiction. Since all numbers are either rational or irrational, this will lead to the conclusion that root 2 must then be irrational.

Assume root 2 is rational, i.e. root 2 = a/b, where a and b are both integers.

Square both sides and multiply by b^2:

2b^2 = a^2 . This is a contradiction. Before continuing to read, you may want to think about why this may be (hint: prime factorisation).

This is a contradiction because the prime factorisation of number has only even powers iff the number is a perfect square (the square of an integer). E.g. 36 = 3^2 * 2^2: all the powers are even (they are all 2), hence the number is a perfect square. This is because 2 can be factorised out of the powers, leaving (3*2)^2, i.e., 6^2. Apply this to any perfect square to understand why its prime factorisation must have only even powers.

The contradiction comes from of the fact that the left hand side always has an odd amount of 2s in its prime factorisation, whereas the right hand side always has an even amount of 2s in its prime factorisation. Hence the two can never be equal, as the equation would suggest. Hence:

2b^2 ≠ a^2. Therefore root 2 ≠ a/b (where a and b are integers) and root 2 must be irrational, as required.

Exercise for the reader: can you apply the same chain of reasoning to all numbers of the form root x, where x is not a perfect square, to prove that these too are irrational?