Proving √2 is irrational

A number is rational if it can be written as a fraction p/q of whole numbers, with q ≠ 0. If it can't, it is irrational. You met √2 in Class 9, but nobody proved it was irrational. Now we will.

One tool first (Theorem 1.2): if a prime p divides a², then p divides a. Example: 5 divides 100 = 10², and 5 also divides 10. Why it always works: a² is made of the same primes as a, just each one twice. So if p is among the primes of a², it was already among the primes of a.

The method: proof by contradiction. Assume the opposite of what you want. Follow it step by step. If you reach something impossible, the assumption must have been wrong.

Proof that √2 is irrational:

• Assume √2 is rational: √2 = a/b. Cancel any common factor first, so a and b are coprime (their only common factor is 1).
• Square both sides: 2 = a²/b², so 2b² = a². That means 2 divides a², and so 2 divides a. Write a = 2c.
• Put it back: 2b² = 4c², so b² = 2c². Now 2 divides b², so 2 divides b too.
• So 2 divides both a and b. But we made them coprime! Impossible.
• The assumption was wrong: √2 is irrational.

The same steps prove √3 and √5 irrational. Just write 3 or 5 wherever you see 2. In fact √p is irrational for every prime p.

A word you need here: coprime is about two numbers whose only common factor is 1, like 8 and 9. Two different primes are always coprime, but coprime numbers need not be prime: 8 and 9 aren't.