The root that never fits

The idea

√2, and in general √p for a prime p, is irrational: assuming √p = a/b with a, b coprime gives pb² = a², and Theorem 1.2 then makes p divide both a and b, a contradiction.

Is √2 a fraction?

Suppose √2 = a/b, written in lowest terms: a and b have no common factor except 1.

Can such whole numbers a and b exist?

• Yes, we just have to find them
• No, and we can prove it
• Not sure

Climb the proof

Climb the proof of Theorem 1.3 rung by rung. It uses Theorem 1.2. Then find what is wrong.

√p is irrational

√2 is irrational, and in general √p is irrational for every prime p. Assuming √p = a/b with a, b coprime gives pb² = a²; Theorem 1.2 then makes p divide both a and b, a contradiction.

Replace 2 by any prime: √3, √5, √7, √11 … are all irrational.

Be careful: √4 = 2 and √9 = 3 are rational. 4 and 9 are not primes.

Notes

√2, and in general √p for a prime p, is irrational. Assume √p = a/b with a, b coprime: pb² = a², and Theorem 1.2 makes p divide both a and b. That is a contradiction.

Check yourself

Which of these is shown irrational by this theorem?

For √5 we get 5b² = a². What is the next conclusion?

• √4. √4 = 2 is rational. 4 is not a prime.
• √7 — correct. Yes! 7 is a prime, so √7 is irrational.
• √9. √9 = 3 is rational. 9 is not a prime.
• √16. √16 = 4 is rational. 16 is not a prime.
• 5 divides a — correct. Yes! 5 divides a², and 5 is prime, so 5 divides a.
• 5 divides b. Not yet. First Theorem 1.2 gives 5 divides a.
• a = b. Nothing says a = b.