9 min readPathway sheetA 3-mark proof, sometimes with a 'hence' part

Proving √2, √3 and √5 are irrational

Assume √p is rational, write it as a/b in lowest terms, and show p divides both a and b. This contradiction proves √p is irrational.

Rational numberCan be written as p/q, where p and q are integers and q ≠ 0

Irrational numberCannot be written as p/q; its decimal never ends and never repeats

Key theoremIf a prime p divides a², then p divides a (a is a positive integer)

MethodProof by contradiction

Rational and irrational numbers

A rational number can be written as p/q, where p and q are integers and q ≠ 0. Examples: 3/4, −5 and 0.25. An irrational number cannot be written this way. Examples: √2, √3, √5 and π. Co-prime integers have no common factor other than 1. Proof by contradiction means: assume the opposite, reach something impossible, so the assumption must be wrong.

The theorem you need

Theorem: let p be a prime number. If p divides a², then p divides a, where a is a positive integer. Why: by the Fundamental Theorem of Arithmetic, the primes in a² are exactly the primes in a, each used twice. So if p is a prime factor of a², it must already be a prime factor of a.

The pattern for any √p, where p is prime

1
Assume √p is rational: √p = a/b, where a and b are co-prime integers and b ≠ 0.
2
Square both sides: p b² = a².
3
So p divides a². By the theorem, p divides a. Write a = pc.
4
Substitute: p b² = p² c², so b² = p c².
5
So p divides b², and by the theorem p divides b.
6★
Now p divides both a and b. This contradicts that a and b are co-prime. So √p is irrational.

★ key step

Example: prove that √5 is irrational

Problem

Prove that √5 is irrational.

Working

Assume, to the contrary, that √5 is rational.
Then √5 = a/b, where a and b are co-prime integers and b ≠ 0.
So √5 b = a. Squaring both sides: 5b² = a².
So 5 divides a². Since 5 is prime, 5 divides a.
Write a = 5c for some integer c. Then 5b² = 25c², so b² = 5c².
So 5 divides b², and therefore 5 divides b.
Now 5 divides both a and b. This contradicts the fact that a and b are co-prime.
So our assumption is wrong.

Answer

Therefore √5 is irrational.

Where students lose marks

Not saying that a and b are co-prime. The contradiction needs it.
Not quoting the theorem: if a prime p divides a², then p divides a.
Squaring wrongly. (a/b)² = a²/b², so √5 = a/b gives 5 = a²/b².
Stopping without the words contradiction and therefore √5 is irrational.
Using the method for a number like √9 = 3, which is rational. The proof works because 5 is prime.

In real life

A square floor tile of side 1 foot has a diagonal of √2 feet, about 1.414 feet. A tile cutter in a Jaipur workshop can only ever measure it approximately, because √2 never ends or repeats.

Exam angle

1 marks: “Which of these is irrational: √16, 22/7, √7 or 1.25?” √7, because 7 is not a perfect square. The others can be written as p/q (√16 = 4).

2 marks: “State the theorem used to prove that √3 is irrational.” If a prime p divides a², then p divides a, where a is a positive integer.

3 marks: “Prove that √2 is irrational.” Assume √2 = a/b (co-prime, b ≠ 0). Then 2b² = a², so 2 divides a; write a = 2c. Then b² = 2c², so 2 divides b. Contradiction, so √2 is irrational.

Remember it

Assume, square, divide, divide again, contradict. Five words for every √p proof.