Book A Free Math Class

Prove That Root 7 Is Irrational — Proof by Contradiction

TL;DR
To prove that root 7 is irrational, assume the opposite — that 7=( \frac{p}{q} ) in lowest terms — and derive a contradiction from the fact that 7 must then divide both p and q. This article walks the full proof by contradiction, explains the prime-divisibility step it depends on, and shows the common errors.

What Is an Irrational Number?

An irrational number is a real number that cannot be written as ( \frac{p}{q} ) for any two integers p and q with ( q \neq 0 ). Its decimal expansion never terminates and never settles into a repeating block. Values like ( \sqrt{7} ), π, and e are irrational; every integer, every terminating decimal, and every repeating decimal is rational.

The square root of 7 is a natural candidate because 7 is not a perfect square — no whole number squares to 7, since ( 2^2 = 4 ) and ( 3^2 = 9 ). To prove ( \sqrt{7} ) is irrational we must rule out every fraction at once, which is exactly what proof by contradiction does.

Proof That Root 7 Is Irrational

Here is the full argument laid out cleanly in one place, one step per line. The worked examples below unpack each step in more detail.

Claim: ( \sqrt{7} ) is irrational.

Assume the opposite — that ( \sqrt{7} ) is rational.

( 7 = \frac{p}{q}, \quad \gcd(p, q) = 1, \quad q \neq 0 )

Square both sides.

( 7 = \frac{p^2}{q^2} )

Multiply through by ( q^2 ).

( p^2 = 7q^2 )

So 7 divides ( p^2 ). Because 7 is prime, 7 divides p.

Write ( p = 7k ).

Substituting:

( (7k)^2 = 7q^2 )
( 49k^2 = 7q^2 )

Divide both sides by 7.

( 7k^2 = q^2 )

So 7 divides ( q^2 ), and since 7 is prime, 7 divides q as well.

Now 7 divides both p and q, which contradicts ( \gcd(p, q) = 1 ). The assumption is false, so ( \sqrt{7} ) is irrational.

The General Method — Why It Works for Any Prime

Nothing in the proof used the specific value 7 except the fact: 7 is prime. That single property drives both divisibility steps through the prime-divisibility rule:

If a prime r divides n^2, then r divides n.

Swap 7 for any other prime and the argument runs unchanged. The square roots of 2, 3, 5, and every other prime are irrational for exactly this reason.

Common Mistakes When Proving Root 7 Is Irrational

Mistake 1: Forgetting the "lowest terms" assumption

Assuming ( 7 = \frac{p}{q} ) without stating that ( \gcd(p, q) = 1 ).

Mistake 2: Using the rule for a composite number

Assuming "r divides n^2 implies r divides n" holds for any r.

Mistake 3: Trying to prove it from the decimal

Treating a long non-repeating decimal from a calculator as a finished proof.

Conclusion

To prove that root 7 is irrational, assume ( 7 = \frac{p}{q} ) in lowest terms and derive a contradiction. The same proof by contradiction works for the square root of any prime, but not for perfect squares like 4.