Prove That Root 2 (√2) is Irrational

A rational number is any number that can be written as p/q where p and q are integers and q ≠ 0. Examples: 1/2, 3/4, 7, −5/3.

An irrational number is a real number that CANNOT be written in this form. Its decimal expansion is non-terminating and non-repeating.

Examples of irrational numbers: √2, √3, √5, π, e, √7.

To prove √2 is irrational, we assume the opposite — that it IS rational — and derive a logical contradiction, which forces us to reject our assumption. This technique is called proof by contradiction (reductio ad absurdum).

Assume, for the sake of contradiction, that √2 is rational.

Then we can write: √2 = p/q, where p and q are integers with no common factors (i.e., p/q is in its lowest terms), and q ≠ 0.

Step 1: Square both sides. 2 = p²/q² p² = 2q²

Step 2: This means p² is even (it equals 2 times something). If p² is even, then p must be even. (Because if p were odd, p² = odd × odd = odd, which contradicts p² being even.)

Step 3: Since p is even, write p = 2k for some integer k. Substitute into p² = 2q²: (2k)² = 2q² 4k² = 2q² q² = 2k²

Step 4: This means q² is even. By the same reasoning as Step 2, q must also be even.

Step 5: We now have:

• p is even (from Step 2)
• q is even (from Step 4)

But this means p and q share 2 as a common factor. This CONTRADICTS our assumption that p/q is in its lowest terms (no common factors).

Step 6: Since our assumption leads to a contradiction, the assumption must be false. Therefore, √2 is NOT rational. Hence, √2 is irrational. Q.E.D.

The proof relies on the lemma: 'If p² is even, then p is even.'

Proof of the lemma by contrapositive: The contrapositive of 'if p² is even then p is even' is 'if p is odd then p² is odd.'

If p is odd, then p = 2m+1 for some integer m. p² = (2m+1)² = 4m² + 4m + 1 = 2(2m² + 2m) + 1

This is of the form 2k+1 (odd), so p² is indeed odd.

Since the contrapositive is true, the original statement 'if p² is even then p is even' is also true.

This lemma is crucial: it lets us conclude in Step 2 that p being squared-even implies p itself is even.

The proof uses two fundamental logical tools:

1. Proof by Contradiction (Reductio ad Absurdum): Assume the negation of what you want to prove. If this assumption leads to a logical impossibility (contradiction), then the assumption is false and the original statement is true.

2. Fundamental Theorem of Arithmetic: Every integer has a unique prime factorisation. If p/q is in lowest terms, gcd(p,q) = 1, meaning they share no common prime factor. Showing both p and q are even (both divisible by 2) violates this.

The proof was known to the ancient Greeks and is often attributed to the Pythagoreans. It showed, shockingly at the time, that not all lengths can be expressed as ratios of whole numbers — a revolutionary mathematical discovery.

General result: √n is irrational for any positive integer n that is not a perfect square.

The same method proves √3 and √5 (and generally √p for any prime p) are irrational:

Proof that √3 is irrational: Assume √3 = p/q in lowest terms. Then p² = 3q², so p² is divisible by 3, so p is divisible by 3 (write p = 3k). Substituting: 9k² = 3q², so q² = 3k², so q is divisible by 3. Contradiction — both p and q are divisible by 3.

Proof that √5 is irrational: Assume √5 = p/q in lowest terms. Then p² = 5q², so 5|p², so 5|p (write p = 5k). Then 25k² = 5q², so q² = 5k², so 5|q. Contradiction.

General pattern: The proof works for any prime number under the square root by using the property that if a prime number divides n², it also divides n.