irrational number

An irrational number is a real number that is not rational, that is, cannot be expressed as a fraction (or ratio) of the form p/q, where p and q are integers.

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proofs that the square root of 2 is irrational

Pythagorean proof, as given by Euclid in his Elements

proof by contradiction:

1. Assume that 2 is rational, that is, there exists integers p and q such that 2 = p/q; take the irreducible form of this fraction, so that p and q have no factors in common
2. square both sides, to give 2 = p²/q²
3. rearrange, to give 2q² = p²
4. hence p² is even
5. hence p is even (trivial proof left as an excercise for the reader); write p = 2m
6. substitute for p in (3), to give 2q² = (2m)² = 4m²
7. divide through by 2, to give q² = 2m²
8. hence q² is even
9. hence q is even

(1) assumes that p and q have no factors in common; (5) and (9) show they they both have 2 as a factor. This is a contradiction. Hence the assumption (1) is false, and 2 is not rational.

proof based on unique factorisation

[I first saw this rather elegant little proof in Chaitin's Meta Math!, p98. It uses the result of unique factorisation into powers of primes (the fundamental theorem of arthimetic).]

1. Assume that 2 is rational, that is, there exists integers p and q such that 2 = p/q; take the irreducible form of this fraction, so that p and q have no factors in common
2. square both sides, and rearrange, to give 2q² = p²
3. factorise p and q into their unique prime factoriastions, p = 2^a3^b5^c..., q = 2^x3^y5^z...
4. substitute into (2), and simplify, to give 2^2x+13^2y5^2z... = 2^2a3^2b5^2c...

Since the LHS and the RHS refer to the same number, and since such factorisation is unique, we have 2x+1 = 2a (or odd = even). This is a contradiction. Hence the assumption (1) is false, and 2 is not rational.