proof that 2 is irrational

Assume that the square root of 2 ( is rational. Then we can write


where a,b and a and b are relatively prime. Then 2=(2)2=(ab)2=a2b2. Thus, a2=2b2. Therefore, 2a2. Since 2 is prime, it must divide a. Then a=2c for some c. Thus, 2b2=a2=(2c)2=4c2, yielding that b2=2c2. Therefore, 2b2. Since 2 is prime, it must divide b.

Since 2a and 2b, we have that a and b are not relatively prime, which contradicts the hypothesisMathworldPlanetmath. Hence, the initial assumptionPlanetmathPlanetmath is false. It follows 2 is irrational.

With a little bit of work, this argument can be generalized to any positive integer that is not a square. Let n be such an integer. Then there must exist a prime p and k,m such that n=pkm, where pm and k is odd. Assume that n=a/b, where a,b and are relatively prime. Then pkm=n=(n)2=(ab)2=a2b2. Thus, a2=pkmb2. From the fundamental theorem of arithmeticMathworldPlanetmath, it is clear that the maximum powers of p that divides a2 and b2 are even. Since k is odd and p does not divide m, the maximum power of p that divides pkmb2 is also odd. Thus, the same should be true for a2. Hence, we have reached a contradictionMathworldPlanetmathPlanetmath and n must be irrational.

The same argument can be generalized even more, for example to the case of nonsquare irreducible fractions and to higher order roots.

Title proof that 2 is irrational
Canonical name ProofThatsqrt2IsIrrational
Date of creation 2013-03-22 12:39:13
Last modified on 2013-03-22 12:39:13
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 11
Author Wkbj79 (1863)
Entry type Proof
Classification msc 11J72
Related topic Irrational
Related topic Surd