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Unformatted text preview: SOME IRRATIONAL NUMBERS PETE L. CLARK Proposition 1. The square root of 2 is irrational. Proof. Suppose not: then there exist integers a and b 6 = 0 such that 2 = a b , meaning that 2 = a 2 b 2 . We may assume that a and b have no common divisor if they do, divide it out and in particular that a and b are not both even. Now clear denominators: a 2 = 2 b 2 . So 2 | a 2 . It follows that 2 | a . Notice that this is a direct consequence of Euclids Lemma if p | a 2 , p | a or p | a . On the other hand, we can simply prove the contrapositive: if a is odd, then a 2 is odd. By the Division Theorem, a number is odd iff we can represent it as a = 2 k + 1, and then we just check: (2 k + 1) 2 = 4 k 2 + 4 k + 1 = 2(2 k 2 + 2 k ) + 1 is indeed again odd. So a = 2 A , say. Plugging this into the equation we get (2 A ) 2 = 4 A 2 = 2 b 2 , b 2 = 2 A 2 , so 2 | b 2 and, as above, 2 | b . Thus 2 divides both a and b : contradiction. Comment: This is a truly classical proof. In G.H. Hardys A Mathematicians Apology , an extended rumination on the nature and beauty of pure mathematics, he gives just two examples of theorems: this theorem, and Euclids proof of the infinitude of primes. As he says, this is inevitably a proof by contradiction (unlike Euclids proof, which constructs new primes in a perfectly explicit way). The orig- inal statement is logically more complicated than what we actually prove in that it takes for granted that there is some real number 2 characterized by being positive and having square equal to 2 and then shows a property of this real number, namely it not being a fraction. But the essence of the matter is that a certain mathematical object does not exist namely a rational number a b such that ( a b ) 2 = 2. This was the first impossibility proof in mathematics....
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