les091203 - THREE PROBLEMS PROBLEM 1: Prove that 2 is...

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THREE PROBLEMS PROBLEM 1 : Prove that 2 is irrational. Proof : Assume to the contrary that 2 is rational, that is 2 = a b , where a and b are integers and b 6 = 0 . Moreover, let a and b have no common divisor > 1 . Then 2 = a 2 b 2 2 b 2 = a 2 . (1) Since 2 b 2 is even, it follows that a 2 is even. Then a is also even (in fact, if a is odd, then a 2 is odd). This means that there exists q Z such that a = 2 q. Substituting this into (1), we get 2 b 2 = (2 q ) 2 b 2 = 2 q 2 . Since 2 q 2 is even, it follows that b 2 is even. Then b is also even by the arguments above. This is a contradiction. ± THEOREM (DIVISION ALGORITHM): For any integers a and b with a 6 = 0 there exist unique integers q and r such that b = aq + r, where 0 r < | a | . The integers q and r are called the quotient and the reminder respectively. PROBLEM 2 : Prove that 3 is irrational. Proof : Assume to the contrary that 3 is rational, that is 3 = a b , where a and b are integers and b 6 = 0 . Moreover, let a and b have no common divisor > 1 . Then 3 = a 2 b 2 3 b 2 = a 2 . (2) Since 3 b 2 is divisible by 3, it follows that a 2 is divisible by 3. Then a is also divisible by 3. In fact, if a is not divisible by 3, then by the Division Algorithm there exists q Z such that a = 3 q + 1 or a = 3 q + 2 . Suppose a = 3 q + 1 , then a 2 = (3 q + 1) 2 = 9 q 2 + 6 q + 1 = 3(3 q 2 + 2 q | {z } q 0 ) + 1 = 3 q 0 + 1 , which is not divisible by 3. We get a contradiction. Similarly, if a = 3 q + 2 , then a 2 = (3 q + 2) 2 = 9 q 2 + 12 q + 4 = 3(3 q 2 + 4 q + 1 | {z } q 00 ) + 1 = 3 q 00 + 1 , which is not divisible by 3. We get a contradiction again.
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les091203 - THREE PROBLEMS PROBLEM 1: Prove that 2 is...

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