# Example let x be an integer if x 1 then x 3 1 is

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Example. Let x be an integer. If x > 1, then x 3 + 1 is composite.

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January 10, 2011 Lecture Outline 9 Solution. Let x be an arbitrary but specific integer such that x > 1. We can rewrite x 3 + 1 as ( x + 1)( x 2 x + 1). Note that since x is an integer both ( x + 1) and ( x 2 x + 1) are integers. Hence ( x + 1) | x 3 + 1 and ( x 2 x + 1) | x 3 + 1. We now need to show that x + 1 > 1 and x 2 x + 1 > 1. Since x > 1, clearly, x + 1 > 1. x 2 x + 1 > 1 by the following reasoning. x > 1 x 2 > x (Multiplying both sides by x .) x 2 x > 0 (Subtracting both sides by x .) x 2 x + 1 > 1 (Adding 1 to both sides.) We can also argue that x 2 x + 1 > 1 by showing that x + 1 < x 3 + 1. Since x > 1 we have x 2 > x and hence x 2 > 1. Multiplying both sides by x again we get x 3 > x . This means that x + 1 < x 3 + 1 and since ( x + 1) | x 3 + 1, we conclude that x 3 + 1 is composite. Note: One student asked the question that why can’t we write x 3 + 1 as x 3 (1 + 1 x 3 ). The reason is that for an integer x > 1, (1 + 1 x 3 ) is not an integer and the proof breaks down. Example. Prove that, for all real numbers x and all integers m , x + m = x + m Solution. Let x = y + ǫ , where y is the largest integer with value at most x and 0 ǫ < 1. Then, x + m = y + ǫ + m x + m = y + m + ǫ = y + m = x + m Example. Prove that if x and y are integers where x + y is even, then x and y are both odd or both even. Solution. To prove the above claim we will prove its contrapositive which is “if exactly one of x or y is even then x + y is odd”. Without loss of generality, for some integers k and l , let x = 2 k be even and y = 2 l + 1 be odd. Then, x + y = 2 k + 2 l + 1 = 2( k + l ) + 1 Since k and l are integers so is k + l and 2( k + l ) is even and hence x + y is odd. Example. Prove that the product of two odd numbers is an odd number.
10 Lecture Outline January 10, 2011 Solution. Let x and y be particular but arbitrarily chosen odd numbers. Then, x = 2 k +1 and y = 2 l + 1, for some integers k and l . We have x · y = (2 k + 1) · (2 l + 1) = 4 kl + 2( k + l ) + 1 = 2(2 kl + k + l ) + 1 Let p = 2 kl + k + l . Since k and l are integers, p is an integer and x · y = 2 p + 1 is odd. Example. Prove that 2 is irrational. Solution. For the purpose of contradiction, assume that 2 be a rational number. Then there are numbers a and b with no common factors such that 2 = a b Squaring both sides of the above equation gives 2 = a 2 b 2 a 2 = 2 b 2 (1) From (1) we conclude that a 2 is even. This fact combined with the result of Example 1 implies that a is even. Then, for some integer k , let a = 2 k (2) Combining (1) and (2) we get 4 k 2 = 2 b 2 2 k 2 = b 2 The above equation implies that b 2 is even and hence b is even. Since we know a is even this means that a and b have 2 as a common factor which contradicts the assumption that a and b have no common factors. We will now give a very elegant proof for the fact that “ 2 is irrational” using the unique factorization theorem which is also called the fundamental theorem of arithmetic . The unique factorization theorem states that every positive number can be uniquely repre- sented as a product of primes. More formally, it can be stated as follows. Given any integer n > 1, there exist a positive integer k , distinct prime numbers p 1 , p 2 , . . . , p k , and positive integers e 1 , e 2 , . . . , e k such that n = p e 1 1 p e 2 2 p e 3 3 · · · p e k k and any other expression of n as a product of primes is identical to this except, perhaps, for the order in which the factors are written.

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January 10, 2011 Lecture Outline 11 Example. Prove that 2 is irrational using the unique factorization theorem.
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