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Unformatted text preview: MATH 2400: PRACTICE PROBLEMS FOR EXAM 1 PETE L. CLARK 1) Find all real numbers x such that x 3 = x . Prove your answer! Solution: If x 3 = x , then 0 = x 3 − x = x ( x + 1)( x − 1). Earlier we showed using the field axioms that if if a product of numbers is equal to zero, then at least one of the factors must be zero. So we have x = 0, x + 1 = 0 or x − 1 = 0, i.e., x = 0 , ± 1. Conversely, these three numbers do indeed satisfy x 3 = x . 2) a) Prove that √ 6 is an irrational number. You may use the fact that if an integer x 2 is divisible by 6, then also x is divisible by 6. (For “extra credit”, prove the fact of the previous sentence using the uniqueness of prime factorizations.) Solution: let N be a positive integer. We claim that if N 2 is divisible by 6, then so is N . Indeed by the uniqueness of prime factorization, since N 2 is divisible by 2 and 3, 2 and 3 must each appear in the prime factorization of N . Now we show that √ 6 is irrational. Seeking a contradiction, we suppose there are integers a,b , b ̸ = 0, such that √ 6 = a b . We may assume that a and b have no common prime divisor. Squaring both sides and clearing denominators gives 6 b 2 = a 2 , so a 2 is divisible by 6. By the above paragraph this means a is divisible by 6: say a = 6 A for some integer A . Thus 6 b 2 = a 2 = (6 A ) 2 = 36 A 2 , so 6 A 2 = b 2 . Thus b 2 is divisible by 6 and b is divisible by 6, contradicting the fact that a and b are relatively prime. Comment: It is not in fact necessary to establish the fact that N 2 divisible by 6 implies N divisible by 6. Indeed, we can use the corresponding fact for divisi bility by 2: if 6 b 2 = a 2 then a 2 is even, so a is even, so a = 2 A for some A ∈ Z . Thus 6 b 2 = a 2 = (2 A ) 2 = 4 A 2 , and so 3 b 2 = 2 A 2 . So 3 b 2 is even. Since 3 is odd and the product of two odd numbers is odd, b 2 must be even and thus b is even. So a and b are both even, contradicting our assumption that they are relatively prime. b) Show that if x 2 is an irrational number, so is x . First Solution: Seeking a contradiction we suppose that x is rational, so x = a b . Then x 2 = a b a b = a 2 b 2 would be rational, contradiction. Second Solution: Rather than presenting the above argument as a proof by contra diction, we may also phrase it as a proof by contrapositive . Namely, the statement 1 2 PETE L. CLARK A = ⇒ B is logically equivalent to its contrapositive statement not A = ⇒ not B . In this case, the contrapositive of the statement to be proved is: if x is rational, so is x 2 . Notice that this is exactly what we showed above. c) Show that √ 2 + √ 3 is irrational. Solution: Let x = √ 2+ √ 3. By part b), it suﬃces to show that x 2 is irrational. Now x 2 = ( √ 2+ √ 3) 2 = 2+2 √ 6+3 = 5+2 √ 6. If 5+2 √ 6 = a b , then 2 √ 6 = a b − 5 = a − 5 b b and thus √ 6 = a − 5 b 2 b would be a rational number, contradicting part a)....
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This note was uploaded on 10/26/2011 for the course MATH 2400 taught by Professor Clark during the Fall '11 term at UGA.
 Fall '11
 Clark
 Calculus, Real Numbers

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