2400practice1SOLS

2400practice1SOLS - MATH 2400: PRACTICE PROBLEMS FOR EXAM 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

This note was uploaded on 10/26/2011 for the course MATH 2400 taught by Professor Clark during the Fall '11 term at UGA.

Page1 / 6

2400practice1SOLS - MATH 2400: PRACTICE PROBLEMS FOR EXAM 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online