Math245Sols4Sum00

# Math245Sols4Sum00 - MATH 245 QUIZ 4 SOLUTIONS 1 Let c be...

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MATH 245: QUIZ 4 SOLUTIONS 1) Let c be the conjecture " is irrational". Let's use a proof by contradiction to prove c . Assume ¬ c , the conjecture " is rational", is true. Then, there exist relatively prime a Z and b Z b 0 ( ) such that = a b . (The "relatively prime" condition ensures that the fraction is in lowest terms.) 3 = a b 3 = a 2 b 2 3 b 2 = a 2 b 2 Z , so 3 divides a 2 and is, therefore, a prime factor of a 2 . Any prime factor of a 2 must also be a prime factor of a , so 3 must also divide a . That means that c Z : a = 3 c . 3 b 2 = 3 c ( ) 2 3 b 2 = 9 c 2 b 2 = 3 c 2 c 2 Z , so 3 divides b 2 and is, therefore, a prime factor of b 2 . Any prime factor of b 2 must also be a prime factor of b , so 3 must also divide b . We've shown that 3 must divide both a and b , which contradicts the relative primality of a and b . Therefore, ¬ c is false, and c is true. Another proof: If were rational, then it would be a rational root of the polynomial x 2 3 . However, the Rational Roots Theorem suggests that the only

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Math245Sols4Sum00 - MATH 245 QUIZ 4 SOLUTIONS 1 Let c be...

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