This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Math 347-G1. (Milin) Solutions to Exam 1 1 1. (20 points) For both claims below, rewrite them in logical notation, negate the logical expressions and then write out an English sentence which expresses the negated claim without using words of negation. (a) There exists a > , so that for all > and all real numbers x , whenever | x | < we have | f ( x ) | < . Solution. In notation: ( > 0)( > 0)( x R ) | x | < | f ( x ) | < Negation: [( > 0)( > 0)( x R ) | x | < | f ( x ) | < ] ( > 0)( > 0)( x R ) | x | < | f ( x ) | The negation is: For all positive there exists a positive and a real number x so that | x | < and | f ( x ) | . (b) For all real numbers a , if a 2 + 5 a + 6 > , then we must have a >- 2 or a <- 3 . Solution. In notation: ( a R ) a 2 + 5 a + 6 > ( a >- 2 a <- 3) Negation: ( a R ) a 2 + 5 a + 6 > ( a >- 2 a <- 3) ( a R ) a 2 + 5 a + 6 > ( a >- 2 a <- 3) ( a R ) a 2 + 5 a + 6 > a - 2 a - 3 The negation is: There exists a real number a so that a 2 + 5 a + 6 > 0, a - 2 and a - 3. 2. (20 points) Give brief proofs to the following two claims. (a) If f : R R is a bounded function, then g ( x ) = f ( x ) + f (- x ) is bounded as well....
View Full Document