a2s - CONCORDIA UNIVERSITY DEPARTMENT OF COMPUTER SCIENCE...

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Unformatted text preview: CONCORDIA UNIVERSITY DEPARTMENT OF COMPUTER SCIENCE AND SOFTWARE ENGINEERING COMP232 MATHEMATICS FOR COMPUTER SCIENCE ASSIGNMENT 2 SOLUTIONS FALL 2011 1. Prove that if x 5 is irrational then x is irrational, by proving the contrapositive. PROOF: The contrapositive is If x is rational then x 5 is rational . To prove this assume that x = p q , where p and q are integers. Then x 5 = p 5 q 5 , which is a rational number. 2. Give a proof by cases to show that there are no integer solutions to the equation 2 x 2 + 5 y 2 = 14 . PROOF: We have 2 x 2 14, i.e. , x 2 7, i.e. , | x | 2 (since x is an integer), and 5 y 2 14, i.e. , y 2 2 . 8, i.e. , | y | 1. There remain 6 cases to be considered, namely, x = 0 , 1 , 2, combined with y = 0 , 1. (Negative x and y need not be considered separately.) It is easily checked that none of these 6 combinations of x and y satisfy the equation. 3. Give a proof by contradiction to show that the cube root of 3 is an irrational number. PROOF: Suppose 3 1 / 3 is rational, i.e. , 3 1 / 3 = p/q , where p and q are relatively prime ( i.e. , p and q only have 1 as common factor.) Then 3 = p 3 /q 3 , or, p 3 = 3 q 3 . Thus 3 | p 3 . Hence 3 | p , say, p = 3 k for some integer k . Then 3 q 3 = (3 k ) 3 = 3 3 k 3 , or, q 3 = 3 2 k 3 . Thus 3 | q 3 , hence 3 | q . So p and q have 3 as common factor. Contradiction. NOTE: We used the fact that if 3 | n 3 then 3 | n , whose contrapositive is easily proved by cases: If 3 negationslash | n then n = 3 k + 1 or n = 3 k + 2, hence n 3 = (3 k + 1) 3 = 3(9 k 3 + 9 k 2 + 3 k ) + 1 or n 3 = (3 k + 2) 3 = 3(9 k 3 + 18 k 2 + 12 k + 2) + 2, both of which are not divisible by 3. 4. Give a proof by contradiction to show that if the integers 1, 2, , 99, 100 , are placed randomly around a circle (without repetition), then there must exist three adjacent numbers along the circle whose sum is greater than 152. PROOF: Suppose the sum of any three adjacent numbers is less than or equal to 152. The number 1 must be somewhere along the circle. The remaining 99 positions can be grouped in 33 nonintersecting groups of 3 adjacent positions. Then the total sum will be less than or equal to 33 152 + 1 = 5017. However, we know that 1 + 2 + ... + 100 = 100...
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This note was uploaded on 01/05/2012 for the course COMP 232 taught by Professor Tba during the Spring '11 term at Concordia Canada.

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a2s - CONCORDIA UNIVERSITY DEPARTMENT OF COMPUTER SCIENCE...

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