# a2sfall2009 - CONCORDIA UNIVERSITY DEPARTMENT OF COMPUTER...

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CONCORDIA UNIVERSITY DEPARTMENT OF COMPUTER SCIENCE AND SOFTWARE ENGINEERING COMP232-Section DD MATHEMATICS FOR COMPUTER SCIENCE ASSIGNMENT 2 SOLUTIONS FALL 2009 1. A perfect number is a positive integer that equals the sum of all of its divisors, except the number itself. Give a proof by contradiction that there is no perfect prime number. PROOF: Suppose p is a perfect number as well as a prime number. Since p is prime, its only divisers are 1 and p itself. Since p is perfect it is the sum of its divisors, not including p itself. Thus p = 1. But 1 not a prime number. Contradiction. 2. Prove that if x 3 is irrational then x is irrational, by proving the contrapositive. PROOF: The contrapositive is “ If x is rational then x 3 is rational ”. To prove this assume that x = p q ,where p and q are integers. Then x 3 = p 3 q 3 , which is a rational number. 3. Give a proof by cases to show that there are no integer solutions to the equation 2 x 2 +5 y 2 =1 4 . 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 , 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. 4. Give a proof by contradiction to show that the cube root of 2 is an irrational number. PROOF: Suppose 2 1 / 3 is rational, i.e. ,2 1 / 3 = p/q p and q are relatively prime ( i.e. , p and q only have 1 as common factor.) Then 2 = p 3 /q 3 ,or , p 3 =2 q 3 .T h u s p 3 is even. Hence p is even, say, p k for some integer k h e n2 q 3 =(2 k ) 3 =8 k 3 ,o r , q 3 =4 k 3 . Thus q 3 is even, hence q is even. So p and q have 2 as common factor. Contradiction. NOTE: We used the fact that if n 3 is even then n is even, whose contrapositive is easily proved: If n is odd then n k + 1, hence n 3 =2(4 k 3 +6 k 2 +3 k ) + 1 is odd. 5. 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.

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a2sfall2009 - CONCORDIA UNIVERSITY DEPARTMENT OF COMPUTER...

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