Microsoft Word ans wks5 06.doc

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1 Answer to worksheet 5 – Proofs 1) The sum of any two even integers is even. Proof: a and b be even integers. By definition of even we have that a = 2n and b = 2m. Let Consider the sum a + b = 2n + 2m = 2(n + m) = 2k where k = n + m is an integer. have shown that my hypothesis is true. Therefore by definition of even we 2) The negative of any even integer is even. Proof: Let a be an arbitrary even integer a nd by definition a = 2n for some integer n. Now -n is an integer. Ther efore by definition -2n consider the -(2n) = 2(-n) = 2k where k = is also an even integer. 3) The sum of any two odd integers is even. Proof: Let a and b be odd integers. By definition of odd we have that a = 2n + 1 and b = 2m + 1. Consider the sum a + b = (2n + 1) + (2m +1) = 2n + 2m +2 = 2k, where k = n + m + 1 is an integer. Therefore by definition of even we have shown that hypothesis is true. a + b is even and my 2 4) If n is odd then n is also even. 2 Counter example: Let n = 3 then (3) = 9 is also odd 5) The product of two odd integers is odd. Proof: Let n and m be two odd integers. By defi nition of odd we have that n = 2a + 1 and m = 2b + 1. Consider the product nm = (2a + 1)(2b +1) = 4ab + 2a + 2b +1= 2( 2ab + a + b) + 1 = 2k + 1, where k = (2 ab +a +b ) is an integer. Therefore by definition of odd we have shown that the product of two odd integers is also odd. 2 is prime. 6) If n is prime then n 2 Counter example: Let n = 5 then n = 25 which is not prime 7) If 2n + 1 is odd then 2n -1 is also odd Proof: Consider 2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1 where k = n-1 is an integer and 2n -1 is odd by definition. 8) The difference of an even integer minus an odd integer is odd. Proof: Let a = 2n be an even integer by definition of even and b = 2k + 1 by definition of odd and n and k are integers. Consider a – b = 2n –(2k + 1) = 2(n – k) -1 = 2p-1 where p = n – k is an integer And where 2p – 1 is odd by proof in problem 7).

2 9) The sum of two ratio nal numbers is rational: tional numbers. By definition of rational Proof: Let r and w be two ra ac where a,b,c,d are integers and b 0 and d ≠ 0 ≠ == randw bd acadbc + 0 is an integer and ad +bc is an Consider where bd ≠ +=+= rw bdbd integer. Therefore the sum is rational by definition. 10) The product of two ra tional numbers is rational. Proof: Let r and w be two rational numbers. By definition of rational ac 0 ≠ where a,b,c,d are integers and b ≠ 0 and d == randw bd acac Consider * is an ac 0 is an integer and where bd ≠ == rw bdbd integer. Therefore th e product is rational by definition.

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