Question

3) [3 marks] Use a proof by cases that for all real number x, xs]x]. You may need this definition. For any real numbers x, [x]= x, if x2 0, -x, otherwise. 4) [3 marks] Give a direct proof that If x is an odd integer and y is an even integer, then x + y is odd. 5) [3 marks] Give a proof by contradiction for the proposition in Q4, above. That is, give a proof by contraction for the proposition: that If x is an odd integer and y is an even integer, then x + y is odd.

Answer 1

To prove x < 1x1 Case I When negative) IN will be a positive ater than all xco (when x is Then positive a real Number And Real Numbers are Real Number . greater Here X<) [when en X is a case-I When x70 cm When xis greated or equal to o) Here negative negative Number 1x1 = X from Case -1 and casez con that x Elx 19 soy integer form integer 2m+) where mis х is an odd in Therefore x can written in Y is an even y = 2n Where integer x+y = (2m+)) +2n 2 (m+n)+) it is an odd Therefore integer x+y is odd 5 let even x+y is an y=zm (y is even) integet Where I is 2 I = x+y = 21 X+ 2m >> X = 2(1-5) 2(1-m) from calculation we come to Result that X is even But in Question integet. Ouh Assumption is wrong Therefore xty is X is an odd Here on odd .