Chapter 8 - Relations - Exercise Set 8.1 - Page 448: 2

To prove this we need to show that the 2 statements (even(m)-even(n) = even(m-n) and odd(m) - odd(n) = odd(m-n) are true, and show that the rest is false. Case 1 ( m and n both are even) let m = 2k , n = 2p m-n = 2k - 2p = 2(k-p) let k-p = j $\in$ Z, because the difference of 2 integers is an integer hence m-n = 2j, by definition of even m-n is even Case 2 ( m and n both are odd) let m = 2k + 1 , n = 2p + 1. m-n = 2k +1 - (2p + 1) = 2k-2p = 2(k - p) let k-p = j $\in$ Z, because the difference of 2 integers is an integer hence m-n = 2j, by definition of even m-n is even. Case 3( m even , n odd) let m = 2k, n = 2p + 1 m-n = 2k -2p - 1 = 2(k-p) - 1 let k-p = j $\in$ Z, because the difference of 2 integers is an integer hence m-n = 2j - 1, which by definition of odd m-n is odd Case 4 ( m odd n even) let m = 2k+1 , n =2p m-n = 2k + 1 -2p = 2(k-p) + 1 let k-p = j $\in$ Z, because the difference of 2 integers is an integer m-n = 2j + 1, which by definition of odd is odd.

To prove this we need to show that the 2 statements (even(m)-even(n) = even(m-n) and odd(m) - odd(n) = odd(m-n) are true, and show that the rest is false. Case 1 ( m and n both are even) let m = 2k , n = 2p m-n = 2k - 2p = 2(k-p) let k-p = j $\in$ Z, because the difference of 2 integers is an integer hence m-n = 2j, by definition of even m-n is even Case 2 ( m and n both are odd) let m = 2k + 1 , n = 2p + 1. m-n = 2k +1 - (2p + 1) = 2k-2p = 2(k - p) let k-p = j $\in$ Z, because the difference of 2 integers is an integer hence m-n = 2j, by definition of even m-n is even. Case 3( m even , n odd) let m = 2k, n = 2p + 1 m-n = 2k -2p - 1 = 2(k-p) - 1 let k-p = j $\in$ Z, because the difference of 2 integers is an integer hence m-n = 2j - 1, which by definition of odd m-n is odd Case 4 ( m odd n even) let m = 2k+1 , n =2p m-n = 2k + 1 -2p = 2(k-p) + 1 let k-p = j $\in$ Z, because the difference of 2 integers is an integer m-n = 2j + 1, which by definition of odd is odd.