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we can write odd no = even no + 1

A) odd + odd = even + 1 + even + 1 = (even + even) + (1+1) = even + 2 = even; A is TRUE

B) odd + even = even + even + 1 = even + 1 = odd; B is TRUE

​​​​​​​C) even + even = even; C is TRUE

​​​​​​​D) odd*odd = odd;

odd*even = odd*(2* some odd or even no) = 2 * some no = even 

even*even = even

D is TRUE

​​​​​​​ALL options are TRUE

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These definitions help in solving this problem

Even Number:
A number m is even iff there exists an integer k such that m = 2k

Odd Number:
A number m is odd iff there exists an integer k such that m = 2k+1

 

(a) If n and m are both odd, then n + m is even.
       n = 2p+1 and m = 2q+1 (p,q are integers)
       n+m = 2p+1+2q+1 = 2(p+q+1) = 2(some integer) = even no

(b) If n is odd and m is even, then n + m is odd.
       n = 2p+1 and m = 2q (p,q are integers)
       n+m = 2p+1+2q = 2(p+q)+1 = 2(some integer)+1 = odd no

(c) If n and m are both even, then n + m is even
      n = 2p and m = 2q (p,q are integers)
      n+m = 2p+2q = 2(p+q) = 2(some integer) = even no

(d) If n and m are both odd, then nm is odd; otherwise, nm is even.

     Both n & m are odd
     n = 2p+1 and m = 2q+1 (p,q are integers)
     n.m = (2p+1).(2q+1) = 4pq+2p+2q+1 = 2(2pq+p+q)+1 = 2(some integer)+1 = odd no

    Both n & m are even
    n = 2p and m = 2q (p,q are integers) 
    n.m = 4pq = even number     

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