Prove or disprove if k | (n + m) then k|(n or k |m
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Let x ∈ N, x odd, then x = 2k + 1 for some k ∈ N. Suppose x = 2k +1 = n(n + 1) for some n ∈ N. • Case 1 n even. Therefore 2k+1 = 2m(2m+1) for some m ∈ NPrinciple of Mathematical Induction: To prove that P(n) is true for all positive integers n, we ... integer k holds, P(m) must also be true. This contradicts P(m) being.
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