2 In class Monday Jan 12th we were trying got prove this bic

2. In class Monday Jan 12th, we were trying got prove this biconditional (goes both ways) statement: If n is a positive integer, then n is even if and only if 3n2+8 is even James proved: if n is even then 3n2+8 is even and if n is odd, then 3n2+ 8 is odd. Zach said we needed to prove: if n is even then 3n2+8 is even and if 3n2+8 is even, then n is even. Is Zach right or James right? (Or both or neither?) Why?

Solution

If n is even, prove that 3n²+8 is even.

step 1 - if n is even, n = 2k for some integer k

step 2 - 3n²+8 = 3(2k)²+8 = 12k²+8

step 3 - If k is even, then :

k²   is even (square of even = even)

2k² is even (product with even = even)

12k²+8 is even (additin with even = even)

So, 3n²+8 is even.

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If n is odd, prove that 3n²+8 is odd.

step 1 - if n is even, n = (2k + 1) for some integer k

step 2 - 3n²+8 = 3(2k + 1)²+8 = 12k² + 12k + 11

step 3 - If k is odd, then :

k²   is odd (square of odd = odd)

12k² is even (product with even = even)

12k² + 12k is even (addition with even = even)

12k² + 12k + 11 is odd (addition with odd = odd)

So, 3n²+8 is off if n is odd.

So, both are right.