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prove that for all integers n, it is the case that n is even if an only if 3n is even. that is, prove both implications: if n is even, then 3n is even, and if 3n is even, then n is even. discrete math

Respuesta :

If n is even, then 3n is even, and if 3n is even, then n is even.Ifn is even, it means that we can find k integer

such that n = 2* k. Then n+1=2*k+1 is odd.

3n+1=6k+1=2"(3k)+1 where 3k is an integer so 3n+1

is odd.

3n =2*(3k) is even.

Besides, if n+1 is odd we can find an integer p so

that n+1-2p+1 so n = 2p is even.

3n+1 is odd means that we can find an integer

q such that 3n+1=2q+1 so 3n=2q as 3 is not

dividable by 2, it means that n is a multiple of 2,

then n is even.

3n is even means that we can write 3n = 22 where

z is an integer and again it means that n is a

multiple of 2 and then n is even.

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