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25)\\  Prove that if n is an integer and 3n + 2 is even, then n is even using  \item[a)] a proof by contraposition.  \paragraph{Solution} Assume that ``If If  $3n + 2$ is even, then n is even" even  is fall; We n is odd, so n = $2k + 1$, k $\in$ $\mathbb{Z}$. Substituting $3n + 2$ = $3(2k + 1) + 2$ = $6k + 5$ = $2(3k + 2) + 1$ is odd. Because the negation of the conclusion of the conditional statement implies that the hypothesis is false, Q.E.D \item[b)] a proof by contradiction.  \paragraph{Solution} Let p is the proposition ``$n$ that $n$  is even". even.  Suppose $\neg$p is true, which means ``$n$ is odd". If we add subtract an odd number from an even number, we get an odd number, so $3n + 2 - n$ = $4n + 2$ = $2(2n + 1)$ is odd. But this is not true. Therefore our supposition was wrong, and the proof by contradiction is complete.\\  27)\\  we need to prove that all these equivalent to x being even. Then x=2k( k is integer). So, 3x+2=6x+2 is even.Similarly, x+5=2x+5=2x+4+1 is odd. $x^2$=4$k^2