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I know that we assume x is even. So, as x is even, x = 2k for some integer k.

Then, that would make for 5(2k)+7 = 10k + 7. And this is where I'm stuck. I know that it isn't complete at 10k+7 to prove that it is odd, but I do not have a clue as to what to do to bring out the 2k+1 to show that it is odd. How do I get there?

Any help is appreciated.

JCMcRae
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2 Answers2

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Note that $$10k+7=10k+6+1=2(5k+3)+1.$$ Since $5k+3$ is an integer, we know that $2(5k+3)$ is even and that $2(5k+3)+1$ is odd.

So, this leads that "$x\in\mathbb Z$ is even $\Rightarrow 5x+7$ is odd" is true.

mathlove
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We've to prove that $5x+7$ is even $\implies$ $x$ is odd.

The contrapositive of above is: $x$ is even $\implies$ $5x+7$ is odd.

Now, since $x$ is even $x = 2k$ for some $k$.

$x = 2k \implies 5x = 10k \implies 5x+7 = 10k + 7 = \text{even number} + \text{odd number}$. Now, it is well known fact that sum of an even number and an odd number is an odd number.

Now, since the truth values of an implication and its contrapositive are equal...

taninamdar
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