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The Definition that I was given for an odd number is: $m$ is odd if there exist an integer $k$ such that $m =2k+1$.

I'm lost on how to prove this, what I have so far is this

$4m+7 = 4i+7$

$=(4i+6)+1$

When giving an explanation I'm not sure if I should say $4i+6$ is an integer (even) and $+1$ is odd.

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    $\begingroup$ $4m+7=2(2m+3)+1$, $2m+3$ is an integer so it fits the form $2k+1$ $\endgroup$ – Harry Alli Jan 31 '18 at 0:22
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When giving an explanation I'm not sure if I should say $4i+6$ is an integer (even) and $+1$ is odd.

No, this is not what you want to say! You are told to use the definition, so you need to use the the definition. That is, the only way you can prove a number is odd (in this context) is by proving it satisfies the definition: that is, there exists an integer $k$ such that the number is $2k+1$.

That mean you need to find some integer $k$ such that $4m+7=2k+1$. In other words, you want to solve that equation for $k$ and show that the value of $k$ you obtain is indeed an integer. Can you see a way to do that?

(Also, when writing a proof, you want to be very clear and precise about all your notation. What do you mean when you write $4n+7=4i+7$? Where did these variables $n$ and $i$ come from? The problem statement, at least in the title of your post, uses $m$ instead, so you should probably use $m$. If you use a different variable, you should explain what it means and how it is related to $m$.)

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