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I need to show that if $3n-1$ is odd then $n$ must be even. I'm doing this in cases.

For the first case I am saying: $$n = 2k \Rightarrow 3n-1 = 6k-1$$ Let $$j = 3k \Rightarrow (3n-1) = (6k-1) = (2j - 1)$$ therefore if $n$ is even then $3n-1$ is odd: $$\forall x,y \in \mathbb{Z}:\quad n = 2x \Rightarrow 3n-1 = 2y + 1$$

Based on how I've seen the use of definitions to write proofs, I feel like this is an acceptable method of direct proof, but what is stopping me from defining: $$u = j-1$$ which would imply $3n-1 = 2u$ which by using definitions would make it even. I feel like I must have an intrinsic misunderstanding of how to do proofs. So I feel I should ask, is the proof above valid? Or if not then why?


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@Bernard

what if, just for the sake of argument, you defined $2(3k+1)$ to be equal to $2u-1$.

therefore $3n-1 = 2u-1$ (odd function).

You might then argue that

$$n=\frac 23u + \frac 23$$ is not an integer,

but couldn't I define $u$ as some real number s.t.

$$u\cdot\frac 23\equiv\frac 23$$ more than some integer $z$?

then n would be an integer and $3n-1 = 2u-1$ making it odd.

Is there any reason why one couldn't make this substitution?

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3 Answers 3

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Actually you proved the converse: if $n$ is even, then $3n-1$ is odd. To complete the proof you must show that, if $n$ is odd, then $3n-1$ is even.

Indeed, if $n$ is odd, it can be written as $n=2k+1$ for some $k$. Then $$3n-1=3(2k+1)-1=6k+2=2(3k+1),$$ which proves $3n-1$ is even.

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  • $\begingroup$ I think I understand. So I'm guessing proof of the converse isn't proof of the original theorem? I'm trying to rewrite the 2(3k+1) in your proof as some odd function but find it to be a bit more difficult $\endgroup$ Sep 23, 2015 at 20:23
  • $\begingroup$ You can have a quite simple proof if you know about congruences. $\endgroup$
    – Bernard
    Sep 23, 2015 at 20:28
  • $\begingroup$ I appreciate your help. I am looking into congruences now, but I want to make sure I understand the concept so if you have a moment, would you be able to respond to my edit of the original question? $\endgroup$ Sep 23, 2015 at 20:50
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Yes, I don't see any wrong in your method. But it can be prettier:

Consider $k\in \Bbb Z$ such that, $$3n-1=2k+1\\\implies 3n=2(k+1)$$

Hence $$3n\mid 2m, \text {where m=k+1}$$

But $\gcd(2,3)=1$, hence $n\mid2$ or $n\mid 2m$.

In both the cases, $n$ must have an even factor. So, $n$ is even.

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Other answers have lead OP to the correct way of proving the proposition, which is proving the forward implication instead of its converse. For what it's worth, I would like to address questions OP directly posed that have been left unanswered

what is stopping me from defining:

$u=j-1$

which would imply $3n−1=2u$ which by using definitions would make it even.

This is an incorrect substitution. If we let $u=j-1$, then $2u=2(j-1)=2j-2$, which is not equal to $2j-1$. But even if we let $u=j-\frac{1}{2}$ such that $2u=2(j-\frac{1}{2})=2j-1$ is correct, $u$ is not an integer, hence we cannot conclude that $2u$ is an even integer.

what if, just for the sake of argument, you defined $2(3k+1)$ to be equal to $2u−1$. therefore $3n−1=2u−1$ (odd function).

This is not an incorrect substitution per se, but OP did not explicitly write what value $u$ is equal to and thus mistakenly believed that $2u$ is twice an integer. If we let $2(3k+1)=2u-1$, then $6k+2=2u-1\implies6k+3=2u\implies3k+\frac{3}{2}=u$. Thus, $u$ is not necessarily an integer since $k$ is an arbitrary integer (example: $k=1$). Therefore, we cannot conclude that $2u-1$ is an odd integer.

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