0
$\begingroup$

I'm trying to proof that a number can't be both odd and even at the same time. The definitions are the follows:

  • A number is called even if there's $m\in\mathbb{Z}$ that satisfies $k = 2m$.
  • A number is called odd if there's $m\in \mathbb{Z}$ that satisfies $k = 2m - 1$.

I wrote a proof by contradiction that uses the field axioms and I wonder if I haven't missed anything. The proof goes as follows: Lets assume by contradiction that this state is invalid, which means let $k\in\mathbb{Z}$ be a number that is both odd and even. That means: $2m = k = 2k - 1$ and by the transitivtiy it means $2m = 2m - 1$. In that case, I can add the negative number of $2m$ to both side of the equation and get:

$2m + (-2m) = 2m - 1 + (-2m)$

$(2m + (-2m)) = (2m + (-2m)) - 1$

$0 = -1$

And since we already proofed $0\ne1$ we get a contradiction. $Q.E.D.$

I wonder if my proof is correct? or do I miss anything?

$\endgroup$
7
  • $\begingroup$ We can't use that definition. The question asks to proof that based on the given odd and even definitions. $\endgroup$
    – OzB
    Commented Apr 14, 2018 at 14:28
  • 5
    $\begingroup$ You have an error. Suppose that $k$ is simultaneously odd and even. Then there is some $\color{blue}{m}\in\Bbb Z$ such that $k=2\color{blue}{m}$ and there is some potentially different $\color{red}{n}\in\Bbb Z$ such that $k=2\color{red}{n}+1$. Your proof assumed that $m=n$ but that need not necessarily be the case. $\endgroup$
    – JMoravitz
    Commented Apr 14, 2018 at 14:28
  • 1
    $\begingroup$ from JMoravitz comment: Clearly, $k=k$, meaning $2m = 2n+1$ for integers m, n. So $2m-2n = 1 \iff 2(m-n) = 1$ Contradiction. The left hand side is even, the right-hand side is odd. $\endgroup$
    – amWhy
    Commented Apr 14, 2018 at 14:30
  • 3
    $\begingroup$ @amWhy your final statement "the lefthand side is even, the right is odd" is circular reasoning. We are trying to prove exactly why that cannot occur. A cleaner approach would probably be to show that every multiple of $2$ is either zero or of magnitude strictly greater than $1$. $\endgroup$
    – JMoravitz
    Commented Apr 14, 2018 at 14:33
  • 1
    $\begingroup$ Alternatively, if some work has been done with prime numbers and divisibility in your course so far, you could use that $2$ is a divisor of $2(m-n)$ which should imply that $2$ is a divisor of $1$ which yields the desired contradiction as well. $\endgroup$
    – JMoravitz
    Commented Apr 14, 2018 at 14:40

1 Answer 1

Reset to default
1
$\begingroup$

That is not correct because you are assuming the same $m$ for your odd and even integers.

You may approach the problem with $$ 2m=2n-1$$ and see where you can take it.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .