0
$\begingroup$

Can someone please revise my proof.

(->)

Let $a$ and $x$ be arbitrary integers. Assume $a$ is odd so there exists an integer $k$ s.t $a = 2k + 1$. $a = 2k + 1 = k + k + 1= k + (k+1)$ , evidently $a$ is the sum of two consecutive integers.

(<-)

Let $a$ be the sum of two consecutive integers. $a = x + (x+ 1) = 2x + 1$. by def. of odd, $a$ must be odd.

$\endgroup$
  • $\begingroup$ What is your definition of "odd"? $\endgroup$ – user7530 Feb 19 '15 at 0:56
  • $\begingroup$ Write your assumptions for reverse direction. Say let $a\in\mathbb{Z}$ and assume that $a$ can be written as the sum of two consecutive integers. It follows that $a=k+(k+1)$ where $k\in\mathbb{Z}$... And from there the rest of your proof is correct. The foward direction looks fine to me. $\endgroup$ – user60887 Feb 19 '15 at 0:57
  • $\begingroup$ Assuming the definition of "odd" is "of the form $2k+1$." If it's "not divisible by two" there is a small step missing, as for instance not every element of $\mathbb{Z}[x]$ that's indivisible by 2 is of the form $2k+1$. $\endgroup$ – user7530 Feb 19 '15 at 0:59
1
$\begingroup$

It's logically fine, but stylistically, it doesn't look nice to define the term $x$ in your first direction, and then not use it until your second direction, and then use two different uses of $a$ among both directions.

If you want to redefine the terms within each direction, that's fine, they are separate proofs. If you want to define the terms once and refer to them within each direction, that's OK too, but don't do a mixture of both.

Furthermore, x is not really arbitrary, in that it is dependent on $a$. I would write "suppose that $a$ is the sum of two consecutive integers. Then there exists an integer $x$ such that $a = x + (x + 1)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.