And similarly the sum and difference of two odd numbers is even, but the product of odd numbers is odd

Can someone help me out here? I'm really stuck on how to start this proof after making my theorem. Thank you!!


closed as off-topic by Daniel W. Farlow, Adam Hughes, HK Lee, user91500, TheGeekGreek Feb 14 '17 at 8:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Daniel W. Farlow, Adam Hughes, HK Lee, user91500, TheGeekGreek
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ What are your thoughts on the problem? Have you tried to approach it on your own? $\endgroup$ – Omnomnomnom Feb 13 '17 at 19:33
  • $\begingroup$ An odd question ? $\endgroup$ – Jean Marie Feb 13 '17 at 19:38
  • $\begingroup$ I tried approaching it by using 2m+1 since that is an odd number and i would just start doing 2(2m+1) and Im not even sure. cause we have to find the difference and i really wasnt sure about it $\endgroup$ – Pikaninja Feb 13 '17 at 19:38

if you have two even numbers, say $$2m,2n$$ with integers $m,n$ then the sum is given by $$2n+2m=2(m+n)=2k$$ thus our sum is even, where we have $$k=m+n$$ and $k$ is an integer number. analogously we have $$2m-2n=2(m-n)=2k$$ where $k=m-n$ is an integer number. and last but not least we have $$2n\cdot 2m=2(2mn)=2k$$ where $k=2mn$


Hint: If $a$ and $b$ are even numbers, then they are of the form $a = 2n$ and $b = 2m$ for $m,n$ integers.


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