Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question I am working on is:

Prove that if $m+n$ and $n+p$ are even integers, where $m$, $n$,and $p$ are integers, then $m+p$ is even. What kind of proof did you use?

I was thinking--and I aware that this may not be the most efficient method--of proving four different cases: $m$ and $p$ are both even; $m$ and $p$ are both odd; or $m$ and $p$ are opposite parity. Would this work? Or is there a better way?

share|improve this question
    
It should work. –  A.P. Feb 12 '13 at 18:05
    
What you mention should work perfectly. As for the name of the method, I don't know, but I've seen that sometimes be called prove by inspection. Which is what you do when you analise all possible cases... –  MyUserIsThis Feb 12 '13 at 18:32

3 Answers 3

up vote 1 down vote accepted

Let $m+n=2a,n+p=2b$ where $m,n,a,b$ are integers

$\implies m+n+n+p=2(a+b)\implies m+p=2(a+b-n)$

share|improve this answer
    
Oh, that's clever. However, I'd like to know if my method is valid. –  Mack Feb 12 '13 at 18:03
1  
@EliMackenzie, it will definitely work. But, playing with all the combinations of parities is error prone. –  lab bhattacharjee Feb 12 '13 at 18:05

This is quite easy way : $(m+p)=((m+n)+(n+p))-2n$

share|improve this answer

Hint $\, $ If $\rm\,m\,$ and $\rm\,p\,$ have same parity as $\rm\,n\,$ then $\rm\,m+p\,$ has same parity as $\rm\,n+n,\,$ which is even.

Said in congruence language: $\rm\ mod\ 2\!:\ m\equiv n,\ p\equiv n\:\Rightarrow\: m+p\equiv n+n\equiv 2n\equiv 0$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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