Question

Let m,n,p be any integers.

If [(m+n) and (n+p) are even integers],

then [(m+p) is even].

Answer 1

another proof :)

Answer 2

What I wrote in 10 pages or less was:

Spose m,n,p are even integers;

2m + 2n = 2(m+n), satisfies the definition of an even integer.

Since the addition of any arbitrary even integers is even, then all the results are true.

Spose m,n,p are odd integers;

2m+1 + 2n+1 = 2(m+n+1), satisfies the definition of an even integer.

Since the addition of any arbitrary odd integers is even, then all the results are true.

Spose m,n are odd integers, and p is any arbitrary integer;

2m + 2n+1 = 2(m+n)+1, makes the premise false; therefore the truth value is undetermined.

Answer 3

which i am sure is unsound

Answer 4

I think id only need the "if even then even" statement instead of trying to prove the round in a shape

Answer 5

...to prove that round is a shape . youknow; these would be funnier if i could spell them correctly

Answer 6

\[m+n=2k+1,n+p=2j+1\]
\[m+n+n+p=2k+1+2j+1=2k+2j+2\]
\[m+2n+p=2k+2j+2\]
\[m+p=2j+2k+2-3n=2(k+j+1-n)\] is even

Answer 7

man did i mess that up. it says "even" and i did it as if odd! but even is the same

Answer 8

:) now that odd is proven ... lol

Answer 9

\[m+n=2k,n+p=2j\]
\[m+n+n+p=2k+2j\]
\[m+p=2j+2k-2n\] there

Answer 10

if i don't get to work...

Answer 11

So, argument by contradiction again.

Suppose that m + n and n + p are even AND m + p is odd.

Then either m or p is odd, but not both. Suppose first that m is odd and p is even.

If m + n is even, then n is is odd. But if n is odd and p is even, then p + n is odd. But p + n is even.

Hence it must be that m is even and p is odd.

But if p is odd and p + n is even, then n is odd. But n is odd and m is even, then m + n is odd. But m + n is even.

Contradiction.

Thus if m + n and n + p are even, then m + p is even.

Answer 12

Yeah, no matter how i look at it is just seems to be too wordy.

But then a proof is only as good as those who agree on it :)

Answer 13

No. It's good if it's correct. And it is. :-) Read it!

Answer 14

i read it, and i agree with it, and i like it

Answer 15

m+n = 2k
n+p = 2j
-----------------
m+n+n+p = 2k + 2j

m+p+2n = 2(k+j)

m+p = 2(k+j) - 2n

m+p = 2(k+j-n)

Answer 16

oh .... thats satellites proof

Answer 17

That's good too.