Question

Given m and n are in the set of integers, prove directly:

If mn=even, then m=even or n=even

Answer 1

hmm.. I would start with the thought that since mn is even -- that means there exists a number, k, such that 2 * k = mn.

Answer 2

good; or in general; mn = 2\(k_1\)

Answer 3

More accurately, k is an integer. Therefore,

k = mn/2 --> which means mn/2 is an integer.

Answer 4

An integer is easily defined as even if it can be divided by 2 evenly.
Proof:
Given all even integers are multiples of 2.
Let mn = even , mn = x.
Assume that n = 2a+1.
2x (2x+1) = 4x+1 (not even).

That was really messy, but you get the idea. It can be easier to prove through contradiction.

Answer 5

good, the direct proof they say may be a bit difficult to do; but i think it can be done :)

Answer 6

It's easy to show if m or n is even, mn/2 is even
now show if both m and n are odd, mn/2 has a remainder

Answer 7

my thought went;

\[(2k_1){2k_2+1}=2k_3\]
\[4k_1k_2+2k_1=2k_3\]
\[2(2(k_1k_2)+k_1)=2k_3\]

Answer 8

mn=2k for some integer k

mn-2k=0
n(m-2k/n)=0
either n=0 or m-2k/n=0
let's assume n doesn't equal to 0
m=2k/n

if n|2k, then m is even and so is n

Answer 9

dropped some paranthesis there, ....

Answer 10

....

Answer 11

dividing by n=0 ?

Answer 12

?

Answer 13

that is why i said assume n does not equal 0

Answer 14

i spoke in no particular order

Answer 15

its my thinking

Answer 16

I don't see how m must even. k could be odd

Answer 17

you guys are my digital scratch paper

Answer 18

\[(2k_1)(2k_2+1)=2k_3\]
\[either;\ m=2k_1, or\ n=2k_1\]

Answer 19

yes you are right phi

if we have
2(3)/2=3 we have odd m

Answer 20

assuming n is even (n=2)

Answer 21

and some integer k is 3

Answer 22

i should had said or instead of and

Answer 23

I think a direct proof would go like this
both m and n even implies mn is even
m even, n odd implies mn is even
m odd, n even ibid
m odd, n odd mn is odd

Answer 24

teacher did the last one, proof by contraposition

Answer 25

How about arguing m and n can be broken down into their prime factors, and then argue one of m or n must have 2 as a prime factor. We need an if and only if argument