Question

Prove that if m and n are integers and mn is even, then m is even or n is even. Use the contrapositive (indirect proof).

Answer 1

Have you found the contrapositive?

Answer 2

No

Answer 3

We are trying to write the contrapositive of:
For all integers m and n, if mn is even, then m is even or n is even.

Instead of writing if p then q we need to write if not q then not p.

So this means we to negate m is even or n is even
We also need to negate mn is even

---
hint for negating that or statement:
not(a or b)=not a and not b

Answer 4

n is odd or m is odd

mn is odd?

Answer 5

look at the hint for negating an or statement
you need to change or to and

Answer 6

yeah I just saw that so n is odd and m is odd?

Answer 7

So you need to prove:
If n and m are odd integers, then mn is odd.

Answer 8

n = 2x and m = (2y +1)
n + m = (2x + (2y +1)) = 2(x + (y + 1))

Answer 9

since n is odd then you need to represent n as a odd integer not an even integer
so n=2x+1 and m=2y+1

Answer 10

and also you are asked to find nm not n+m

Answer 11

umm im thinking

Answer 12

nm means n times m

Answer 13

n=2x+1 and m=2y+1

nm = (2x +1)(2x +1)

Answer 14

ok then multiply the (2x+1) and (2y+1)
I assume you meant nm=(2x+1)(2y+1)

Answer 15

I get 4xy +2x +2y + 1. Not sure if that is odd

Answer 16

see if you can put an the odd looking form which is 2*some integer+1
or 2k+1
where k is some integer

Answer 17

look at those first three terms you wrote

Answer 18

all of them have a factor 2 in common

Answer 19

which terms? I am sorry I am lost

Answer 20

4xy,2x,2y

Answer 21

those first three terms of 4xy+2x+2y+1

Answer 22

they all have a 2 in common (the first terms do)

Answer 23

three terms*

Answer 24

so how do you use 2k + 1?

Answer 25

you want to write 4xy+2x+2y+1 in the form that is 2k+1

Answer 26

I am not sure how to do that?

Answer 27

Do you know how to factor?

Answer 28

There is a factor 2 in the first three terms

Answer 29

2(2x^2 + y) + 1?

Answer 30

4xy+2x+2y+1
2*2xy+2*x+2*y+1
2(2xy+x+y)+1
Isn't this in the form 2k+1
2xy+x+y is an integer?

Answer 31

Not sure how you got the square

Answer 32

I am sorry Im a little lost

Answer 33

4xy,2x,and 2y all have a common factor 2
so I factored a 2 from all three them in the expression
4xy+2x+2y+1
2(2xy+x+y)+1

Answer 34

ok so that is in form of 2k +1?

Answer 35

k = 2xy + x + y?

Answer 36

\[4xy+2x+2y+1 \\ 2 \cdot 2xy+ 2 \cdot x +2 \cdot y +1 \\ 2(2xy+x+y)+1\]

Do you think it is in the form
2k+1?

Answer 37

Is 2xy+x+y an integer?

Answer 38

Is the set of integers closed under addition and multiplication?

Answer 39

x,y are integers

Answer 40

When you mutliply x and y are multiply them with any other integer you will get an integer and when you add them with any other integer you will get an integer
2xy+x+y is an integer
2(2xy+x+y)+1
is in the form
2k+1

Answer 41

Therefore mn is odd

Answer 42

Oh ok so you plug (2xy + x + y) into k?

Answer 43

You don't have to plug into k
You are just trying to show you can write mn in the form that is 2k+1
We did that by writing mn as 2(2xy+x+y)+1

Answer 44

oh I see it. Thank you very much!

Answer 45

np

Answer 46

Suppose neither of them are even, that means both m, n are odd, and mn is even
so that m, n can be written as
m = 2k +1
n = 2s +1 for some k, s in Z
mn = (2k+1)(2s +1) = 4ks +2k+2s +1= 2(2ks +k +s) +1,
let 2ks +k +s = l, then mn = 2l +1 form the form of an odd number (contradict)
--> It is not the case that both them are odd or one of them must be even