Thread: Direct Proof

If x is an odd integer and y is an odd integer then xy is an odd integer. Use direct proof to prove the following.

I am stumped on where to go from here. All I know is that the definition of an odd integer is n=2k+1 where n is the integer.

So I assume that I need x=2n+1 and y=2m+1 but I really don't know where to go from here. This seems really simple but where to start is stubborn to me.

Hello,

Originally Posted by premierplayer

If x is an odd integer and y is an odd integer then xy is an odd integer. Use direct proof to prove the following.

I am stumped on where to go from here. All I know is that the definition of an odd integer is n=2k+1 where n is the integer.

So I assume that I need x=2n+1 and y=2m+1 but I really don't know where to go from here. This seems really simple but where to start is stubborn to me.

Then xy=(2n+1)(2m+1)=4mn+...
expand and see if it can be written in the form 2k+1, for some integer k

Originally Posted by Moo

Hello,

Then xy=(2n+1)(2m+1)=4mn+...
expand and see if it can be written in the form 2k+1, for some integer k

xy=(2n+1)(2m+1)=4mn+2 because of distributive property?

Originally Posted by premierplayer

xy=(2n+1)(2m+1)=4mn+2 because of distributive property?

Is it university level ?

read this : FOIL rule - Wikipedia, the free encyclopedia

Yeah it is, its just I haven't taken a math class in like 5 years. Go me.

ok so I foiled the following...

4nm+2n+2m+1

I am now sure where to go from here tho.

ok so ive been working on this..

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

xy= (2n+1)(2m)+1

xy= 4nm +2n +2m +1 using FOIL

2(nm+n+m) +1 using factoring by 2

n & m are real numbers so...

nm + n + m = k (real number) using addition/multiplication of real numbers

2k+1 is the same as the definition of odd number...

is this right? Can anyone add to this.

Yes, that works. It is a bit long.
Note that an even number plus one is odd.