Prove that the product of two odd integers is odd

- lgbasallote

Prove that the product of two odd integers is odd

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- lgbasallote

i suppose the first step is to let the integers be a and b

- ParthKohli

I'd probably use proof by contrapositive.

- klimenkov

Let the first one be \(2n+1\), the second - \(2k+1\). Multiply -\((2n+1)(2k+1)=2(2nk+k+n)+1\) - odd.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

note thta the product of two even is even

- anonymous

contrad

- ParthKohli

...or the direct proof suggested by @klimenkov

- lgbasallote

no one spoil the steps yer......

- lgbasallote

let the asker finish what he's typing first...im just too lagged...

- lgbasallote

anyway...
so i let
m = 2k + 1
n = 2x + 1
now... solving for the product
mn = (2k + 1)(2x + 1)
mn = 4kx + 2k + 2x + 1
then...
mn = 2(2kx + k + x) + 1
so it's odd

- lgbasallote

but this is just direct proof and too simple and dull....how to prove by contradiction?

- lgbasallote

i suppose i do
mn = 2y
where m and n are odd
(2k + 1)(2x+1) = 2y
4kx + 2k + 2x + 1 = 2y
then...
4kx + 2k + 2x - 2y = -1
2(2kx + k + x - y) = -1
then...
2kx + k + x - y = -1/2
since k, x and y are integers, the right side should also be an integer. thus, contradiction.
is that how it's done?

- anonymous

That's how it's done =)
The only thing to point out is that this is not a formal proof. To formalize it, you would need to clearly state all of your assumptions, plus give a reasoning for all of your steps. Then at the end, you would have to explicitly state what contradiction you have uncovered.

- lgbasallote

care to elaborate?

- anonymous

\[m*n=(2k)(2x)=2(2kx)\]

- anonymous

the product of two non odd integers is non odd

- anonymous

In this case, the contradiction is that k,x, and y are all integers. The product of integers is an integer, therefore 2kx is an integer. The sum or difference of integers is an integer, therefore
2kx + k + x -y is an integer.
You conclude that
2kx + k +x -y = -1/2
So that is a contradiction.

- anonymous

(2k-1)(2k+1) = 4k^2 - 1

- lgbasallote

isn't that what i just said?

- anonymous

And the part where I say to state your assumptions, it would simply be what you have there, except you would say
"Assume that m and n are odd numbers such that their product is even. M is odd, therefore m = 2k+1 for some integer k. N is odd, therefore n = 2x+1 for some integer x. Their product is even, therefore mn = 2y for some integer y."

- anonymous

It's a slightly neater version of what Klimenkov said...

- lgbasallote

@estudier please don;t mix your complicated proofs here......

- anonymous

:-)

- anonymous

LGBA, I'm only nitpicking in the way a college professor would nitpick. Your proof is great, it's just not formal or rigorous the way it is required to be in a college course.

- lgbasallote

im not in college...i'm a sophomore high school...

- anonymous

Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...

- anonymous

Haha I know, buddy. And it's incredible that you can write proofs like this already. You're a champion.

- lgbasallote

are analytic proofs the ones i do and prefer?

- anonymous

Analytic is fancy calculus, real numers, continuity, blah.....

- lgbasallote

so what am i doing?

- anonymous

Estudier, I was never allowed to do any purely algebraic proofs. I was required formal paragraph proofs pretty much 100% of the time.

- anonymous

But my study was certainly skewed towards the analytic side.

- lgbasallote

like those geometric proofs with all the tables?

- anonymous

(2k-1)(2k+1) = 4k^2 - 1 QED is an acceptable proof.
You can dress it up with some words and put in a paragraph if u like....

- anonymous

CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS

- anonymous

Ew...
I totally disagree. You've made so many assumptions there.

- lgbasallote

@estudier you love skipping steps a lot, don't you

- anonymous

What, like 2k-1 is odd, u mean?

- lgbasallote

like proving square of even is even

- anonymous

or that 4k^2 is even...

- anonymous

For one thing, you'd like for it to be true with any two odd integers. Your proof assumes that the integers are two apart. 2k+1 and 2k-1

- lgbasallote

it's the principle you used for (2k)^2 - 1 after all

- lgbasallote

i think "square of even is even" is what they call conjecture

- anonymous

And I mean... call me old fashioned, but I really hate that you're not stating your assumptions at all. You have 1 line. Gawd that gives me the jibblies in a real bad way.

- anonymous

In number theory class, I can assure u that will not be required to prove that odd numbers are of the form 2k plus/minus 1 or that odd numbers are 2 apart....

- lgbasallote

like i said...im just in primary school...not in number theory class

- anonymous

Ah well, this is all a question of what u are allowed to assume....

- anonymous

...
Estudier, you've proved, for example, that the product of 5 and 7 is even, because you could choose k=6.
However, you haven't proved that the product of 5 and 9 is even, because you can not choose a k such that (2k+1)(2k-1) = 5*9

- anonymous

WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY

- lgbasallote

primary school -> the educational level before high school and after preschool

- anonymous

Do you see my point, Estudier? The task is to prove that the product of ANY two odd integers is even. You've only proven that the product of consecutive odd integers is even.

- anonymous

are you in primary

- anonymous

product is odd*

- lgbasallote

yes

- anonymous

why are you doing discreet mathematics which i am doing at college

- anonymous

Because LGBA is a boss.

- lgbasallote

i believe that education belongs to no level

- anonymous

lol he really is i thought you are a teacher at some school,

- anonymous

I'm with Klimenkov (but we are not in primary school, I guess)

- lgbasallote

^he's smart

- lgbasallote

i meant the one above estudier.......

- anonymous

what grade are you guys

- anonymous

I'm a teacher, if that's what you were asking. I teach algebra 2 and geometry.

- anonymous

@klimenkov and @estudier @lgbasallote

- anonymous

prodigees

- lgbasallote

i'm actually a liberal arts teacher in primary school. i was just kidding about that primary school thing

- anonymous

lol i was just about to say you shuld take the IMO

- lgbasallote

IMO?

- anonymous

but it is still true ,you are in primary

- klimenkov

I'm 20. I just like math.

- lgbasallote

somehow yes

- anonymous

IMO-international math olympiad

- anonymous

IMO-international math olympiad

- lgbasallote

no thanks....as you can see from my picture...i don't really like math....

- anonymous

lol you guys are funny,if you hated maths you wont be on openstudy
is that a true proposition

- lgbasallote

openstudy is not just about math

- anonymous

Nay.

- lgbasallote

p: i like math
q: i'm on openstudy
if p then q
since p is F and q is T..the implication holds true

- anonymous

p=you hate math
q=you are not on OS
what is the contrapositive

- anonymous

por q or p xor q or?

- anonymous

not clear,i dont get it @estudier

- anonymous

Just me not being funny....:)

- anonymous

\[¬(p→q) \]expand

- anonymous

x,y odd -> x-1, y-1 even (I don't have to prove that as well, do I?)
x-1 = 2p, y-1 =2q -> x = 2p+1, y =2q+1
(2p+1)(2q+1) = 4pq +2p +2q +1 QED

- anonymous

ohh thats true but i was talking abut
\[¬(p→q) ≡ p ⋀ (¬q)\]

- anonymous

p->q is not p or t so not that...

- anonymous

q, I mean, not t

- anonymous

yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.