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lgbasallote
 2 years ago
Prove that the product of two odd integers is odd
lgbasallote
 2 years ago
Prove that the product of two odd integers is odd

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lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2i suppose the first step is to let the integers be a and b

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I'd probably use proof by contrapositive.

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.4Let the first one be \(2n+1\), the second  \(2k+1\). Multiply \((2n+1)(2k+1)=2(2nk+k+n)+1\)  odd.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0note thta the product of two even is even

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0...or the direct proof suggested by @klimenkov

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2no one spoil the steps yer......

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2let the asker finish what he's typing first...im just too lagged...

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2anyway... 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
 2 years ago
Best ResponseYou've already chosen the best response.2but this is just direct proof and too simple and dull....how to prove by contradiction?

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2i 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?

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2That'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.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0the product of two non odd integers is non odd

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2In 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.

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0(2k1)(2k+1) = 4k^2  1

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2isn't that what i just said?

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2And 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."

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0It's a slightly neater version of what Klimenkov said...

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2@estudier please don;t mix your complicated proofs here......

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2LGBA, 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
 2 years ago
Best ResponseYou've already chosen the best response.2im not in college...i'm a sophomore high school...

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2Haha I know, buddy. And it's incredible that you can write proofs like this already. You're a champion.

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2are analytic proofs the ones i do and prefer?

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Analytic is fancy calculus, real numers, continuity, blah.....

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2so what am i doing?

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2Estudier, I was never allowed to do any purely algebraic proofs. I was required formal paragraph proofs pretty much 100% of the time.

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2But my study was certainly skewed towards the analytic side.

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2like those geometric proofs with all the tables?

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0(2k1)(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....

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2Ew... I totally disagree. You've made so many assumptions there.

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2@estudier you love skipping steps a lot, don't you

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0What, like 2k1 is odd, u mean?

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2like proving square of even is even

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0or that 4k^2 is even...

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2For 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 2k1

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2it's the principle you used for (2k)^2  1 after all

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2i think "square of even is even" is what they call conjecture

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2And 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.

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0In 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
 2 years ago
Best ResponseYou've already chosen the best response.2like i said...im just in primary school...not in number theory class

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Ah well, this is all a question of what u are allowed to assume....

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2... 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)(2k1) = 5*9

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2primary school > the educational level before high school and after preschool

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2Do 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.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0why are you doing discreet mathematics which i am doing at college

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2Because LGBA is a boss.

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2i believe that education belongs to no level

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0lol he really is i thought you are a teacher at some school,

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0I'm with Klimenkov (but we are not in primary school, I guess)

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2i meant the one above estudier.......

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.2I'm a teacher, if that's what you were asking. I teach algebra 2 and geometry.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0@klimenkov and @estudier @lgbasallote

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2i'm actually a liberal arts teacher in primary school. i was just kidding about that primary school thing

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0lol i was just about to say you shuld take the IMO

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0but it is still true ,you are in primary

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.4I'm 20. I just like math.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0IMOinternational math olympiad

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0IMOinternational math olympiad

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2no thanks....as you can see from my picture...i don't really like math....

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0lol you guys are funny,if you hated maths you wont be on openstudy is that a true proposition

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2openstudy is not just about math

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.2p: i like math q: i'm on openstudy if p then q since p is F and q is T..the implication holds true

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0p=you hate math q=you are not on OS what is the contrapositive

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0not clear,i dont get it @estudier

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Just me not being funny....:)

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0x,y odd > x1, y1 even (I don't have to prove that as well, do I?) x1 = 2p, y1 =2q > x = 2p+1, y =2q+1 (2p+1)(2q+1) = 4pq +2p +2q +1 QED

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0ohh thats true but i was talking abut \[¬(p→q) ≡ p ⋀ (¬q)\]

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0p>q is not p or t so not that...
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