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lgbasalloteBest ResponseYou've already chosen the best response.2
i suppose the first step is to let the integers be a and b
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I'd probably use proof by contrapositive.
 one year ago

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

JonaskBest ResponseYou've already chosen the best response.0
note thta the product of two even is even
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
...or the direct proof suggested by @klimenkov
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
no one spoil the steps yer......
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
let the asker finish what he's typing first...im just too lagged...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
but this is just direct proof and too simple and dull....how to prove by contradiction?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
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?
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
care to elaborate?
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[m*n=(2k)(2x)=2(2kx)\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
the product of two non odd integers is non odd
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
(2k1)(2k+1) = 4k^2  1
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
isn't that what i just said?
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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."
 one year ago

estudierBest ResponseYou've already chosen the best response.0
It's a slightly neater version of what Klimenkov said...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
@estudier please don;t mix your complicated proofs here......
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
im not in college...i'm a sophomore high school...
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...
 one year ago

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

lgbasalloteBest ResponseYou've already chosen the best response.2
are analytic proofs the ones i do and prefer?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Analytic is fancy calculus, real numers, continuity, blah.....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
so what am i doing?
 one year ago

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

SmoothMathBest ResponseYou've already chosen the best response.2
But my study was certainly skewed towards the analytic side.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
like those geometric proofs with all the tables?
 one year ago

estudierBest 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....
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
Ew... I totally disagree. You've made so many assumptions there.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
@estudier you love skipping steps a lot, don't you
 one year ago

estudierBest ResponseYou've already chosen the best response.0
What, like 2k1 is odd, u mean?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
like proving square of even is even
 one year ago

estudierBest ResponseYou've already chosen the best response.0
or that 4k^2 is even...
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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 2k1
 one year ago

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

lgbasalloteBest ResponseYou've already chosen the best response.2
i think "square of even is even" is what they call conjecture
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
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....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
like i said...im just in primary school...not in number theory class
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Ah well, this is all a question of what u are allowed to assume....
 one year ago

SmoothMathBest 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
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
primary school > the educational level before high school and after preschool
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
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.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
why are you doing discreet mathematics which i am doing at college
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.2
Because LGBA is a boss.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
i believe that education belongs to no level
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
lol he really is i thought you are a teacher at some school,
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I'm with Klimenkov (but we are not in primary school, I guess)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
i meant the one above estudier.......
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
what grade are you guys
 one year ago

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

JonaskBest ResponseYou've already chosen the best response.0
@klimenkov and @estudier @lgbasallote
 one year ago

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

JonaskBest ResponseYou've already chosen the best response.0
lol i was just about to say you shuld take the IMO
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
but it is still true ,you are in primary
 one year ago

klimenkovBest ResponseYou've already chosen the best response.4
I'm 20. I just like math.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
IMOinternational math olympiad
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
IMOinternational math olympiad
 one year ago

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

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

lgbasalloteBest ResponseYou've already chosen the best response.2
openstudy is not just about math
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
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
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
p=you hate math q=you are not on OS what is the contrapositive
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
not clear,i dont get it @estudier
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Just me not being funny....:)
 one year ago

estudierBest ResponseYou've already chosen the best response.0
x,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
 one year ago

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

estudierBest ResponseYou've already chosen the best response.0
p>q is not p or t so not that...
 one year ago
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