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

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

klimenkov Group TitleBest 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.
 2 years ago

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

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

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

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

lgbasallote Group TitleBest 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
 2 years ago

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

lgbasallote Group TitleBest 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?
 2 years ago

SmoothMath Group TitleBest 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.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
care to elaborate?
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[m*n=(2k)(2x)=2(2kx)\]
 2 years ago

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

SmoothMath Group TitleBest 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.
 2 years ago

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

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

SmoothMath Group TitleBest 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."
 2 years ago

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

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

SmoothMath Group TitleBest 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.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
im not in college...i'm a sophomore high school...
 2 years ago

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

SmoothMath Group TitleBest 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.
 2 years ago

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

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

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

SmoothMath Group TitleBest 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.
 2 years ago

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

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

estudier Group TitleBest 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....
 2 years ago

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

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

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

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

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

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

SmoothMath Group TitleBest 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
 2 years ago

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

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

SmoothMath Group TitleBest 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.
 2 years ago

estudier Group TitleBest 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....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
like i said...im just in primary school...not in number theory class
 2 years ago

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

SmoothMath Group TitleBest 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
 2 years ago

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

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

SmoothMath Group TitleBest 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.
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
are you in primary
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.2
product is odd*
 2 years ago

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

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

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

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

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

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
^he's smart
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
what grade are you guys
 2 years ago

SmoothMath Group TitleBest 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.
 2 years ago

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

lgbasallote Group TitleBest 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
 2 years ago

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

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

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

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
somehow yes
 2 years ago

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

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

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

Jonask Group TitleBest 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
 2 years ago

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

lgbasallote Group TitleBest 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
 2 years ago

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

estudier Group TitleBest ResponseYou've already chosen the best response.0
por q or p xor q or?
 2 years ago

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

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[¬(p→q) \]expand
 2 years ago

estudier Group TitleBest 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
 2 years ago

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

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

estudier Group TitleBest ResponseYou've already chosen the best response.0
q, I mean, not t
 2 years ago
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