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i suppose the first step is to let the integers be a and b
I'd probably use proof by contrapositive.
Let the first one be \(2n+1\), the second - \(2k+1\). Multiply -\((2n+1)(2k+1)=2(2nk+k+n)+1\) - odd.
note thta the product of two even is even
...or the direct proof suggested by @klimenkov
no one spoil the steps yer......
let the asker finish what he's typing first...im just too lagged...
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
but this is just direct proof and too simple and dull....how to prove by contradiction?
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?
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.
care to elaborate?
the product of two non odd integers is non odd
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.
(2k-1)(2k+1) = 4k^2 - 1
isn't that what i just said?
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."
It's a slightly neater version of what Klimenkov said...
@estudier please don;t mix your complicated proofs here......
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.
im not in college...i'm a sophomore high school...
Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...
Haha I know, buddy. And it's incredible that you can write proofs like this already. You're a champion.
are analytic proofs the ones i do and prefer?
Analytic is fancy calculus, real numers, continuity, blah.....
so what am i doing?
Estudier, I was never allowed to do any purely algebraic proofs. I was required formal paragraph proofs pretty much 100% of the time.
But my study was certainly skewed towards the analytic side.
like those geometric proofs with all the tables?
(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....
CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS
Ew... I totally disagree. You've made so many assumptions there.
@estudier you love skipping steps a lot, don't you
What, like 2k-1 is odd, u mean?
like proving square of even is even
or that 4k^2 is even...
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
it's the principle you used for (2k)^2 - 1 after all
i think "square of even is even" is what they call conjecture
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.
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....
like i said...im just in primary school...not in number theory class
Ah well, this is all a question of what u are allowed to assume....
... 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
WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY
primary school -> the educational level before high school and after preschool
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.
are you in primary
product is odd*
why are you doing discreet mathematics which i am doing at college
Because LGBA is a boss.
i believe that education belongs to no level
lol he really is i thought you are a teacher at some school,
I'm with Klimenkov (but we are not in primary school, I guess)
i meant the one above estudier.......
what grade are you guys
I'm a teacher, if that's what you were asking. I teach algebra 2 and geometry.
@klimenkov and @estudier @lgbasallote
i'm actually a liberal arts teacher in primary school. i was just kidding about that primary school thing
lol i was just about to say you shuld take the IMO
but it is still true ,you are in primary
I'm 20. I just like math.
IMO-international math olympiad
IMO-international math olympiad
no thanks....as you can see from my picture...i don't really like math....
lol you guys are funny,if you hated maths you wont be on openstudy is that a true proposition
openstudy is not just about math
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
p=you hate math q=you are not on OS what is the contrapositive
por q or p xor q or?
not clear,i dont get it @estudier
Just me not being funny....:)
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
ohh thats true but i was talking abut \[¬(p→q) ≡ p ⋀ (¬q)\]
p->q is not p or t so not that...
q, I mean, not t