lgbasallote
  • lgbasallote
Prove that the product of two odd integers is odd
Mathematics
chestercat
  • chestercat
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lgbasallote
  • lgbasallote
i suppose the first step is to let the integers be a and b
ParthKohli
  • ParthKohli
I'd probably use proof by contrapositive.
klimenkov
  • klimenkov
Let the first one be \(2n+1\), the second - \(2k+1\). Multiply -\((2n+1)(2k+1)=2(2nk+k+n)+1\) - odd.

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anonymous
  • anonymous
note thta the product of two even is even
anonymous
  • anonymous
contrad
ParthKohli
  • ParthKohli
...or the direct proof suggested by @klimenkov
lgbasallote
  • lgbasallote
no one spoil the steps yer......
lgbasallote
  • lgbasallote
let the asker finish what he's typing first...im just too lagged...
lgbasallote
  • 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
  • lgbasallote
but this is just direct proof and too simple and dull....how to prove by contradiction?
lgbasallote
  • 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
  • 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
  • lgbasallote
care to elaborate?
anonymous
  • anonymous
\[m*n=(2k)(2x)=2(2kx)\]
anonymous
  • anonymous
the product of two non odd integers is non odd
anonymous
  • 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
  • anonymous
(2k-1)(2k+1) = 4k^2 - 1
lgbasallote
  • lgbasallote
isn't that what i just said?
anonymous
  • 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
  • anonymous
It's a slightly neater version of what Klimenkov said...
lgbasallote
  • lgbasallote
@estudier please don;t mix your complicated proofs here......
anonymous
  • anonymous
:-)
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
  • lgbasallote
im not in college...i'm a sophomore high school...
anonymous
  • anonymous
Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...
anonymous
  • anonymous
Haha I know, buddy. And it's incredible that you can write proofs like this already. You're a champion.
lgbasallote
  • lgbasallote
are analytic proofs the ones i do and prefer?
anonymous
  • anonymous
Analytic is fancy calculus, real numers, continuity, blah.....
lgbasallote
  • lgbasallote
so what am i doing?
anonymous
  • 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
  • anonymous
But my study was certainly skewed towards the analytic side.
lgbasallote
  • lgbasallote
like those geometric proofs with all the tables?
anonymous
  • 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
  • anonymous
CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS
anonymous
  • anonymous
Ew... I totally disagree. You've made so many assumptions there.
lgbasallote
  • lgbasallote
@estudier you love skipping steps a lot, don't you
anonymous
  • anonymous
What, like 2k-1 is odd, u mean?
lgbasallote
  • lgbasallote
like proving square of even is even
anonymous
  • anonymous
or that 4k^2 is even...
anonymous
  • 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
  • lgbasallote
it's the principle you used for (2k)^2 - 1 after all
lgbasallote
  • lgbasallote
i think "square of even is even" is what they call conjecture
anonymous
  • 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
  • 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
  • lgbasallote
like i said...im just in primary school...not in number theory class
anonymous
  • anonymous
Ah well, this is all a question of what u are allowed to assume....
anonymous
  • 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
  • anonymous
WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY
lgbasallote
  • lgbasallote
primary school -> the educational level before high school and after preschool
anonymous
  • 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
  • anonymous
are you in primary
anonymous
  • anonymous
product is odd*
lgbasallote
  • lgbasallote
yes
anonymous
  • anonymous
why are you doing discreet mathematics which i am doing at college
anonymous
  • anonymous
Because LGBA is a boss.
lgbasallote
  • lgbasallote
i believe that education belongs to no level
anonymous
  • anonymous
lol he really is i thought you are a teacher at some school,
anonymous
  • anonymous
I'm with Klimenkov (but we are not in primary school, I guess)
lgbasallote
  • lgbasallote
^he's smart
lgbasallote
  • lgbasallote
i meant the one above estudier.......
anonymous
  • anonymous
what grade are you guys
anonymous
  • anonymous
I'm a teacher, if that's what you were asking. I teach algebra 2 and geometry.
anonymous
  • anonymous
@klimenkov and @estudier @lgbasallote
anonymous
  • anonymous
prodigees
lgbasallote
  • lgbasallote
i'm actually a liberal arts teacher in primary school. i was just kidding about that primary school thing
anonymous
  • anonymous
lol i was just about to say you shuld take the IMO
lgbasallote
  • lgbasallote
IMO?
anonymous
  • anonymous
but it is still true ,you are in primary
klimenkov
  • klimenkov
I'm 20. I just like math.
lgbasallote
  • lgbasallote
somehow yes
anonymous
  • anonymous
IMO-international math olympiad
anonymous
  • anonymous
IMO-international math olympiad
lgbasallote
  • lgbasallote
no thanks....as you can see from my picture...i don't really like math....
anonymous
  • anonymous
lol you guys are funny,if you hated maths you wont be on openstudy is that a true proposition
lgbasallote
  • lgbasallote
openstudy is not just about math
anonymous
  • anonymous
Nay.
lgbasallote
  • 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
  • anonymous
p=you hate math q=you are not on OS what is the contrapositive
anonymous
  • anonymous
por q or p xor q or?
anonymous
  • anonymous
not clear,i dont get it @estudier
anonymous
  • anonymous
Just me not being funny....:)
anonymous
  • anonymous
\[¬(p→q) \]expand
anonymous
  • 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
  • anonymous
ohh thats true but i was talking abut \[¬(p→q) ≡ p ⋀ (¬q)\]
anonymous
  • anonymous
p->q is not p or t so not that...
anonymous
  • anonymous
q, I mean, not t
anonymous
  • anonymous
yes

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