Question

Prove that the product of two odd integers is odd

Answer 1

i suppose the first step is to let the integers be a and b

Answer 2

I'd probably use proof by contrapositive.

Answer 3

Let the first one be \(2n+1\), the second - \(2k+1\). Multiply -\((2n+1)(2k+1)=2(2nk+k+n)+1\) - odd.

Answer 4

note thta the product of two even is even

Answer 5

contrad

Answer 6

...or the direct proof suggested by @klimenkov

Answer 7

no one spoil the steps yer......

Answer 8

let the asker finish what he's typing first...im just too lagged...

Answer 9

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

Answer 10

but this is just direct proof and too simple and dull....how to prove by contradiction?

Answer 11

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?

Answer 12

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.

Answer 13

care to elaborate?

Answer 14

\[m*n=(2k)(2x)=2(2kx)\]

Answer 15

the product of two non odd integers is non odd

Answer 16

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.

Answer 17

(2k-1)(2k+1) = 4k^2 - 1

Answer 18

isn't that what i just said?

Answer 19

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."

Answer 20

It's a slightly neater version of what Klimenkov said...

Answer 21

@estudier please don;t mix your complicated proofs here......

Answer 22

:-)

Answer 23

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.

Answer 24

im not in college...i'm a sophomore high school...

Answer 25

Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...

Answer 26

Haha I know, buddy. And it's incredible that you can write proofs like this already. You're a champion.

Answer 27

are analytic proofs the ones i do and prefer?

Answer 28

Analytic is fancy calculus, real numers, continuity, blah.....

Answer 29

so what am i doing?

Answer 30

Estudier, I was never allowed to do any purely algebraic proofs. I was required formal paragraph proofs pretty much 100% of the time.

Answer 31

But my study was certainly skewed towards the analytic side.

Answer 32

like those geometric proofs with all the tables?

Answer 33

(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....

Answer 34

CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS

Answer 35

Ew...
I totally disagree. You've made so many assumptions there.

Answer 36

@estudier you love skipping steps a lot, don't you

Answer 37

What, like 2k-1 is odd, u mean?

Answer 38

like proving square of even is even

Answer 39

or that 4k^2 is even...

Answer 40

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

Answer 41

it's the principle you used for (2k)^2 - 1 after all

Answer 42

i think "square of even is even" is what they call conjecture

Answer 43

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.

Answer 44

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....

Answer 45

like i said...im just in primary school...not in number theory class

Answer 46

Ah well, this is all a question of what u are allowed to assume....

Answer 47

...
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

Answer 48

WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY

Answer 49

primary school -> the educational level before high school and after preschool

Answer 50

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.

Answer 51

are you in primary

Answer 52

product is odd*

Answer 53

yes

Answer 54

why are you doing discreet mathematics which i am doing at college

Answer 55

Because LGBA is a boss.

Answer 56

i believe that education belongs to no level

Answer 57

lol he really is i thought you are a teacher at some school,

Answer 58

I'm with Klimenkov (but we are not in primary school, I guess)

Answer 59

^he's smart

Answer 60

i meant the one above estudier.......

Answer 61

what grade are you guys

Answer 62

I'm a teacher, if that's what you were asking. I teach algebra 2 and geometry.

Answer 63

@klimenkov and @estudier @lgbasallote

Answer 64

prodigees

Answer 65

i'm actually a liberal arts teacher in primary school. i was just kidding about that primary school thing

Answer 66

lol i was just about to say you shuld take the IMO

Answer 67

IMO?

Answer 68

but it is still true ,you are in primary

Answer 69

I'm 20. I just like math.

Answer 70

somehow yes

Answer 71

IMO-international math olympiad

Answer 72

IMO-international math olympiad

Answer 73

no thanks....as you can see from my picture...i don't really like math....

Answer 74

lol you guys are funny,if you hated maths you wont be on openstudy
is that a true proposition

Answer 75

openstudy is not just about math

Answer 76

Nay.

Answer 77

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

Answer 78

p=you hate math
q=you are not on OS
what is the contrapositive

Answer 79

por q or p xor q or?

Answer 80

not clear,i dont get it @estudier

Answer 81

Just me not being funny....:)

Answer 82

\[¬(p→q) \]expand

Answer 83

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

Answer 84

ohh thats true but i was talking abut
\[¬(p→q) ≡ p ⋀ (¬q)\]

Answer 85

p->q is not p or t so not that...

Answer 86

q, I mean, not t

Answer 87

yes