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lgbasallote

  • 2 years ago

Prove that the product of two odd integers is odd

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  1. lgbasallote
    • 2 years ago
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    i suppose the first step is to let the integers be a and b

  2. ParthKohli
    • 2 years ago
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    I'd probably use proof by contrapositive.

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

  4. Jonask
    • 2 years ago
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    note thta the product of two even is even

  5. Jonask
    • 2 years ago
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    contrad

  6. ParthKohli
    • 2 years ago
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    ...or the direct proof suggested by @klimenkov

  7. lgbasallote
    • 2 years ago
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    no one spoil the steps yer......

  8. lgbasallote
    • 2 years ago
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    let the asker finish what he's typing first...im just too lagged...

  9. lgbasallote
    • 2 years ago
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    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

  10. lgbasallote
    • 2 years ago
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    but this is just direct proof and too simple and dull....how to prove by contradiction?

  11. lgbasallote
    • 2 years ago
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    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?

  12. SmoothMath
    • 2 years ago
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    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.

  13. lgbasallote
    • 2 years ago
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    care to elaborate?

  14. Jonask
    • 2 years ago
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    \[m*n=(2k)(2x)=2(2kx)\]

  15. Jonask
    • 2 years ago
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    the product of two non odd integers is non odd

  16. SmoothMath
    • 2 years ago
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    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.

  17. estudier
    • 2 years ago
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    (2k-1)(2k+1) = 4k^2 - 1

  18. lgbasallote
    • 2 years ago
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    isn't that what i just said?

  19. SmoothMath
    • 2 years ago
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    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."

  20. estudier
    • 2 years ago
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    It's a slightly neater version of what Klimenkov said...

  21. lgbasallote
    • 2 years ago
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    @estudier please don;t mix your complicated proofs here......

  22. estudier
    • 2 years ago
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    :-)

  23. SmoothMath
    • 2 years ago
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    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.

  24. lgbasallote
    • 2 years ago
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    im not in college...i'm a sophomore high school...

  25. estudier
    • 2 years ago
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    Number theory proofs are allowed to be a bit more casual than , say , analytic proofs...

  26. SmoothMath
    • 2 years ago
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    Haha I know, buddy. And it's incredible that you can write proofs like this already. You're a champion.

  27. lgbasallote
    • 2 years ago
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    are analytic proofs the ones i do and prefer?

  28. estudier
    • 2 years ago
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    Analytic is fancy calculus, real numers, continuity, blah.....

  29. lgbasallote
    • 2 years ago
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    so what am i doing?

  30. SmoothMath
    • 2 years ago
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    Estudier, I was never allowed to do any purely algebraic proofs. I was required formal paragraph proofs pretty much 100% of the time.

  31. SmoothMath
    • 2 years ago
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    But my study was certainly skewed towards the analytic side.

  32. lgbasallote
    • 2 years ago
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    like those geometric proofs with all the tables?

  33. estudier
    • 2 years ago
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    (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....

  34. Jonask
    • 2 years ago
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    CALCULUS LIMIT,AND CONT. PROOFS ARE TYPICAL ANALYTICAL PROOFS

  35. SmoothMath
    • 2 years ago
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    Ew... I totally disagree. You've made so many assumptions there.

  36. lgbasallote
    • 2 years ago
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    @estudier you love skipping steps a lot, don't you

  37. estudier
    • 2 years ago
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    What, like 2k-1 is odd, u mean?

  38. lgbasallote
    • 2 years ago
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    like proving square of even is even

  39. estudier
    • 2 years ago
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    or that 4k^2 is even...

  40. SmoothMath
    • 2 years ago
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    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

  41. lgbasallote
    • 2 years ago
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    it's the principle you used for (2k)^2 - 1 after all

  42. lgbasallote
    • 2 years ago
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    i think "square of even is even" is what they call conjecture

  43. SmoothMath
    • 2 years ago
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    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.

  44. estudier
    • 2 years ago
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    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....

  45. lgbasallote
    • 2 years ago
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    like i said...im just in primary school...not in number theory class

  46. estudier
    • 2 years ago
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    Ah well, this is all a question of what u are allowed to assume....

  47. SmoothMath
    • 2 years ago
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    ... 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

  48. Jonask
    • 2 years ago
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    WHAT DO LITERALLY MEAN YOU ARE IN PRIMARY

  49. lgbasallote
    • 2 years ago
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    primary school -> the educational level before high school and after preschool

  50. SmoothMath
    • 2 years ago
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    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.

  51. Jonask
    • 2 years ago
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    are you in primary

  52. SmoothMath
    • 2 years ago
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    product is odd*

  53. lgbasallote
    • 2 years ago
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    yes

  54. Jonask
    • 2 years ago
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    why are you doing discreet mathematics which i am doing at college

  55. SmoothMath
    • 2 years ago
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    Because LGBA is a boss.

  56. lgbasallote
    • 2 years ago
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    i believe that education belongs to no level

  57. Jonask
    • 2 years ago
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    lol he really is i thought you are a teacher at some school,

  58. estudier
    • 2 years ago
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    I'm with Klimenkov (but we are not in primary school, I guess)

  59. lgbasallote
    • 2 years ago
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    ^he's smart

  60. lgbasallote
    • 2 years ago
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    i meant the one above estudier.......

  61. Jonask
    • 2 years ago
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    what grade are you guys

  62. SmoothMath
    • 2 years ago
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    I'm a teacher, if that's what you were asking. I teach algebra 2 and geometry.

  63. Jonask
    • 2 years ago
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    @klimenkov and @estudier @lgbasallote

  64. Jonask
    • 2 years ago
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    prodigees

  65. lgbasallote
    • 2 years ago
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    i'm actually a liberal arts teacher in primary school. i was just kidding about that primary school thing

  66. Jonask
    • 2 years ago
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    lol i was just about to say you shuld take the IMO

  67. lgbasallote
    • 2 years ago
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    IMO?

  68. Jonask
    • 2 years ago
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    but it is still true ,you are in primary

  69. klimenkov
    • 2 years ago
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    I'm 20. I just like math.

  70. lgbasallote
    • 2 years ago
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    somehow yes

  71. Jonask
    • 2 years ago
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    IMO-international math olympiad

  72. Jonask
    • 2 years ago
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    IMO-international math olympiad

  73. lgbasallote
    • 2 years ago
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    no thanks....as you can see from my picture...i don't really like math....

  74. Jonask
    • 2 years ago
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    lol you guys are funny,if you hated maths you wont be on openstudy is that a true proposition

  75. lgbasallote
    • 2 years ago
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    openstudy is not just about math

  76. SmoothMath
    • 2 years ago
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    Nay.

  77. lgbasallote
    • 2 years ago
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    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

  78. Jonask
    • 2 years ago
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    p=you hate math q=you are not on OS what is the contrapositive

  79. estudier
    • 2 years ago
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    por q or p xor q or?

  80. Jonask
    • 2 years ago
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    not clear,i dont get it @estudier

  81. estudier
    • 2 years ago
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    Just me not being funny....:)

  82. Jonask
    • 2 years ago
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    \[¬(p→q) \]expand

  83. estudier
    • 2 years ago
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    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

  84. Jonask
    • 2 years ago
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    ohh thats true but i was talking abut \[¬(p→q) ≡ p ⋀ (¬q)\]

  85. estudier
    • 2 years ago
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    p->q is not p or t so not that...

  86. estudier
    • 2 years ago
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    q, I mean, not t

  87. Jonask
    • 2 years ago
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    yes

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