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Loser66

  • one year ago

Please, explain me.

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  1. Loser66
    • one year ago
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    I understand 3, but 2, how??

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  2. ParthKohli
    • one year ago
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    Hey, you can probably use induction for proving II.

  3. Loser66
    • one year ago
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    The given information doesn't include \(n* (x \oplus y)\), hence we can't define the product of it.

  4. ParthKohli
    • one year ago
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    OK, one method to do this problem is to recognize the operators as \(+\) and \(\cdot\). Here is how you use induction. It holds true for \(n = 1\). Let \(k(x\oplus y) =kx \oplus ky\). We have to prove that \((k+1)(x \oplus y) = (k+1) x \oplus (k+1)y\). Now we can use the recursive definition. \((k+1)(x\oplus y) = k (x\oplus y) \oplus (x\oplus y) \)

  5. Loser66
    • one year ago
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    Look at 1) we have if \(x^k\) defined, then ..... but to \((x\oplus y)\), we don't have any thing to define \((x\odot y)^k\), hence 1) is not the correct one. And if it is so, the same logic for 2) which indicates to \(\oplus\)

  6. Loser66
    • one year ago
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    We don't have \(k(x\oplus y) = kx\oplus ky\)

  7. ParthKohli
    • one year ago
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    But that's the inductive step!

  8. Loser66
    • one year ago
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    But first off, the operation must be defined, right?

  9. ParthKohli
    • one year ago
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    It has been defined recursively...

  10. ParthKohli
    • one year ago
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    \[(k+1)x = kx \oplus x\]

  11. ParthKohli
    • one year ago
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    Hey, you could observe that the operators are + and * though, right?

  12. Loser66
    • one year ago
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    ok, you can say if \(k (x\oplus y)\), then \((k+1) (x\oplus y)= k(x\oplus y)+(x\oplus y)\)

  13. ParthKohli
    • one year ago
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    That \(+\) should be \(\oplus\). Now \(k(x\oplus y)=kx \oplus ky\).

  14. Loser66
    • one year ago
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    hihihi... I don't see anything allows me to distribute like that.

  15. ParthKohli
    • one year ago
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    That is the inductive step, man.

  16. ParthKohli
    • one year ago
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    We have assumed that \(k(x\oplus y) = kx \oplus ky\). We have to show that the same holds for \(k+1\)

  17. Loser66
    • one year ago
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    Ok, I temporarily "pretend" that I agree with you on that step, then??

  18. ParthKohli
    • one year ago
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    \[kx \oplus ky \oplus x \oplus y\]Now collect the terms and you get \((k+1)x \oplus (k+1)y\) which was to be shown.

  19. ParthKohli
    • one year ago
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    Anyway, I gotta go. So you see how this induction works, right?

  20. Loser66
    • one year ago
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    I am sorry for being dummy. I don't get it still. I rearrange what you said \(Assume\) \(n(x\oplus y ) = nx \oplus ny\) now prove \((n+1) (x\oplus y) = (n+1) x \oplus (n+1) y\) Given condition gives us \((n+1) (x\oplus y) = n(x\oplus y) + (x\oplus y)\\nx\oplus ny \color {red}{+} x\oplus y \)

  21. Loser66
    • one year ago
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    I don't know what given condition allows me pick like terms to add up. :(

  22. ParthKohli
    • one year ago
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    Hey, that is the definition of \(\cdot\).\[(k+1)x = kx\oplus x\]given in your question

  23. ParthKohli
    • one year ago
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    That \(\color{red}+\) sign should be \(\oplus\).

  24. ParthKohli
    • one year ago
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    OK, let's start from scratch. Sorry for being a little casual the last time.

  25. Loser66
    • one year ago
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    Do steps please.

  26. Loser66
    • one year ago
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    Question: is it a Set theory problem??

  27. ParthKohli
    • one year ago
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    OK, so we have to prove \(n\cdot (x\oplus y)\) for all positive integers \(n\). As a matter of fact, this holds for all real numbers (which I'm not going to cover), so proving it for positive integers is going to be simper than that. What tells us to use induction is that the definition of \(\cdot\) is given recursively in terms of \(k\) and \(k+1\), so we're going to have to prove in such a way that we only have to work with \(k\) and \(k+1\). That method of proof is called induction. Let \(P(n)\) be a given statement that you have to prove for all positive integers \(n\). We first prove that \(P(1)\) (known as the base-case) is true. Now comes the "inductive step". We first assume that \(P(k)\), where \(k\) is an arbitrary positive integer is true. We have to show that \(P(k+1)\) is true assuming \(P(k)\). This has a "domino-effect".

  28. Loser66
    • one year ago
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    I know induction.

  29. zzr0ck3r
    • one year ago
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    me too, hes a jerk sometimes

  30. zzr0ck3r
    • one year ago
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    I am proctoring a test, so I will look at this problem and come back, not sure if you solved it yet because it looks like fin

  31. Loser66
    • one year ago
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    @zzr0ck3r who is your "he" ??

  32. zzr0ck3r
    • one year ago
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    fun

  33. zzr0ck3r
    • one year ago
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    induction is my he

  34. Loser66
    • one year ago
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    ok, :) good luck (for your final)

  35. zzr0ck3r
    • one year ago
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    I am giving it, not taking it :)

  36. ParthKohli
    • one year ago
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    Here is the given statement: \(P(n): n\cdot(x\oplus y) = n x \oplus ny\). Vital assumptions: 1. commutativity, associativity and closure under \(\oplus\). 2. \((k+1)x = kx \oplus x\). Let us first show \(P(1)\), which is trivial. Now we assume that \(P(k): k\cdot(x\oplus y)= kx \oplus ky\) is true for \(k\). Let us use this to prove \(P(k+1)\).\[P(k+1): (k+1)\cdot (x\oplus y) = (k+1)\cdot x \oplus (k+1) \cdot y\]Now\[(k+1)\cdot (x\oplus y)\]\[=k\cdot (x \oplus y) \oplus (x\oplus y)\tag{definition of dot}\]\[= kx \oplus ky \oplus x \oplus y \tag{inductive assumption}\]\[= kx \oplus x \oplus ky \oplus y \tag{commutativity}\]\[= (k+1)\cdot x \oplus (k+1)\cdot y \tag{using definition of dot}\]

  37. ParthKohli
    • one year ago
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    We've thus proven \(P(k+1)\).

  38. ParthKohli
    • one year ago
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    Try proving all those properties for rational numbers as an exercise.

  39. Loser66
    • one year ago
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    the line " definition of dot" !!!

  40. Loser66
    • one year ago
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    oh, got it. hihihih

  41. Loser66
    • one year ago
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    Thanks a ton.............

  42. ParthKohli
    • one year ago
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    \[(k+1) \cdot x = k\cdot x \oplus x\]\[(k+1)\cdot \heartsuit = k \cdot \heartsuit \oplus \heartsuit\]\[(k+1)\cdot \spadesuit = k \cdot \spadesuit \oplus\spadesuit \]

  43. ParthKohli
    • one year ago
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    Oh - never mind then...

  44. Loser66
    • one year ago
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    Finally, you can crack my skull to put the logic in. Thanks for being patient to me.

  45. ParthKohli
    • one year ago
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    Anyway, if you ever receive these questions on a multiple-choice test, spend a minute thinking if these operators are related to any that we already know. That makes things easier.

  46. Loser66
    • one year ago
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    The problem is: I have to have the answer in about 2 minutes. I need tricks to figure out right after seeing the problem. To this problem, option 3 is trivial, but option 2 made no sense to me. Whatever, I got it now. oh, yeah... life is beautiful now. lalalalala...

  47. Loser66
    • one year ago
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    Again, thank you so much.:)

  48. ParthKohli
    • one year ago
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    Which is why it is useful to know that these operators are just addition and multiplication. lol :P No problem!

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