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anonymous

  • one year ago

using induction, prove that for all integers n > 0, n3 + 2n is divisible by 3. please help.

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  1. anonymous
    • one year ago
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    first of all maybe we can see why it is true without induction

  2. anonymous
    • one year ago
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    or just do it by induction either way have you ever done a proof by induction?

  3. anonymous
    • one year ago
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    i'm supposed to solve it using induction. it's obvious that it's true. even if i plug in 0 it works. but, i don't understand understand how to set this up. i know i need a base case. would that be n=0?

  4. anonymous
    • one year ago
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    no, base case is \(n=1\)

  5. anonymous
    • one year ago
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    in which case you get 3, so done with the base case

  6. anonymous
    • one year ago
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    next you assume it is true if \(n=k\) in other words you say "we assume \(k^3+2k\) is divisible by \(3\)" and using that, prove it is true if \(n=k+1\)

  7. anonymous
    • one year ago
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    most of it now is algebra from here on in replace \(n\) by \(k+1\) and separate out a part of the result that looks like \(k^3+2k\) and the rest will be obviously divisible by \(3\)

  8. anonymous
    • one year ago
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    one sec let me try to absorb what you said. thank you for helping me. i will reply in a min with questions

  9. anonymous
    • one year ago
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    take your time, do the algebra

  10. anonymous
    • one year ago
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    what is the significance of substituting n for k?

  11. anonymous
    • one year ago
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    you substitute \(k\) for \(n\) to say what you get to assume is true

  12. anonymous
    • one year ago
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    not \(n\) for \(k\)

  13. anonymous
    • one year ago
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    ok, so by saying k we are just saying we assume it's true. this signifies induction?

  14. anonymous
    • one year ago
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    and if k+1 is true then the proof is complete?

  15. anonymous
    • one year ago
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    yes that is the "induction hypothesis" assume it is true if \(n=k\) then you use that to prove it is true if \(n=k+1\)

  16. anonymous
    • one year ago
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    some people cheat and assume it is true for \(n\) then show it is true for \(n+1\) but really we should change the variable

  17. anonymous
    • one year ago
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    are you working on the algebra?

  18. anonymous
    • one year ago
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    i am trying to work on the algebra. this is what i have to solve: (k+1)^3+2(k+1)=3

  19. anonymous
    • one year ago
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    oh no not \(=3\) no one says it is equal to 3

  20. anonymous
    • one year ago
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    just take \[(k+1)^2+2(k+1)\] and expand it, then take out the part that looks like \(k^3+2k\) and separate it from the rest

  21. anonymous
    • one year ago
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    oops i meant \[(k+1)^3+2(k+1)\]

  22. anonymous
    • one year ago
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    go ahead and expand it, let me know what you get

  23. anonymous
    • one year ago
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    still doing the algebra?

  24. anonymous
    • one year ago
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    i am.. i feel retarded. thanks again for helping :\ i am trying to figure out what else i do after (k+1)((k+1)^2)+2(k+1)

  25. anonymous
    • one year ago
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    multiply it all out there is no way around it

  26. anonymous
    • one year ago
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    do you know how to cube \(k+1\) easily?

  27. anonymous
    • one year ago
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    in general \[(a+b)^3=a^2+3a^2b+3ab^2+b^2\] in your case \(a=k,, b=1\)

  28. anonymous
    • one year ago
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    and of course \(2(k+1)=2k+2\)

  29. anonymous
    • one year ago
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    okay. i have to do (k+1)(k+1)(k+1) and multiply all of it.

  30. anonymous
    • one year ago
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    yes, but i showed you how to do it in one step

  31. anonymous
    • one year ago
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    yeah i got it. sorry about that

  32. anonymous
    • one year ago
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    take your time, let me know what you get when you do all this algebra then we will be done in two steps after

  33. anonymous
    • one year ago
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    what did you mean by separate it from the rest? it's all multiplied out. k^2+3k^2(1)+3k(1)^2+1^2 + 2k+1 but there is no right side

  34. anonymous
    • one year ago
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    ok you should have \[k^3+3k^2+3k+1+2k+2\]

  35. anonymous
    • one year ago
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    you forgot to distribute with \(2(k+1)\) i think

  36. anonymous
    • one year ago
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    and also you wrote \(1^2\) but i am sure you didn't mean to

  37. anonymous
    • one year ago
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    now only two steps separate out the \(k^3+2k\) part, that we get to assume is divisible by 3

  38. anonymous
    • one year ago
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    you have \[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\]

  39. anonymous
    • one year ago
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    the blue part is divisible by 3 "by induction" we got to assume that the red part is divisible by 3 because... well because it is you can factor a 3 out of it if you want to be formal

  40. anonymous
    • one year ago
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    if the red wasn't divisible would it not be a proof?

  41. anonymous
    • one year ago
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    yeah you need that part to be divisible by 3 the blue part is the induction hypotheses the red part is what you have to show is divisible by 3, but there is not much to show since each term has a 3 in it

  42. anonymous
    • one year ago
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    \[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\] \[\overbrace{\color{blue}{k^3+2k}}^{\text{ div by 3 by induction}}+\overbrace{\color{red}{3(k^2+k+1)}}^{\text{ has a factor of 3}}\]

  43. anonymous
    • one year ago
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    and now we are done hope this helped clarify how to do a proof by induction they all pretty much work this way

  44. anonymous
    • one year ago
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    i can't believe i found someone to walk me through that

  45. anonymous
    • one year ago
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    yeah it is usually a thankless task because most people have no idea about what it means so you have to explain every step of the way, and they get frustrated really it is something you have to see done a few times to understand the logic glad it helped

  46. anonymous
    • one year ago
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    if i can find out how to donate to you i will :)

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