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KingGeorge

[SOLVED] George's problem of the [insert arbitrary time unit] Define a function \(f: \mathbb{Z}^+ \longrightarrow \mathbb{Z}^+\) such that \(f\) is strictly increasing, is multiplicative, and \(f(2)=2\). Show that \(f(n) =n\) for all \(n\). Hint 1: You need to find an upper and a lower bound for a certain \(n\) and show that the bounds are the same. Hint 2: Find the upper and lower bound for \(n=18\). Using this show that \(f(3)=3\). Now deduce that \(f(n)=n\) for all \(n\). [EDIT: It should be noted that this problem is relatively difficult (but only if you don't see the right process)]

  • one year ago
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  1. KingGeorge
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    Strictly increasing means that if \(a>b\), then \(f(a)>f(b)\). Multiplicative means that if \(a, b\) are coprime, then \(f(ab)=f(a)f(b)\).

    • one year ago
  2. beginnersmind
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    Do you need help with this or is it more of a puzzle? (I managed to prove that f(3)=3)

    • one year ago
  3. KingGeorge
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    More of a puzzle. I know how to get the answer. BTW, how did you prove f(3)=3?

    • one year ago
  4. beginnersmind
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    2=f(2)<f(3)<f(4)=f(2)*f(2)=4

    • one year ago
  5. KingGeorge
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    2 and 2 are not coprime, so that doesn't work.

    • one year ago
  6. beginnersmind
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    oh

    • one year ago
  7. nbouscal
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    Any ring automorphism on Z that takes 2 to itself will by definition take every n to itself? (Just a guess, I'm not that far along yet).

    • one year ago
  8. KingGeorge
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    That's certainly not the solution I have.

    • one year ago
  9. nbouscal
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    Well it's kind of part of the definition of Z, isn't it, that what you describe in the problem would be the case? I mean, it should be a pretty straightforward proof by induction, I would think. But I might be misunderstanding the question.

    • one year ago
  10. KingGeorge
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    I'm using \(\mathbb{Z}^+\) which is equivalent to \(\mathbb{N}\) with out 0. I'm just repeating the problem the way it was given to me.

    • one year ago
  11. nbouscal
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    Hey! You can answer a question I have! I just discovered this site today. How do you do inline TeX? And yeah, I follow what ring you're referring to, I just don't see what makes the problem interesting, I guess.

    • one year ago
  12. beginnersmind
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    I need some help, I know almost no number theory. To define a multiplicative function I need to define it on all powers of all primes, right?

    • one year ago
  13. nbouscal
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    Sorry, not a ring at all, just a set. Even so, it seems like a pretty simple induction proof to show that f(n)=n if f(2)=2. Still could be missing something though.

    • one year ago
  14. KingGeorge
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    Use "\ (" instead of "\ [" (without the spaces to do inline \(\LaTeX\). Defining it on all powers of primes would certainly be helpful. Also, this is not just a simple induction proof.

    • one year ago
  15. KingGeorge
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    I will return to answer questions after dinner.

    • one year ago
  16. nbouscal
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    Awesome. Most sites I use just use the $ notation, haven't seen it with parentheses before. Also, yes, I see where the problem is with this, will think about it for some time and try to work it out.

    • one year ago
  17. nbouscal
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    1 is coprime to every integer, so \(f(n)=n\) by definition.

    • one year ago
  18. joemath314159
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    That just means:\[f(n\cdot 1)=f(n)f(1)\]first you need to show f(1)=1, which can be done like this:\[f(2)=f(2\cdot 1)=f(2)f(1)\Longrightarrow 2=2f(1)\Longrightarrow f(1)=1\]

    • one year ago
  19. beginnersmind
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    I think that only proves f(n)=f(n)*f(1)=f(n)

    • one year ago
  20. nbouscal
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    Right, what Joe said.

    • one year ago
  21. beginnersmind
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    Anyway, I'm stuck, I'll read KG's solution later.

    • one year ago
  22. nbouscal
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    Or alternatively, the way that I thought about it, if f is strictly increasing, the only element left in Z+ that is less than 2 is 1, so f(1) has to be 1

    • one year ago
  23. nbouscal
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    I spent a good ten minutes thinking about rebuilding the series of primes, should have remembered, always start with the identity :P

    • one year ago
  24. beginnersmind
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    George Polya said, if you can't solve the problem, try to solve an easier one that's similar. Any suggestions?

    • one year ago
  25. beginnersmind
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    Looking only at the set of primes and numbers that are the product of n distinct primes what constraint do we get for f, if we want to keep it strictly increasing?

    • one year ago
  26. nbouscal
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    I tried building it up from primes that way and hit a dead end, which is when I went back to basics and realized you could just force the definition via the identity. Building it by primes seems a bit more difficult, and I'm not positive how you would go about it.

    • one year ago
  27. beginnersmind
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    "and realized you could just force the definition via the identity." What do you mean by that?

    • one year ago
  28. experimentX
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    f(2*1) = f(2)f(1) f(1) = 1 f(3*2) = f(3) 2 f(3) = f(6)/2 f(5) = f(10)/2 f(n) = f(2n)/2 for odd n

    • one year ago
  29. KingGeorge
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    @experimentX That is correct, but it still doesn't solve the original problem.

    • one year ago
  30. KingGeorge
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    I have posted a hint for those who had previously participated.

    • one year ago
  31. Ishaan94
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    \[f(2) = f(1\cdot2) = f(1)\cdot f(2) = 2 \implies f(1) = 1\] Lets assume \(f(k-1) = k-1\) and same follows up from 2. \[f(k) = f\left(\frac k 2 \cdot 2\right) = f\left(\frac k2\right)\cdot f(2)\] \[1 < \frac{k}2< k-1\] Is this enough lol? :/ :(

    • one year ago
  32. KingGeorge
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    If you knew that \(f(n)=n\) for at least one value of \(n\geq3\) that would work (with a little bit of fixing). But first you need to show that \(f(n)=n\) for some value of \(n\) that's not 1 or 2.

    • one year ago
  33. Ishaan94
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    2=f(2)<f(3)<f(4)=f(2)*f(2)=4 what's wrong with this one?

    • one year ago
  34. KingGeorge
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    2 is not coprime to 2. So \(f(2)f(2)\) does not necessarily equal \(f(4)\)

    • one year ago
  35. experimentX
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    f(2) = f(1+1) = f(1) + f(1) ???

    • one year ago
  36. Ishaan94
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    I don't understand it. Can you give me an example which proves f(2)f(2) isn't necessarily f(4). Sorry If I sound dumb.

    • one year ago
  37. KingGeorge
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    @experimentX Not necessarily. This is not necessarily a homomorphism. @Ishaan94 Suppose \(f(3) =10\) and \(f(6)=f(2)f(3)=20\). \(f(4)\) could be anywhere in between. The function is given as multiplicative (see first post for definition), so if the two arguments don't have a gcd of 1, you can't say anything about the product.

    • one year ago
  38. KingGeorge
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    @experimentX It would probably be simpler to say that the function is just multiplicative. Not additive, and what you posted was the definition of additive.

    • one year ago
  39. Ishaan94
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    \[f(3)>f(2) \implies f(3) \ge f(2) + 1 \implies f(3) \ge 3\]\[f(4) \ge 4 \implies f(n) \ge n \]Can we assume \(f(n)=n+k\)? but k may not be necessarily a constant term right?

    • one year ago
  40. Ishaan94
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    But this doesn't help either :/

    • one year ago
  41. KingGeorge
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    That's the right direction. If you follow that direction correctly, you'll be able to get the lower bound you need. To get the upper bound you need, you'll need to change things a little bit first.

    • one year ago
  42. KingGeorge
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    I'll post a second, more helpful hint in half an hour or so if you think you need it.

    • one year ago
  43. Ishaan94
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    \[f(3) =3 +k, \quad f(6) =f(3)f(2) = 6+2k, \quad f(5)< 6 +2k, \quad f(4)> 3+k \]\[f(5) \le 5 + 2k, \quad f(4) \ge 4+ k\]\[f(5) \ge f(4) + 1 \implies f(4)+1\le f(5)\le5+2k \] If I prove \(f(4) +1 = 5+2k\) then this might work out, but \(f(4) +1 \ge 5+k\). It's just more and more and more inequalities. :/

    • one year ago
  44. Ishaan94
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    From the above inequalities, \[5+ k\le f(5) \le 5+2k\]

    • one year ago
  45. KingGeorge
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    Have posted 2nd hint. I would also suggest letting \(f(3)=k\). It makes the manipulations easier

    • one year ago
  46. Ishaan94
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    wait... \(f(4) + 1 \le 5+ 2k \implies f(4) \le 4 +2k \implies f(3) <(4) \le 4+2k\) \(\implies 4 + k \le f(4) \le 4 + 2k\) \[\implies n + k \le f(n) \le n + 2k\] woow, nice. 4 + 2k< 5 + k k< 1 Woow, Wow. K<1. Yay! I am so happy! K can not be less than Zero as f(n) is supposed to be an increasing function. So, K has to be Zero. Yay! Yayyy! But I am not sure if it's right. :/

    • one year ago
  47. KingGeorge
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    I don't think that's quite right. It's almost correct, but in your last expression, 4+2k>5+k, I'm pretty sure the second k is not the same k.

    • one year ago
  48. KingGeorge
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    If you could convince me that the k's are in fact the same, you would indeed have a proof.

    • one year ago
  49. Ishaan94
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    No, it's same. Really?\[f(3) >2 \implies f(3) \ge f(2) + 1 \implies f(3) \ge 3\]\[\implies f(n) \ge n\] Let \(f(3) = 3 + k\). \[f(6) = f(2) \cdot f(3) = 6 + 2k\] \[f(5) < f(6) \implies f(5) < 6+ 2k \implies f(5) \le 5 + 2k \tag1\] \[f(4) < f(5) \implies f(4) + 1 \le f(5) \tag2\] From1 and 2.\[f(4) +1 \le f(5) \le 5+2k \tag3\] \[f(3) < f(4) \implies f(3) + 1 \le f(4) \implies 4 + k \le f(4) \tag 4\] From 4 and 3 \[5 + k \le f(5) \le 5 + 2k \tag5\]Also from 3\[f(4) + 1\le 5+2k \implies f(4) \le 4 + 2k\tag6\]From 4 and 6 \[4+k \le f(4) \le 4+ 2k\] \[\implies n+k \le f(n) \le n +2k\tag7 \] From 5 and 6 \[4+2k < 5+k \implies k<1 \implies k=0\tag8\]From 7 and 8 \[n\le f(n)\le n \implies f(n) = n\]

    • one year ago
  50. Ishaan94
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    Can I type QED now? :D

    • one year ago
  51. KingGeorge
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    I think I see the actual problem now. You're trying to compare an upper bound for 4, and a lower bound for 5. Your claim is that those bounds can't overlap, but I don't see anything that shows those bounds can't overlap. It is true of course, that the best bounds don't overlap, but these bounds are not as small as possible. If \(f(4)=4+2k\), then of course \(4+2k<f(5)<6+2k\). And we're done. But what if \(f(4)<4+2k\)? That is still possible and we no longer get the same implication.

    • one year ago
  52. KingGeorge
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    I must sleep now. I will return to see any corrections you have made.

    • one year ago
  53. Ishaan94
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    Ohh hmm :( I will try again.

    • one year ago
  54. Ishaan94
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    Hmm but I still do have the bounds hehehe \[n+k \le f(n) \le n+2k\] Wait... for \(n=2\),\[2+k \le f(2) \le 2+2k \implies 2+ k \le 2\le 2+2k \implies k=0\]

    • one year ago
  55. Ishaan94
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    How about this?

    • one year ago
  56. Ishaan94
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    Shall I type QED now? :D

    • one year ago
  57. KingGeorge
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    You have those bounds for \(n \geq3\) so you can't use it for \(n=2\). For \(n=2\), you're given that it's 2. You're method is great, you just need to refine it to get an upper and lower bound for a single number and show that the bounds are the same.

    • one year ago
  58. Ishaan94
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    Oh I knew something was wrong. hmm I will try again. (Y) :D

    • one year ago
  59. KingGeorge
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    It's very helpful to set \(f(3)=a\). From there, get a lower and upper bound for \(f(18)\). If you do it just right, you'll get a quadratic that you can solve for a.

    • one year ago
  60. Ishaan94
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    I can't take \(f(18)=f(3)f(6)\). Can I? I think \(f(2)\cdot f(9)\) is right. Hmm \(2f(9)\).

    • one year ago
  61. KingGeorge
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    That's the right track. Now what's the upper bound of\(f(9)\) in terms of \(f(3)\)?

    • one year ago
  62. KingGeorge
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    Hint: \[f(9)\leq f(10)-1\]

    • one year ago
  63. Ishaan94
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    4k-3 or 4a -3?

    • one year ago
  64. KingGeorge
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    Right. So this means that \(f(18) \leq \text{____}\)

    • one year ago
  65. Ishaan94
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    8k - 6 \(\ge\) f(18) How do I solve it for f(17)?

    • one year ago
  66. KingGeorge
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    Ignore f(17). You found the upper bound of f(18). Now we need to find the lower bound for f(18). Once again, you want to start at f(3)=k.

    • one year ago
  67. KingGeorge
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    In this part, you'll want to get a quadratic. You'll want to look at f(3) and f(5) and use these to get to f(18).

    • one year ago
  68. Ishaan94
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    I can only get f(15) using f(3) and f(5)

    • one year ago
  69. KingGeorge
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    And from f(15) you can get f(18)

    • one year ago
  70. Ishaan94
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    \[2k + k^2 \le f(15) \le 2k^2 -k\]

    • one year ago
  71. KingGeorge
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    Just use \[2k + k^2 \le f(15)\]Now this means that \[\text{______}\le f(18)\]

    • one year ago
  72. Ishaan94
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    \((k+1)^2 + 2 \le f(18) \le 8k+6\)

    • one year ago
  73. KingGeorge
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    \[(k+1)^2 + 2 \le 8k+6\]Now just solve for k

    • one year ago
  74. Ishaan94
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    \((k+1)^2 + 2 \le 8k + 6 \implies k^2 + 2k+3 \le 8k+6 \implies k^2 -6k - 3\le0\) \[ \implies 3 -2\sqrt3 \le k \le 3 + 2\sqrt3 \] But is this what I am supposed to get?

    • one year ago
  75. KingGeorge
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    There was a typo that was making this rather hard. It should be \[(k+1)^2 + 2 \le f(18) \le 8k-6\]You had this right originally, there was just a typo that was perpetuated. Try and solve this one.

    • one year ago
  76. Ishaan94
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    Ohh I can't concentrate at all :( I see now, \[3\le k \le 3 \implies k=3 \implies f(3) =3\]

    • one year ago
  77. Ishaan94
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    So, we finally have f(3)=3. But I don't get it why did we have to go all the way from 4 and 5 to 18. Why a quadratic?\[\]

    • one year ago
  78. KingGeorge
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    We can get the relation \[k^2-6k+9 \le 0\implies (k-3)^2 \le 0\]There is only one value of \(k\) that solves this, \(k=3\). That's why we need a quadratic. It allows us to get an inequality with only 1 solution. If it were linear, we would have a line with infinite solutions satisfying the inequality.

    • one year ago
  79. Ishaan94
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    eh the proof still isn't complete I will have to show it for f(n). I think I will have to use induction. \[f(3) =3\] Lets assume it works up from 3 to 2k-1. \[f(k) = 2\cdot f\left(k\right) = 2k\] I am not sure if it's right. I can only recall strong induction from your previous problem.

    • one year ago
  80. KingGeorge
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    To show it for all n, just take \(f(2)f(3)=f(6)=2*3=6\) So \(f(4)=4\) and \(f(5)=5\) since it has to be strictly increasing. We can just continue this process up to infinity.

    • one year ago
  81. Ishaan94
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    and why wasn't f(1) enough for us to use induction? why did we need to solve it for f(3).

    • one year ago
  82. KingGeorge
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    \(f(1)\) just let us get \(f(2)\). We needed a value greater than 2 to generate larger numbers using the multiplicative rule.

    • one year ago
  83. Ishaan94
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    ohh, thanks. i couldn't have done this without your help.

    • one year ago
  84. KingGeorge
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    You did very well. =D

    • one year ago
  85. KingGeorge
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    I appreciate the amount of time you spent on this. Thanks for actually doing this.

    • one year ago
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