## KingGeorge Group Title [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)] 2 years ago 2 years ago

1. KingGeorge Group Title

Strictly increasing means that if $$a>b$$, then $$f(a)>f(b)$$. Multiplicative means that if $$a, b$$ are coprime, then $$f(ab)=f(a)f(b)$$.

2. beginnersmind Group Title

Do you need help with this or is it more of a puzzle? (I managed to prove that f(3)=3)

3. KingGeorge Group Title

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

4. beginnersmind Group Title

2=f(2)<f(3)<f(4)=f(2)*f(2)=4

5. KingGeorge Group Title

2 and 2 are not coprime, so that doesn't work.

6. beginnersmind Group Title

oh

7. nbouscal Group Title

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).

8. KingGeorge Group Title

That's certainly not the solution I have.

9. nbouscal Group Title

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.

10. KingGeorge Group Title

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.

11. nbouscal Group Title

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.

12. beginnersmind Group Title

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?

13. nbouscal Group Title

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.

14. KingGeorge Group Title

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.

15. KingGeorge Group Title

16. nbouscal Group Title

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.

17. nbouscal Group Title

1 is coprime to every integer, so $$f(n)=n$$ by definition.

18. joemath314159 Group Title

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:$f(2)=f(2\cdot 1)=f(2)f(1)\Longrightarrow 2=2f(1)\Longrightarrow f(1)=1$

19. beginnersmind Group Title

I think that only proves f(n)=f(n)*f(1)=f(n)

20. nbouscal Group Title

Right, what Joe said.

21. beginnersmind Group Title

Anyway, I'm stuck, I'll read KG's solution later.

22. nbouscal Group Title

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

23. nbouscal Group Title

I spent a good ten minutes thinking about rebuilding the series of primes, should have remembered, always start with the identity :P

24. beginnersmind Group Title

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

25. beginnersmind Group Title

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?

26. nbouscal Group Title

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.

27. beginnersmind Group Title

"and realized you could just force the definition via the identity." What do you mean by that?

28. experimentX Group Title

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

29. KingGeorge Group Title

@experimentX That is correct, but it still doesn't solve the original problem.

30. KingGeorge Group Title

I have posted a hint for those who had previously participated.

31. Ishaan94 Group Title

$f(2) = f(1\cdot2) = f(1)\cdot f(2) = 2 \implies f(1) = 1$ 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? :/ :(

32. KingGeorge Group Title

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.

33. Ishaan94 Group Title

2=f(2)<f(3)<f(4)=f(2)*f(2)=4 what's wrong with this one?

34. KingGeorge Group Title

2 is not coprime to 2. So $$f(2)f(2)$$ does not necessarily equal $$f(4)$$

35. experimentX Group Title

f(2) = f(1+1) = f(1) + f(1) ???

36. Ishaan94 Group Title

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.

37. KingGeorge Group Title

@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.

38. KingGeorge Group Title

@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.

39. Ishaan94 Group Title

$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?

40. Ishaan94 Group Title

But this doesn't help either :/

41. KingGeorge Group Title

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.

42. KingGeorge Group Title

I'll post a second, more helpful hint in half an hour or so if you think you need it.

43. Ishaan94 Group Title

$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. :/

44. Ishaan94 Group Title

From the above inequalities, $5+ k\le f(5) \le 5+2k$

45. KingGeorge Group Title

Have posted 2nd hint. I would also suggest letting $$f(3)=k$$. It makes the manipulations easier

46. Ishaan94 Group Title

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. :/

47. KingGeorge Group Title

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.

48. KingGeorge Group Title

If you could convince me that the k's are in fact the same, you would indeed have a proof.

49. Ishaan94 Group Title

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 $n\le f(n)\le n \implies f(n) = n$

50. Ishaan94 Group Title

Can I type QED now? :D

51. KingGeorge Group Title

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.

52. KingGeorge Group Title

53. Ishaan94 Group Title

Ohh hmm :( I will try again.

54. Ishaan94 Group Title

Hmm but I still do have the bounds hehehe $n+k \le f(n) \le n+2k$ Wait... for $$n=2$$,$2+k \le f(2) \le 2+2k \implies 2+ k \le 2\le 2+2k \implies k=0$

55. Ishaan94 Group Title

56. Ishaan94 Group Title

Shall I type QED now? :D

57. KingGeorge Group Title

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.

58. Ishaan94 Group Title

Oh I knew something was wrong. hmm I will try again. (Y) :D

59. KingGeorge Group Title

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.

60. Ishaan94 Group Title

I can't take $$f(18)=f(3)f(6)$$. Can I? I think $$f(2)\cdot f(9)$$ is right. Hmm $$2f(9)$$.

61. KingGeorge Group Title

That's the right track. Now what's the upper bound of$$f(9)$$ in terms of $$f(3)$$?

62. KingGeorge Group Title

Hint: $f(9)\leq f(10)-1$

63. Ishaan94 Group Title

4k-3 or 4a -3?

64. KingGeorge Group Title

Right. So this means that $$f(18) \leq \text{____}$$

65. Ishaan94 Group Title

8k - 6 $$\ge$$ f(18) How do I solve it for f(17)?

66. KingGeorge Group Title

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.

67. KingGeorge Group Title

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).

68. Ishaan94 Group Title

I can only get f(15) using f(3) and f(5)

69. KingGeorge Group Title

And from f(15) you can get f(18)

70. Ishaan94 Group Title

$2k + k^2 \le f(15) \le 2k^2 -k$

71. KingGeorge Group Title

Just use $2k + k^2 \le f(15)$Now this means that $\text{______}\le f(18)$

72. Ishaan94 Group Title

$$(k+1)^2 + 2 \le f(18) \le 8k+6$$

73. KingGeorge Group Title

$(k+1)^2 + 2 \le 8k+6$Now just solve for k

74. Ishaan94 Group Title

$$(k+1)^2 + 2 \le 8k + 6 \implies k^2 + 2k+3 \le 8k+6 \implies k^2 -6k - 3\le0$$ $\implies 3 -2\sqrt3 \le k \le 3 + 2\sqrt3$ But is this what I am supposed to get?

75. KingGeorge Group Title

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.

76. Ishaan94 Group Title

Ohh I can't concentrate at all :( I see now, $3\le k \le 3 \implies k=3 \implies f(3) =3$

77. Ishaan94 Group Title

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?

78. KingGeorge Group Title

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.

79. Ishaan94 Group Title

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.

80. KingGeorge Group Title

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.

81. Ishaan94 Group Title

and why wasn't f(1) enough for us to use induction? why did we need to solve it for f(3).

82. KingGeorge Group Title

$$f(1)$$ just let us get $$f(2)$$. We needed a value greater than 2 to generate larger numbers using the multiplicative rule.

83. Ishaan94 Group Title

ohh, thanks. i couldn't have done this without your help.

84. KingGeorge Group Title

You did very well. =D

85. KingGeorge Group Title

I appreciate the amount of time you spent on this. Thanks for actually doing this.