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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)]
Strictly increasing means that if \(a>b\), then \(f(a)>f(b)\). Multiplicative means that if \(a, b\) are coprime, then \(f(ab)=f(a)f(b)\).
Do you need help with this or is it more of a puzzle? (I managed to prove that f(3)=3)
More of a puzzle. I know how to get the answer. BTW, how did you prove f(3)=3?
2=f(2)<f(3)<f(4)=f(2)*f(2)=4
2 and 2 are not coprime, so that doesn't work.
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).
That's certainly not the solution I have.
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.
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.
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.
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?
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.
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.
I will return to answer questions after dinner.
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.
1 is coprime to every integer, so \(f(n)=n\) by definition.
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:\[f(2)=f(2\cdot 1)=f(2)f(1)\Longrightarrow 2=2f(1)\Longrightarrow f(1)=1\]
I think that only proves f(n)=f(n)*f(1)=f(n)
Anyway, I'm stuck, I'll read KG's solution later.
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
I spent a good ten minutes thinking about rebuilding the series of primes, should have remembered, always start with the identity :P
George Polya said, if you can't solve the problem, try to solve an easier one that's similar. Any suggestions?
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?
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.
"and realized you could just force the definition via the identity." What do you mean by that?
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
@experimentX That is correct, but it still doesn't solve the original problem.
I have posted a hint for those who had previously participated.
\[f(2) = f(1\cdot2) = f(1)\cdot f(2) = 2 \implies f(1) = 1\] 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? :/ :(
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.
2=f(2)<f(3)<f(4)=f(2)*f(2)=4 what's wrong with this one?
2 is not coprime to 2. So \(f(2)f(2)\) does not necessarily equal \(f(4)\)
f(2) = f(1+1) = f(1) + f(1) ???
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.
@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.
@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.
\[f(3)>f(2) \implies f(3) \ge f(2) + 1 \implies f(3) \ge 3\]\[f(4) \ge 4 \implies f(n) \ge n \]Can we assume \(f(n)=n+k\)? but k may not be necessarily a constant term right?
But this doesn't help either :/
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.
I'll post a second, more helpful hint in half an hour or so if you think you need it.
\[f(3) =3 +k, \quad f(6) =f(3)f(2) = 6+2k, \quad f(5)< 6 +2k, \quad f(4)> 3+k \]\[f(5) \le 5 + 2k, \quad f(4) \ge 4+ k\]\[f(5) \ge f(4) + 1 \implies f(4)+1\le f(5)\le5+2k \] 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. :/
From the above inequalities, \[5+ k\le f(5) \le 5+2k\]
Have posted 2nd hint. I would also suggest letting \(f(3)=k\). It makes the manipulations easier
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. :/
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.
If you could convince me that the k's are in fact the same, you would indeed have a proof.
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 \[n\le f(n)\le n \implies f(n) = n\]
Can I type QED now? :D
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.
I must sleep now. I will return to see any corrections you have made.
Ohh hmm :( I will try again.
Hmm but I still do have the bounds hehehe \[n+k \le f(n) \le n+2k\] Wait... for \(n=2\),\[2+k \le f(2) \le 2+2k \implies 2+ k \le 2\le 2+2k \implies k=0\]
Shall I type QED now? :D
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.
Oh I knew something was wrong. hmm I will try again. (Y) :D
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.
I can't take \(f(18)=f(3)f(6)\). Can I? I think \(f(2)\cdot f(9)\) is right. Hmm \(2f(9)\).
That's the right track. Now what's the upper bound of\(f(9)\) in terms of \(f(3)\)?
Hint: \[f(9)\leq f(10)-1\]
Right. So this means that \(f(18) \leq \text{____}\)
8k - 6 \(\ge\) f(18) How do I solve it for f(17)?
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.
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).
I can only get f(15) using f(3) and f(5)
And from f(15) you can get f(18)
\[2k + k^2 \le f(15) \le 2k^2 -k\]
Just use \[2k + k^2 \le f(15)\]Now this means that \[\text{______}\le f(18)\]
\((k+1)^2 + 2 \le f(18) \le 8k+6\)
\[(k+1)^2 + 2 \le 8k+6\]Now just solve for k
\((k+1)^2 + 2 \le 8k + 6 \implies k^2 + 2k+3 \le 8k+6 \implies k^2 -6k - 3\le0\) \[ \implies 3 -2\sqrt3 \le k \le 3 + 2\sqrt3 \] But is this what I am supposed to get?
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.
Ohh I can't concentrate at all :( I see now, \[3\le k \le 3 \implies k=3 \implies f(3) =3\]
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?\[\]
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.
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.
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.
and why wasn't f(1) enough for us to use induction? why did we need to solve it for f(3).
\(f(1)\) just let us get \(f(2)\). We needed a value greater than 2 to generate larger numbers using the multiplicative rule.
ohh, thanks. i couldn't have done this without your help.
You did very well. =D
I appreciate the amount of time you spent on this. Thanks for actually doing this.