KingGeorge
  • KingGeorge
[SOLVED by @eliassaab] Define a function \(f: \mathbb{Z}^+ \longrightarrow \mathbb{Z}^+\) such that \(f\) is strictly increasing, \(f\) is multiplicative, and \(f(2)=2\). Show that \(f(n)=n\) for all \(n\).
Meta-math
katieb
  • katieb
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anonymous
  • anonymous
f'(a) > 0 f(a)f(b) = f(ab) f(2) = 2
anonymous
  • anonymous
\[f(2) = f(1 \times 2) = f(1)f(2)\] hence f(1) = 1
KingGeorge
  • KingGeorge
Be careful with your derivatives. This is not a continuous function.

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anonymous
  • anonymous
ah no it isnt
anonymous
  • anonymous
i should have said x ≥ y, then f(x) ≥ f(y)
anonymous
  • anonymous
i was thinking induction?
KingGeorge
  • KingGeorge
That's what I first thought as well, but for induction to work, you need to show that \(f(3)=3\).
anonymous
  • anonymous
f(2^n) = 2^n doesn't particularly help
anonymous
  • anonymous
ah it does, now we know f(4) cant we just say \[f(2) < f(3) < f(4) \] and since f(3) is an integer...
anonymous
  • anonymous
thats it i think
KingGeorge
  • KingGeorge
Multiplicative means that if \(\gcd(a, b)=1\), then \(f(a\cdot b)=f(a)\cdot f(b)\) \(\gcd(2, 2)=2\) so we can't say that \(f(2\cdot2)=f(2)\cdot f(2)\)
anonymous
  • anonymous
ah
anonymous
  • anonymous
i did not know that
KingGeorge
  • KingGeorge
If it were strictly multiplicative we could say that, but unfortunately, it's not given that it's strictly multiplicative.
anonymous
  • anonymous
not having any luck. i tried using the fact that consecutive integers are coprime
KingGeorge
  • KingGeorge
let me know if you want a hint.
anonymous
  • anonymous
i think i'll sleep on it, im too tired to do maths now lol
anonymous
  • anonymous
f(3)f(5)=f(15)
KingGeorge
  • KingGeorge
That's clever. Significantly more simple than the solution I had.

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