Here's the question you clicked on:
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 \(n\).
f'(a) > 0 f(a)f(b) = f(ab) f(2) = 2
\[f(2) = f(1 \times 2) = f(1)f(2)\] hence f(1) = 1
Be careful with your derivatives. This is not a continuous function.
i should have said x ≥ y, then f(x) ≥ f(y)
i was thinking induction?
That's what I first thought as well, but for induction to work, you need to show that \(f(3)=3\).
f(2^n) = 2^n doesn't particularly help
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...
thats it i think
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 \(f(2\cdot2)=f(2)\cdot f(2)\)
i did not know that
If it were strictly multiplicative we could say that, but unfortunately, it's not given that it's strictly multiplicative.
not having any luck. i tried using the fact that consecutive integers are coprime
let me know if you want a hint.
i think i'll sleep on it, im too tired to do maths now lol
f(3)f(5)=f(15)<f(18) =f(2) f(9)=2f(9)< 2 f(10) =4 f(5) Henc2 f(2)=2<f(3) < 4 f(3) =3
That's clever. Significantly more simple than the solution I had.