Question

[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 \(n\).

Answer 1

f'(a) > 0

f(a)f(b) = f(ab)

f(2) = 2

Answer 2

\[f(2) = f(1 \times 2) = f(1)f(2)\]
hence f(1) = 1

Answer 3

Be careful with your derivatives. This is not a continuous function.

Answer 4

ah no it isnt

Answer 5

i should have said

x ≥ y, then f(x) ≥ f(y)

Answer 6

i was thinking induction?

Answer 7

That's what I first thought as well, but for induction to work, you need to show that \(f(3)=3\).

Answer 8

f(2^n) = 2^n

doesn't particularly help

Answer 9

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...

Answer 10

thats it i think

Answer 11

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 \(f(2\cdot2)=f(2)\cdot f(2)\)

Answer 12

ah

Answer 13

i did not know that

Answer 14

If it were strictly multiplicative we could say that, but unfortunately, it's not given that it's strictly multiplicative.

Answer 15

not having any luck. i tried using the fact that consecutive integers are coprime

Answer 16

let me know if you want a hint.

Answer 17

i think i'll sleep on it, im too tired to do maths now lol

Answer 18

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

Answer 19

That's clever. Significantly more simple than the solution I had.