## KingGeorge Group Title [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 $$n$$. 2 years ago 2 years ago

1. eigenschmeigen Group Title

f'(a) > 0 f(a)f(b) = f(ab) f(2) = 2

2. eigenschmeigen Group Title

$f(2) = f(1 \times 2) = f(1)f(2)$ hence f(1) = 1

3. KingGeorge Group Title

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

4. eigenschmeigen Group Title

ah no it isnt

5. eigenschmeigen Group Title

i should have said x ≥ y, then f(x) ≥ f(y)

6. eigenschmeigen Group Title

i was thinking induction?

7. KingGeorge Group Title

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

8. eigenschmeigen Group Title

f(2^n) = 2^n doesn't particularly help

9. eigenschmeigen Group Title

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...

10. eigenschmeigen Group Title

thats it i think

11. KingGeorge Group Title

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

12. eigenschmeigen Group Title

ah

13. eigenschmeigen Group Title

i did not know that

14. KingGeorge Group Title

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

15. eigenschmeigen Group Title

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

16. KingGeorge Group Title

let me know if you want a hint.

17. eigenschmeigen Group Title

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

18. eliassaab Group Title

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

19. KingGeorge Group Title

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