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KingGeorge

  • 4 years ago

[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 \(n\).

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  1. eigenschmeigen
    • 3 years ago
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    f'(a) > 0 f(a)f(b) = f(ab) f(2) = 2

  2. eigenschmeigen
    • 3 years ago
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    \[f(2) = f(1 \times 2) = f(1)f(2)\] hence f(1) = 1

  3. KingGeorge
    • 3 years ago
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    Be careful with your derivatives. This is not a continuous function.

  4. eigenschmeigen
    • 3 years ago
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    ah no it isnt

  5. eigenschmeigen
    • 3 years ago
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    i should have said x ≥ y, then f(x) ≥ f(y)

  6. eigenschmeigen
    • 3 years ago
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    i was thinking induction?

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

  8. eigenschmeigen
    • 3 years ago
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    f(2^n) = 2^n doesn't particularly help

  9. eigenschmeigen
    • 3 years ago
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    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
    • 3 years ago
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    thats it i think

  11. KingGeorge
    • 3 years ago
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    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
    • 3 years ago
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    ah

  13. eigenschmeigen
    • 3 years ago
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    i did not know that

  14. KingGeorge
    • 3 years ago
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    If it were strictly multiplicative we could say that, but unfortunately, it's not given that it's strictly multiplicative.

  15. eigenschmeigen
    • 3 years ago
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    not having any luck. i tried using the fact that consecutive integers are coprime

  16. KingGeorge
    • 3 years ago
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    let me know if you want a hint.

  17. eigenschmeigen
    • 3 years ago
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    i think i'll sleep on it, im too tired to do maths now lol

  18. eliassaab
    • 3 years ago
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    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
    • 3 years ago
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    That's clever. Significantly more simple than the solution I had.

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