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 March 28th, 2018, 05:33 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 224 Thanks: 26 Function proof If $\displaystyle f(n+1)> f(f(n))$ Show that $\displaystyle f(n)=n$
March 28th, 2018, 06:13 AM   #2
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Quote:
 Originally Posted by idontknow If $\displaystyle f(n+1)> f(f(n))$ Show that $\displaystyle f(n)=n$
Counterexample:
Let f(n) = n + 0.1.

Then
f(f(n)) = f(n + 0.1) = (n + 0.1) + 0.1 = n + 0.2

f(n + 1) = (n + 1) + 0.1 = n + 1.1

Thus f(n + 1) > f(f(n)) but f(n) is not equal to n.

-Dan

 March 28th, 2018, 07:00 AM #3 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 187 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry I'm assuming this is meant to be a function $\mathbb{N} \to \mathbb{N}$. Have you tried anything so far? With most maths problems, and this sort of problem especially, there's very little to gain from seeing someone else's solution without having a proper go at it yourself. I'll give one hint for now. With these problems, it often helps to think about certain special values of the function. For example, you could try firstly to prove that $f(0) = 0$ (or, if $0$ isn't an element of $\mathbb{N}$ for you, that $f(1) = 1$).

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