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March 28th, 2018, 05:33 AM   #1
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Function proof

If $\displaystyle f(n+1)> f(f(n))$
Show that $\displaystyle f(n)=n$
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March 28th, 2018, 06:13 AM   #2
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Quote:
Originally Posted by idontknow View Post
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
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March 28th, 2018, 07:00 AM   #3
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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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