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 March 28th, 2018, 05:33 AM #1 Senior Member   Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 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: 311 Thanks: 109 Math Focus: Number Theory, Algebraic 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 $f(1) = 1$). Tags function, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pedr0bessa Abstract Algebra 0 July 11th, 2017 01:42 PM millychoochoo Number Theory 2 March 9th, 2014 08:21 AM page929 Number Theory 0 September 29th, 2010 09:29 AM geyikrali Real Analysis 2 July 19th, 2008 04:29 AM page929 Abstract Algebra 0 December 31st, 1969 04:00 PM

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