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Let f:N→N be strictly increasing function such that f(f(n))=3n then f(11)=?

Let f:N→N be strictly increasing function such that f(f(n))=3n then f(11)=?

Grade:12

2 Answers

mycroft holmes
272 Points
14 years ago

We first note that we must have f(n) >n. To prove this first its easy to see that for no n can we have f(n) = n as f([f(n)] = 3n would then yield f(n) = 3n which is a contradiction. Suppose f(n)<n, then since f is strictly increasing we have f[(f(n)] < f(n) so that 3n<f(n)<n which is again a contradiction. Thus we have established that f(n)>n for all n

 

Now, if f(1) = k>1 (since f(1) = k>1 as proved above), we have f(k) = 3 which means 3>k so that k =1 or 2. Since k>1, this gives k = 2

Hence f(2) =3 and so f(3) = f(f(2)) = 6 and similarly f(6) = 9. Now, we will attempt to plug the gap by finding f(4) and f(5). This is pretty much immediate since f(4) and f(5) must lie between f(3) = 6 and f(6) = 9 and since f(5)>f(4) we must have f(4) = 7 and f(5) = 8

 

Now f(7) = 12 and f(8) = 15, f(9) = f(f(6)) = 18

 

f(12) = f(f(7)) = 21. From a similar argument as previous, we must have f(10) = 19 and so f(11) = 20

Winner forever
26 Points
2 years ago
f(f(n))=3n
Apply f on the both sides of the equation:
f(3n)=f(f(f(n)))
This simply intend to say that replacing n by f(n) :
f(3n)=3f(n)
It follows that f(3)=3f(1). If f(1)=1, then we obtain.
3=3.1=3.f(1)=f(3.1)=f(f(f(1)))=f(f(1))=f(1)=1
This employees 3=1 which is a contradiction:
So f(1)> 1 and hence
Let us assume
f(k)=3 for a natural number k
Now since k< 3
k=1,2
Since k> 1 it follows that k=2
This employees f(2)=3
Using the fact that f is an increasing function:
1< f(1)< f(2)=3
Since function gives only natural number outputs A natural number between 1 and 3 is the value of the function at 1
 

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