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# help again

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We define a function f(x) such that f(11)=34, and if there exists an integer a such that f(a)=b, then f(b) is defined

and f(b)=3b+1 if b is odd

$$f(b)=\frac{b}{2}$$ if b is even.

What is the smallest possible number of integers in the domain of f?

Oct 15, 2019

#1
+3

We are given that  f(11)  =  34  so we know that  11  is in the domain.

domain so far  =  {11}

There exists an integer  11  such that  f(11) = 34 ,  so  f(34)  is defined. And so  34  is in the domain.

domain so far  =  {11, 34}

So notice that just like we had to add the result of  f(11)  to the domain, we will now have to add the result of  f(34)  to the domain (then the result of that, and the result of that, etc.).

 f(34)  =  17 _____ domain so far  =  {11, 34, 17} f(17)  =  52 domain so far  =  {11, 34, 17, 52} f(52)  =  26 domain so far  =  {11, 34, 17, 52, 26} f(26)  =  13 domain so far  =  {11, 34, 17, 52, 26, 13} f(13)  =  40 domain so far  =  {11, 34, 17, 52, 26, 13, 40} f(40)  =  20 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20} f(20)  =  10 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10} f(10)  =  5 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5} f(5)  =  16 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16} f(16)  =  8 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8} f(8)  =  4 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4} f(4)  =  2 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2} f(2)  =  1 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} f(1)  =  4 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1}

Finally, we have reached a point where we do not have to add anything else to the domain.

4  is already in the set, so we do not have to add anything else.

So the smallest possible number of integers in the domain is  15

(Thank you CPhill for correcting my previous mistake!)

Oct 15, 2019
edited by hectictar  Oct 16, 2019

#1
+3

We are given that  f(11)  =  34  so we know that  11  is in the domain.

domain so far  =  {11}

There exists an integer  11  such that  f(11) = 34 ,  so  f(34)  is defined. And so  34  is in the domain.

domain so far  =  {11, 34}

So notice that just like we had to add the result of  f(11)  to the domain, we will now have to add the result of  f(34)  to the domain (then the result of that, and the result of that, etc.).

 f(34)  =  17 _____ domain so far  =  {11, 34, 17} f(17)  =  52 domain so far  =  {11, 34, 17, 52} f(52)  =  26 domain so far  =  {11, 34, 17, 52, 26} f(26)  =  13 domain so far  =  {11, 34, 17, 52, 26, 13} f(13)  =  40 domain so far  =  {11, 34, 17, 52, 26, 13, 40} f(40)  =  20 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20} f(20)  =  10 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10} f(10)  =  5 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5} f(5)  =  16 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16} f(16)  =  8 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8} f(8)  =  4 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4} f(4)  =  2 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2} f(2)  =  1 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} f(1)  =  4 domain so far  =  {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1}

Finally, we have reached a point where we do not have to add anything else to the domain.

4  is already in the set, so we do not have to add anything else.

So the smallest possible number of integers in the domain is  15

(Thank you CPhill for correcting my previous mistake!)

hectictar Oct 15, 2019
edited by hectictar  Oct 16, 2019
#2
+1

Nicely done, hectictar    !!!!   Oct 15, 2019