In this coursework I was asked to investigate the Phi Function (f) of a number (n).

In this coursework I was asked to investigate the Phi Function (φ) of a number (n). The Phi Function of a number (n) is delineated as the number of positive integers less than n, which have no factor (other than 1) in common, i.e. co-prime with n. Example: φ(16) = 8. The integers less than 16 that have no factors apart from 1 in common with 16 are 1, 3, 5, 7, 9, 11, 13, and 15. There are 8 altogether. To calculate φ(n) I will list out the numbers from 1 till n−1. I will then cross out all the numbers that have a common factor with n. The remaining numbers will give me the φ of n.

In this part of the coursework will be investigating the phi function of:

φ(p)
φ(p)²

Part 1:

Find the value of:

φ(3):
1 2

3 = 1,2

The number 3 only has 2 positive co-prime integers they are the numbers 1 and 2.

(ii) φ(8):

1 2 3 4 5 6 7

8 = 1,3,5,7

There are 4 positive co-prime integers for the number 8

(iii) φ(11):

1 2 3 4 5 6 7 8 9 10 11

11 = 1,2,3,4,5,6,7,8,9,10

The number 11 has 10 positive co-prime integers, they are shown above.

(iv) φ(24):

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

24 = 1,5,7,11,13,17,19,23

The number 24 has 8 positive co-prime integers, they are shown above.

I will create a table for the Phi values of the numbers from 1 to 30. I have created this table because it would be easier to lookup the phi values of the numbers within this range instead of solving it every time I need to find a value quickly. In this table I have crossed out the factors that are not co-prime and left the co-prime ones. In the end of each row, I counted the uncrossed numbers to find the phi value of that number.

φ(2); 1, = 1

φ(3); 1 2; =2

φ(4); 1 2 3; =2

φ(5); 1 2 3 4; =4

φ (6); 1 2 3 4 5; =2

φ (7); 1 2 3 4 5 6; =6

φ (8); 1 2 3 4 5 6 7; =4

φ (9); 1 2 3 4 5 6 7 8; =6

φ (10); 1 2 3 4 5 6 7 8 9; =4

φ (11); 1 2 3 4 5 6 7 8 9 10; =10

φ (12); 1 2 3 4 5 6 7 8 9 10 11; =4

φ (13); 1 2 3 4 5 6 7 8 9 10 11 12; =12