Ф(7) x Ф(4)=6 x 2=12;
So Ф(7 x 4) = Ф(7) x (4) is true.
2)
Ф(6 x 4)=Ф(6) x Ф(4);
Ф(6)=2;
1 2 3 4 5 6
Ф(4)=2;
I have found Ф(4) already.
Ф(6 x 4)= Ф(24)=8;
I have found Ф(24) already.
Ф(6) x Ф(4)=2 x 2=4;
So Ф(6 x 4)=Ф(6) x Ф(4) is true.
b)
Ф(5 x 2)=Ф(5) x Ф(2);
Ф(5)=4;
I have found Ф(5) already.
Ф(2)=1;
1 2
Ф(5 x 2)= Ф(10)=4;
1 2 3 4 5 6 7 8 9 10
Ф(5) x Ф(2)=4 x 1=4;
So Ф(5 x 2)=Ф(5) x Ф(2) is true.
Ф(7 x 3)=Ф(7) x Ф(3);
Ф(7)=6;
I have found Ф(7) already.
Ф(3)=2;
I have found Ф(3) already.
Ф(7 x 3)= Ф(21)=12;
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Ф(7) x Ф(3)=6 x 2=12;
So Ф(7) x Ф(3)=6 x 2=12 is true.
Ф(4 x 5)=Ф(4) x Ф(5);
Ф(4)=2;
I have found Ф(4) already.
Ф(5)=4;
I have found Ф(5) already.
Ф(4 x 5)= Ф(20)=8;
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ф(4) x Ф(5)=2 x 4=8;
So Ф(4 x 5)=Ф(4) x Ф(5) is true.
As you can see I have checked three separate choices of n and m for
Ф(n x m)=Ф(n) x Ф(m), and all of them were true.
Now I will find the Phi function for all integers from 2 to 30 to make easier my future investigations. I am starting with 2 because I am not sure what Phi of 1 is.
Ф(2)=1; 1 2
Ф(3)=2; 1 2 3
Ф(4)=2; 1 2 3 4
Ф(5)=4; 1 2 3 4 5
Ф(6)=2; 1 2 3 4 5 6
Ф(7)=6; 1 2 3 4 5 6 7
Ф(8)=4; 1 2 3 4 5 6 7 8
Ф(9)=6; 1 2 3 4 5 6 7 8 9
Ф(10)=4; 1 2 3 4 5 6 7 8 9 10
Ф(11)=10; 1 2 3 4 5 6 7 8 9 10 11
Ф(12)=4; 1 2 3 4 5 6 7 8 9 10 11 12
Ф(13)=12; 1 2 3 4 5 6 7 8 9 10 11 12 13
Ф(14)=6; 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ф(15)=8; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ф(16)=8; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Ф(17)=16; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Ф(18)=6; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Ф(19)=18; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Ф(20)=8; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ф(21)=12; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Ф(22)=10; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Ф(23)=22; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
Ф(24)=8; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Ф(25)=20; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ф(26)=12; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
Ф(27)=18; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Ф(28)=12; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
Ф(29)=28; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Ф(30)=8; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Now I will try to find my easier way to find Ф(n). First I will try to find prime factors.
I will do an example. Let the number be 6.
As you can see prime factors of 6 are 2 and 3, which means that it is made of 2 and 3. So if I was to cross all the numbers, which are not co-prime with 6, I would cross each second and each third number, I could find the amount of numbers I would cross.
I have noticed, that ф of numbers which consist of 2 only are halves of this number, now I will explain what I mean. When you want to find prime numbers, which the main number consists of, you take the main number and divide it by prime numbers only, until you get 1 as an answer. For example you want to do that for number 12. So:
The prime numbers for 12 are 2 and 3.
And it explains my method of finding a ф of a number. For example I know that prime numbers of 12 are 2 and 3, so I write down 12 numbers, then I cross every second and third number, and the number of not crossed numbers is an answer.
So, for a number, which consists of 2 only I would cross each second numbers which is a half of all numbers, so the other half is the answer.
Part 3
In this part I am going to investigate a rule, which is Ф(n x m)= Ф(n) x Ф(m).
Fist of all I will try to investigate it with odd and even numbers.
Even-Even.
