Graham Pollock 5P
Introduction
The Phi is described as the number of positive integers less than n (positive integer) which have no factor, other than 1, in common, co-prime, with n.
In my investigation I aim to identify and explain the rules and equations associated with the Phi function. To go about this I will investigate Φ (n). This will be further touched upon and will help me investigate and support any conclusions I hope to gain from investigating whether (a x b) = (a) x (b) in certain cases.
Table of Phi’s from 2-40
In my coursework there are some examples which range up to 60 these were obtained from a reliable website and checked over to ensure they were correct.
Investigation
From the Phi table shown previously we can see that some clear patterns emerged. For example I noticed that for all the prime numbers the Phi is one less than the prime number.
Φ2=1=1 Φ11=1 2 3 4 5 6 7 8 9 10=10
Φ3=1 2=2 Φ13=1 2 3 4 5 6 7 8 9 10 11 12=12
Φ7=1 2 3 4 5 6=6 Φ17=1 2 3 4 5 6 7 ...
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In my coursework there are some examples which range up to 60 these were obtained from a reliable website and checked over to ensure they were correct.
Investigation
From the Phi table shown previously we can see that some clear patterns emerged. For example I noticed that for all the prime numbers the Phi is one less than the prime number.
Φ2=1=1 Φ11=1 2 3 4 5 6 7 8 9 10=10
Φ3=1 2=2 Φ13=1 2 3 4 5 6 7 8 9 10 11 12=12
Φ7=1 2 3 4 5 6=6 Φ17=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16=16
This is because a prime number only has two numbers which divide into it, 1 and itself. The actual number is not included in our counting while the 1 is hence why the number is one less than the prime.
The Phi of 2 is the only odd value while the rest all have even values. This is because only one number is counted in
This can be arranged into an equation like so.
(p is a prime number)
Φ p = p – 1
Phi values of 2
I also looked at some other patterns in order to see if there was a formula which could solve the Phi of any number. Beginning with the number 2.
Φ2=1 Φ18=6
Φ4=2 Φ20=8
Φ6=2 Φ22=10
Φ8=4 Φ24=8
Φ10=4 Φ26=12
Φ12=4 Φ28=12
Φ14=6 Φ30=8
Φ16=8 Φ32=16
From this there seems to be no apparent pattern between all the multiples of two however I did spot a pattern where all the squares of 2 were always half of the Phi number. I will further investigate this by placing them in a table.
This can be expressed in a formula: Φ2x=½ (2x)
We must test this formula to check it is right.
e.g. Φ21=½(21) Φ23=½(23) Φ24=½(24)
Φ2=½(2) Φ8=½(8) Φ16=½(16)
Φ2=1 Φ8=4 Φ16=8
This formula has proven to work with these numbers. However on further inspection the difference between the Phi and its Phi value was the value of the previous square value. For example Φ32 (also wrote as 25) =24
Phi values of 3
This has led me onto my next number which is 3.
Φ3=2 Φ21=12
Φ6=2 Φ24=8
Φ9=6 Φ27=18
Φ12=4 Φ30=8
Φ15=8 Φ33=20
Φ18=6 Φ36=12
All these numbers are divisible by 2
I again did not see any clear pattern emerging. However I did see that as the powers of 3 increased the number was always 2/3 of the Phi value we are intending to work out.
We can see that the Φ of 3x is 2/3 of 3x. Therefore as an individual equation
Φ3x=2/3(3x)
We can see also see that Φ33=33-32 so Φ32=32-31. Therefore the formula should be
Φ3x=3x - 3x-1
To check this theory is correct we must test this.
e.g. Φ32=32-31 Φ33=33-32
Φ9=9-3 Φ27=27-9
Φ9=6 Φ27=18
Phi values of 5
As there seems to be a continuing pattern through the primes I will investigate 5 the next prime number.
Φ5=4 Φ40=16
Φ10=4 Φ45=24
Φ15=8 Φ50=20
Φ20=8 Φ55=40
Φ25=20 Φ60=16
Φ30=8 Φ65=48
Φ35=24 Φ70=24
All these numbers are divisible by 4.
When looking at these numbers there appears to be no clear pattern between all the numbers but after previously looking at the powers of numbers there appears to be no exception to the rule.
We can see that the Phi of 5x is 4/5 of 5x. Therefore as an equation
Φ5x=4/5(5x)
Again looking at these numbers we can see that the Phi value (5x) is always 5x-5x-1
e.g. Φ52=52-51
Φ25=25-5
Φ25=20
7-The next prime number
I used 7 due to it being the next prime and as there seems to be a continuing pattern with primes it would be suitable to continue.
Φ7=6 Φ42=12
Φ14=6 Φ49=42
Φ21=12 Φ56=24
Φ28=12 Φ63=36
Φ35=24 Φ70=60
These are all multiples of 6.
Again there is no clear pattern between all the numbers but the powers of 7x were all 6/7 of 7x. Therefore we can rearrange this into an equation.
Φ7x=6/7(7x)
We must test this to see if this is correct and from the examples below it is.
e.g. Φ71=6/7(71) Φ72=6/7(72)
Φ7=6/7(7) Φ49=6/7(49)
Φ7=6 Φ49=42
After previously looking at the equations for the other prime numbers it would only be reasonable to investigate into this with 7.
For example,
Φ72=72-71
Φ49=49-7
Φ49=42
As the same pattern keeps emerging in all these prime numbers we can formulate this into one equation, as shown below.
This formula below is applicable for the powers of all prime numbers.
Φ px=px-px-1 (where p is prime)
We must test this in order to see if this formula works and from the examples below this formula is applicable.
e.g. Φ24= (24)-(24-1) Φ33=(33)-(33-1) Φ52=(52)-(52-1)
Φ16=16-8 Φ27=27-9 Φ25=25-5
Φ16=8 Φ27=18 Φ25=20
Φ71= (71)-(71-1)
Φ7=7-1
Φ7=6
After looking at these numbers this way I tried a more radical approach. This approach I undertook was by looking at the squares of primes which equal the number, n. All numbers are made up of primes so this has influenced me to check for any relationship.
First of all I must work out the primes that go into a number.
e.g.
This allows me to use the previous formula which I have found, Φ px=px-px-1 (where p is prime), to work out the Phi of any number. To find the Phi of any number the number must be broken down into its smallest prime factors like shown above. We can then use the equation found previously to find out the Phi values of each of the primes to the power of any number. These Phi values multiplied together equal the Phi value of n.
For example:
Φ18= Φ(21) x Φ(32) Φ21= Φ(31) x Φ(71)
Using Φ px=px-px-1 (where p is prime) Φ21=(31-30) x Φ(71-70)
Φ18= (21-20) x (32-31) Φ21=2 x 6
Φ18=1 x 6 Φ21=12
Φ18=6
Φ55= Φ(51) x Φ(111) Φ36= Φ(22) x Φ(32)
Φ55= (51-50) x (111-110) Φ36= (22-21) x (32-31)
Φ55= 4 x 10 Φ36= 2 x 6
Φ55=40 Φ36= 12
Φ15= Φ(51) x Φ(31) Φ19= Φ(191)
Φ15= (51-50) x (31-30) Φ19= (191-190)
Φ15= 4 x 2 Φ19= 18
Φ15=8
Φ12= Φ(31) x Φ(22) Φ26= Φ(21) x Φ(131)
Φ12= (31-30) x (22-21) Φ26= (21-20) x (131-130)
Φ12= 2 x 2 Φ26= 1 x 12
Φ12=4 Φ26= 12
Φ27= Φ(33) Φ32= Φ(25)
Φ27= (33-32) Φ32= (25-24)
Φ27= (27-9) Φ32= (32-16)
Φ27=18 Φ32= 16
Therefore from this we can see that this formula works and can work the Phi of any number. This can be placed into a formula. This formula is
Φ(x)= Φ(y)x Φ(z)
(Where y and z are the breakdowns of x into its lowest prime factors)
Part 2
In ordinary integer equations we are know that (axb) = (a) x (b), however I wish to test this and see if it is applicable in the case of Phi.
We are asked to check if
Φ (7 x 4) = Φ(7) x Φ(4) Φ(4x6)= Φ(4) x Φ(6)
Φ28=6 x 2 Φ24= 2 x 2
12=12 8=4
In this case it does balance. In this case it does not balance.
So in some cases it does balance and in some cases it does not so therefore this has led me into further investigation.
I have decided to investigate a combination of different types of numbers starting with
Even and Odd-Where the odd is prime and is co-prime to the even
Φ (7 x 4) = Φ(7) x Φ(4) Φ (8x5) = Φ(8) x Φ(5)
Φ28=6 x 2 Φ40= 4 x 4
12=12 16=16
Φ (4x9)= Φ(9) x Φ(4) Φ(3x4) = Φ(3) x Φ(4)
Φ36= 6 x 2 Φ12= 2 x 2
12=12 4=4
Φ (6x7) = Φ(6 )x Φ(7) Φ (11x4) = Φ(11) x Φ(4)
Φ42=2 x 6 Φ44= 10 x 2
12=12 20=20
From these we can see that this works but we must further investigate.
Even and odd-Where the odd is prime and is not co-prime to the even
Φ (3x12) = Φ(3) x Φ(12) Φ (3x18) = Φ(3) x Φ(18)
Φ36= 2 x 4 Φ 54= 2 x 6
12=8 18=12
Φ (5x10) = Φ(5) x Φ(10)
Φ50= 4 x 4
20=16
Even and odd –Where the odd is prime and they are co-primes of each other
Φ(9x4)= Φ(9) x Φ(4) Φ(15x4)= Φ(15) x Φ(4)
Φ36= 6 x 2 Φ60= 8 x 2
12=12 16=16
Odd and odd (1 prime 1 non prime)
Φ(3x9)= Φ(3) x Φ(9) Φ(5x9)= Φ(5)x Φ(9)
Φ27= 2 x 6 Φ(45)= 4 x 6
18= 12 24=24
Φ(15x 3)= Φ(15) x Φ(3)
Φ45= 8 x 2
24= 16
Prime and prime
Φ (5x7) = Φ(5) x Φ(7) Φ (3x5) = Φ(3) x Φ(5)
Φ35=4 x 6 Φ15 = 2 x 4
24=24 8=8
Φ (11x2) = Φ(11) x Φ(2) Φ (13x3) = Φ(13) x Φ(3)
Φ22= 10 x 1 Φ39= 12 x 2
22=10 24=24
Φ (7x2) = Φ(7) x Φ(2) Φ (17x3) = Φ(17) x Φ(3)
Φ14 = 6 x 1 Φ51 = 16 x 2
6=6 32=32
Φ (11x5) = Φ(11) x Φ(5) Φ (2x23)= Φ(2) x Φ(23)
Φ55= 10 x 4 Φ 46= 1 x 22
40=40 22=22
Even and even
Φ(4x6)= Φ(4) x Φ(6) Φ(4x10)= Φ(4) x Φ(10)
Φ24= 2 x 2 Φ40= 2 x 4
8=4 16=8
Φ(8x4)= Φ(8) x Φ(4) Φ(12x4)= Φ(12) x Φ(4)
Φ32= 4 x 2 Φ48= 4 x 2
16=8 16=8
Φ(6x10)= Φ(6) x Φ(10)
Φ60= 2 x 4
16=8
In this we can see that the Phi of each of the even numbers is half of n.
(Where m and n are even)
Φ(m x n)=½ Φ(m x n)
In order to get a better view and to see if there is any link between all the numbers that work or do not work it would be better to amalgamate them into a table.
I noticed that inside the ones which do not balanced, i.e. the ones where
(a x b) = (a) x (b), all had more than one common factor other than 1. While the ones which do balance the numbers are co-primes of each other.
Therefore this means Φ(c x d)= Φ(c) x Φ(d) (where c and d are co-primes)
Φ(e x f) = Φ(e) x Φ(f) (where e and f have more than 1 common factors other than 1)
Conclusions
Throughout this coursework I have found at wide range of equations allowing us to work out different aspects of Phi. At the beginning I found the basic formulas, these being
- Φ p = p – 1
A number of minor formulas helped me amalgamate and form a formula for all prime numbers which in turn helped me later. These formulas were:
- Φ2x=½ (2x)
- Φ3x=2/3(3x)
- Φ3x=3x - 3x-1
- Φ5x=4/5(5x)
- Φ7x=6/7(7x)
- Φ px=px-px-1 (where p is prime)
This particular one helped me support, form and follow the equation:
- Φ(x)= Φ(y)x Φ(z)
(Where y and z are the breakdowns of x into its lowest prime factors)
In part two I have found that (a x b) = (a) x (b) only in the cases where a and b are co-primes. These examples can be solved using the equations shown previously. When the numbers are not co-prime and have more than one common factor other than 1 they do not balance this equation may be used.
Φ(m x n)=½ Φ(m x n) (where m and n are even)
These formulas allow me to work out the Phi of any number providing it can be split into its prime factors.