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• Level: GCSE
• Subject: Maths
• Word count: 1261

# My investigation is about the Phi Function .

Extracts from this document...

Introduction

My investigation is about the Phi Function Φ. I am investigating the different ways on how to find the Phi Functions of different numbers and finding easier ways of finding the Phi Functions of large numbers. I will go through four parts for this coursework. I will start from the simplest cases of numbers and will go to more complicated. For any positive integer n, the Phi function Φ(n) is defined as the number of positive integers less than n which have no factor (other than 1) in common (are co-prime) with n:

So Φ(12)=4, because the positive integers less than 10 which have no factors other than 1, in common with 12 are 1, 5, 7, 11 i.e. 4 of them. These four numbers are not factors of 12.

Also Φ(6)=2, because the positive integers less than 6 which have no factors other than 1, in common with 6 are 1, 5 i.e. 2 of them. These two numbers are not factors of two.

For the first part I will find

Middle

“P-1”. “P” stands for Prime number.

I the second part I am going to check whether Φ(n×m)=Φ(n)×Φ(m) or Φ(n×m)≠Φ(n)×Φ(m). I am going to check the first two from the sheet.

1. Φ(7×4)=Φ(7)×Φ(4)

Φ(28)=12. There are 12 numbers which are not factors          of 28.

Φ(7)=6. There are 6 numbers which are not factors of   7.

Φ(4)=2. There are 2 numbers which are not factors of

4.

So Φ(7×4)=Φ(7)×Φ(4)

1. Φ(6×4)≠Φ(6)×(4)

Φ(24)=8

Φ(6)=2

Φ(4)=2

So Φ(6×4)≠Φ(6)×Φ(4)

I am now going to check whether or not Φ(nxm)=Φ(n)xΦ(m) for at least two separate choices of my own of n and m.

1. Φ(2×8)=Φ(2)×Φ(8)

Φ(16)=8. There are 8 positive numbers less than 16 which are not factors of 16.

Φ(2)=1. There is 1 positive number less than 2 which is not a factor of 2.

Φ(8)=4. There are 4 positive numbers less than 8 which are not factors of 8.

So Φ(2×8)≠Φ(2)×Φ(8)

1. Φ(7×3)=Φ(7)×Φ(3)

Φ(21)=12

Φ(7)=6

Φ(3)=2

So Φ(7×3)=Φ(7)×Φ(3)

For the third part I am investigating why Φ(n×m)=Φ(n)×Φ(m) whilst in other cases this is not so. I have found out that it depends on the types of numbers used as n and m. Now I will check whether or not Φ(n×m)=Φ(n)×Φ(m)

Conclusion

I have to investigate “Φ(Pⁿ Q )” if P and Q are prime in the fourth part. I have found two formulas for finding the Phi Functions of large numbers. One of the formulas is  “(Pⁿˉ¹)(P-1)”. The other formula is “Pⁿ(1-1/P)”.

FIRST FORMULA:(Pⁿˉ¹)(P-1)

To find Φ(81):

1. Firstly the 81 has to be split up into prime numbers.

 81 3 27 3 9 3 3 3 1

81 converted into prime numbers=3

So the formula can now be used to find Φ(81):

2. (3 ˉ¹)(3-1)=

3. 3³×2

4. 54

So Φ(81)=54

SECOND FORMULA: Pⁿ(1-1/P)

To find Φ(81):

1. Firstly 81 has to be split into prime numbers.
 81 3 27 3 9 3 3 3 1

81 converted into prime numbers=3

So the formula can now be used to find Φ(81):

1. 3 (1-1/3)=
1. 81×2/3=
1. 54

So Φ(81)=54

The two formulas give out the same results for the Φ(81). The “P” on the formulas stands for prime number. The formulas are faster at finding the Phi rather than writing all the numbers out and cancelling or circling them. There are four steps to work out the Phi’s from the formulas and that is why I have numbered my working out.

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

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