Φ (7 x 4)
= Φ (28)
Φ (28) = 27, 25, 23, 19, 17, 15, 13, 11, 9, 5, 3, 1.
= Φ (28) = 12
Φ (7)
Φ (7) = 6, 5, 4, 3, 2, 1.
Φ (7) = 6
Φ (4)
Φ (4) = 3, 1.
Φ (4) = 2
Φ (7) x Φ (4)
= 6 x 2
= 12
As the answer gotten from Φ (7) x Φ (4) is equal to that of Φ (7 x 4), the equation Φ (7 x 4) = Φ (7) x Φ (4) is correct.
12 integers = 6 integers x 2 integers
12 integers = 12 integers.
- Φ (6 x 4) ≠ Φ (6) x Φ (4)
Solution
2 – Φ (6 x 4) ≠ Φ (6) x Φ (4)
Φ (6 x 4)
= Φ (24)
Φ (24) = 23, 19, 17, 13, 11, 7, 5, 1.
Φ (24) = 8
Φ (6)
Φ (6) = 5, 1.
Φ (6) = 2
Φ (4)
Φ (4) = 3,1.
Φ (4) = 2
As the answer gotten from Φ (6 x 4), is not equal to that of Φ (6) x Φ (4) the equation Φ (6 x 4) ≠ Φ (6) x Φ (4) is correct.
8 integers ≠ 2 integers x 2 integers
8 integers ≠ 4 integers
- Check whether or not Φ (n x m) = Φ (n) x Φ (m) for at least two separate choices of n and m
Solution:
For my first choice I will use the numbers 3 and 9 where 3 = n and 9 = m.
Φ (n x m) = Φ (n) x Φ (m)
= Φ (3 x 9) = Φ (3) x Φ (9)
Φ (27) = Φ (3) x Φ (9)
Φ (27) = 26, 25, 24, 23, 22, 20, 19, 17, 16, 15, 14, 13,
12, 11, 10, 8, 7, 5, 4, 2, 1.
Φ (27) = 22
Φ (3) = 2,1
Φ (3) = 2
Φ (9)
Φ (9) = 8, 7, 5, 4, 2, 1.
Φ (9) = 6
Φ (27) = Φ (3) x Φ (9)
22 = 2 x 6
As the answer gotten from Φ (27) is not equal to that of Φ (3) x Φ (9), the equation Φ (n x m) = Φ (n) x Φ (m), where n = 3 and m = 9 is not correct.
22 integers ≠ 2 integers x 6 integers
22 integers ≠ 12 integers.
Solution
For my second choice I will use the numbers 3 and 7 where 3 = n and 7 = m
Φ (n x m) = Φ (n) x Φ (m)
Φ (3 x 7) = x Φ (3) x Φ (7)
Φ (3 x 7)
= Φ (21)
Φ (21) = 20, 19, 17, 16, 13, 11, 10, 8, 5. 4, 2, 1.
= Φ (21) = 12
Φ (3)
= Φ (3) = 2,1.
Φ (3) = 2
Φ (7)
= Φ (7) = 6, 5, 4, 3, 2, 1.
Φ (7) = 6
As the answer gotten from Φ (21) is equal to that of Φ (3) x Φ (7), the equation Φ (n x m) = Φ (n) x Φ (m), where n = 3 and m = 7 is correct.
12 integers = 2 integers x 6 integers
12 integers = 12 integers.
Solution
For my third choice I will use the numbers 8 and 4 where n = 8 and 4= m.
Φ (n x m) = Φ (n) x Φ (m)
Φ (8 x 4) = Φ (8) x Φ (4)
Φ (8 x 4)
= Φ (32)
Φ (32) = 31, 29, 27, 25, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1.
Φ (32) = 16
Φ (8)
Φ (8) = 7, 5, 3, 1
= Φ (8) = 4
Φ (4)
Φ (4) = 3, 1
Φ (4) = 2
As the answer gotten from Φ (32) is not equal to that of Φ (4) x Φ (8), the equation Φ (n x m) = Φ (n) x Φ (m), where n = 8 and m = 4 is incorrect.
16 integers ≠ 4 integers x 2 integers
16 integers ≠ 8 integers
PART 3.
In some cases Φ (n x m) = Φ (n) x Φ (m) whilst in other cases this is not so. Investigate this situation.
Investigation:
From the results gotten in Part 2 its very obvious that in cases where the variables n and m have common factors the equation:
Φ (n × m) = Φ (n) × Φ (m);
does not work. In cases in which the variables n and m had no common factors the equation:
Φ (n × m) = Φ (n) × Φ (m);
worked. With this information I can draw a very accurate prediction.
Prediction:
With the knowledge gotten from part two I predict that variables with no common variables will prove the equation right.
I will investigate this further showing some more examples of this theory. I will also look for any other patterns that may occur for this equation.
I have chosen to use the sets of numbers below. The first four sets of numbers have factors in common so I expect that the equation Φ (n × m) = Φ (n) × Φ (m) will not work. The second set however have no common factors so I expect the equation Φ (n × m) = Φ (n) × Φ (m) to work.
- 9 , 12
- 8 , 10
- 2 , 6
- 5 , 15
- 4 , 9
- 2 , 3
- 11 , 4
- 5 , 7
Solution 1
In this solution the variable n= 9 and the variable m = 12.
Φ (n × m) = Φ (n) × Φ (m)
Φ (9 × 12) = Φ (9) × Φ (12)
Φ (108) = Φ (9) × Φ (12)
Φ (108) = 107, 103, 101, 97, 95, 91, 89, 85, 83, 79, 77, 73, 71, 67, 65, 61, 59, 55, 53, 49, 47, 43, 41, 37, 35, 31, 29, 25, 23, 19, 17, 13, 11, 7, 5, 1.
Φ (108) = 36
Φ (9) = 8, 7, 5, 4, 2, 1.
Φ (9) = 6
Φ (12) = 11, 7, 5, 1.
Φ (12) = 4
As predicted the equation did not work as:
36 integers ≠ 6 integers x 4 integers
36 integers ≠ 24 integers
Solution 2
In this solution the variable n= 8 and the variable m = 10.
Φ (n × m) = Φ (n) × Φ (m)
Φ (8 × 10) = Φ (8) × Φ (10)
Φ (80)
Φ (80) = 79, 77, 73, 71, 69, 67, 63, 61, 59, 57, 53, 51, 49, 47, 43, 41, 39, 37, 33, 31, 29, 27, 23, 21, 19, 17, 13, 11, 9, 7, 3, 1.
Φ (80) = 32
Φ (10) = 9, 7, 3, 1.
Φ (10) = 4
Φ (8) = 7, 5, 3, 1.
Φ (8) = 4
As predicted the equation did not work as:
32 integers ≠ 4 integers x 4 integers
32 integers ≠ 16 integers
Solution 3
In this solution the variable n= 2 and the variable m = 6.
Φ (n × m) = Φ (n) × Φ (m)
Φ (2 × 6) = Φ (2) × Φ (6)
Φ (12) = 11, 7, 5, 1.
Φ (12) = 4
Φ (2) = 1.
Φ (2) = 1.
Φ (6) = 5, 1.
Φ (6) = 2
As predicted the equation did not work as:
4 integers ≠ 1 integer x 2 integers
4 integers ≠ 2 integers
Solution 4
In this solution the variable n= 5 and the variable m = 15.
Φ (n × m) = Φ (n) × Φ (m)
Φ (5 × 15) = Φ (5) × Φ (15)
Φ (75) = 74, 73, 71, 68, 67, 64, 62, 61, 59, 58, 56, 53, 52, 49, 47, 46, 44, 43, 41, 38, 37, 34, 32, 31, 29, 28, 26, 23, 22, 19, 17, 16, 14, 13, 11, 8, 7, 4, 2, 1.
Φ (75) = 40
Φ (5) = 4, 3, 2, 1.
Φ (5) = 4
Φ (15) = 14, 13, 11, 8, 7, 4, 2, 1.
Φ (15) = 8
As predicted the equation did not work as:
40 integers ≠ 4 integers x 8 integers
40 integers ≠ 32 integers
Solution 5
In this solution the variable n= 4 and the variable m = 9.
Φ (n × m) = Φ (n) × Φ (m)
Φ (4 × 9) = Φ (4) × Φ (9)
Φ (36) = 35, 31, 29, 25, 23, 19, 17, 13, 11, 7, 5, 1.
Φ (36) = 12
Φ (4) = 3, 1.
Φ (4) = 2
Φ (9) = 8, 7, 5, 4, 2, 1.
Φ (9) = 6
As predicted the equation did work as:
12 integers = 2 integers x 6 integers
12 integers = 12 integers
Solution 6
In this solution the variable n= 2 and the variable m = 3.
Φ (n × m) = Φ (n) × Φ (m)
Φ (2 × 3) = Φ (2) × Φ (3)
Φ (6) = 5, 1.
Φ (6) = 2
Φ (2) = 1
Φ (2) = 1
Φ (3) = 2, 1.
Φ (3) = 2
As predicted the equation did work as:
2 integers = 1 integers x 2 integers
2 integers = 2 integers
Solution 8
In this solution the variable n= 11 and the variable m = 4.
Φ (n × m) = Φ (n) × Φ (m)
Φ (11 × 4) = Φ (11) × Φ (4)
Φ (44) = 44, 41, 39, 37, 35, 31, 29, 27, 25, 23, 21, 19, 17, 15, 13, 9, 7, 5, 3, 1.
Φ (44) = 20
Φ (11) = 10, 9, 8, 7, 6, 5, 4, 3, 2, 1.
Φ (11) = 10
Φ (4) = 3, 1.
Φ (4) = 2
As predicted the equation did work as:
20 integers = 10 integers x 2 integers
20 integers = 20 integers
Solution 8
In this solution the variable n= 5 and the variable m = 4.
Φ (n × m) = Φ (n) × Φ (m)
Φ (5 × 7) = Φ (5) × Φ (7)
Φ (35)= 34, 33, 32, 31, 29, 27, 26, 24, 23, 22, 19, 18, 17, 16, 13, 12, 11, 9, 8, 6, 4, 3, 2, 1.
Φ (35) = 24
Φ (5) = 4, 3, 2, 1.
Φ (5) = 4
Φ (7) = 6, 5, 4, 3, 2, 1.
Φ (7) = 6
As predicted the equation did work as:
24 integers = 4 integers x 6 integers
24 integers = 24 integers
Analysis
A multiplicative function is a function f such that f(a × b) = f(a) × f(b). Is phi multiplicative? To an extent it is as in all the sets of numbers that had common factors the equation Φ (n × m) = Φ (n) × Φ (m) did not work.
Conclusion
With all the problems and examples I have solved in relation to this I can conclude that;
Φ (n x m) = Φ (n) x Φ (m)
where n is a prime number and m is any number at all.
PART 4.
If p and q are prime investigate: Φ (pmqn)
The equation Φ (pmqn) = Φ (pm) x Φ (qn) and so with the knowledge gotten from part 3 I can predict that Φ (pmqn) = Φ (pm) x Φ (qn) will be true where p and q are prime. to further illustrate this I will do a few examples.
Example 1.
P= 2, q = 3 ; m = 3, n = 2
Φ (pm x qn)
Φ (pmqn) = Φ (pm) x Φ (qn)
Φ (23 x 32) = Φ (23) x Φ (32)
Φ (8 x 9) = Φ (8) x Φ (9)
Φ (72) = Φ (8) x Φ (9)
24 = 4 x 6
24 = 24
Example 2.
P= 7, q = 5 ; m = 4, n = 6
Φ (pm x qn)
Φ (pmqn) = Φ (pm) x Φ (qn)
Φ (54 x 71) = Φ (54) x Φ (71)
Φ (625 x 7 ) = Φ (625) x Φ (7)
Φ (4375) = Φ (625) x Φ (7)
3000 = 500 x 6
3000 = 3000
Example 3
P= 5, q = 11 ; m = 1, n = 3
Φ (pm x qn)
Φ (pmqn) = Φ (pm) x Φ (qn)
Φ (51 x 113) = Φ (51) x Φ (113)
Φ (5 x 1331) = Φ (5) x Φ (1331)
Φ (6655) = Φ (5) x Φ (1331)
4840 = 4 x 1210
4840 = 4840
Example 4
P= 13, q = 2 ; m = 2, n = 4
Φ (pm x qn)
Φ (pmqn) = Φ (pm) x Φ (qn)
Φ (132 x 24) = Φ (132) x Φ (24)
Φ (169 x 16 ) = Φ (169) x Φ (16)
Φ (2704) = Φ (169) x Φ (16)
1248 = 156 x 8
1248 = 1248