phi of 6?

$$\\\mathbf{How\ to\ calcultate\ the\ \textit{Euler phi function} \ \phi(n):}\\
$ We have the prime factorization of n = p_1\cdot p_2\cdot p_3 \cdots\\
\phi(n) = n \cdot (1-\frac{1}{p_1})\cdot (1-\frac{1}{p_2}) \cdot (1-\frac{1}{p_3}) \cdots$$

$$\\\mathbf{Example\ 1: n = 6 } $\\
The prime factorization of 6 = 2 * 3 = p_1*p_2 \\
\phi(6) = 6 \cdot (1-\frac{1}{2}) \cdot (1-\frac{1}{3}) \\
\phi(6) = 6 \cdot \frac{1}{2} \cdot \frac{2}{3} \\
\phi(6) = \frac{6}{3} \\
\phi(6) = 2$$

$$\\\mathbf{Example\ 2: n = 9 } $\\
The prime factorization of 9 = 3^2 = p_1^2 \\
\phi(9) = 9 \cdot (1-\frac{1}{3}) \\
\phi(9) = 9 \cdot \frac{2}{3} \\
\phi(6) = 3 \cdot 2 \\
\phi(6) = 6$$

$$\\\mathbf{Example\ 3: n = 7 } $\\
The prime factorization of 7 = 7 = p_1 \qquad 7 $ is a prime number!$ \\
\phi(7) = 7 \cdot (1-\frac{1}{7}) \\
\phi(7) = 7 \cdot \frac{6}{7} \\
\phi(7) = 6$$

$$\\\mathbf{Example\ 4: n = 11 } $\\
The prime factorization of 11 = 11 = p_1 \qquad 11 $ is a prime number!$ \\
\phi(11) = 11 \cdot (1-\frac{1}{11}) \\
\phi(11) = 11 \cdot \frac{10}{11} \\
\phi(11) = 10$$

$$\boxed{\text{ In general $ \phi(p) = p-1 $, if p is a prime number }}\\\\
\begin{array}{lr}
p = 2: &\phi(2) = 1 \qquad =(2-1)\\
p = 3: &\phi(3) = 2 \qquad =(3-1)\\
p = 5: &\phi(5) = 4 \qquad =(5-1)\\
p = 7: &\phi(7) = 6 \qquad =(7-1)\\
p = 11: &\phi(11) = 10 \qquad =(11-1)\\
p = 13: &\phi(13) = 12 \qquad =(13-1)\\
\cdots & \phi(p) = p-1
\end{array}$$

The first 99 values of the Phi function are:

	+0	+1	+2	+3	+4	+5	+6	+7	+8	+9
0+		1	1	2	2	4	2	6	4	6
10+	4	10	4	12	6	8	8	16	6	18
20+	8	12	10	22	8	20	12	18	12	28
30+	8	30	16	20	16	24	12	36	18	24
40+	16	40	12	42	20	24	22	46	16	42
50+	20	32	24	52	18	40	24	36	28	58
60+	16	60	30	36	32	48	20	66	32	44
70+	24	70	24	72	36	40	36	60	24	78
80+	32	54	40	82	24	64	42	56	40	88
90+	24	72	44	60	46	72	32	96	42	60

what is Φ6?

$$\!\ \varphi(6) = 2. \qquad 1 \text{ and } 5 \\\\
6 = 2 \cdot 3\\
\!\ \varphi(6)}=6\cdot (1-\frac{1}{2})( 1-\frac{1}{3} )}=2$$

Heureka, could you (or someone else) explain the phi function to me please?

I thought phi was an irrational number like pi.     

Hello Melody,

$$\\\mathbf{De{finition:}}\\
\begin{text}
L{et} $ n \ge 1$ be an integer. Then we de{fine} the \\
\textit{Euler phi function} $\phi$ by\\
$\phi(n)=$ the number of positive integers \\
less than $n$ that are relatively prime to $n$
\end{text}$$

$$\\\mathbf{Relatively Prime:}\\
$
\begin{text}
Describes two numbers for which\\
the only common factor is 1. \\
In other words, relatively prime numbers have\\
a greatest common factor $(gcf)$ of $1$. \\
For example, $6$ and $35$ are relatively prime $(gcf = 1)$.\\
The numers $6$ and $8$ are not relatively prime $(gcf = 2)$.
\end{text}$$

$$\\\mathbf{Example\ 1: n = 6 }
$ and n is not a prime number $\\
\samll{\text{
$
\begin{array}{rr}
\textcolor[rgb]{1,0,0}{1}& gcf(6,1) = \textcolor[rgb]{1,0,0}{1}\\
2& gcf(6,2) = 2\\
3& gcf(6,3) = 3\\
4& gcf(6,4) = 2\\
\textcolor[rgb]{1,0,0}{5}& gcf(6,5) = \textcolor[rgb]{1,0,0}{1}\\
6& gcf(6,6) = 6\\
\end{array}
$
}}\\
6 \text{ has } \textcolor[rgb]{0,0,1}{2} \text{ relative primes } 1 \text{ and } 5 \text { see the red color. So } \phi(6) = \textcolor[rgb]{0,0,1}{2}$$

$$\\\mathbf{Example\ 2: n = 9 }
$ and n is not a prime number $\\
\samll{\text{
$
\begin{array}{rr}
\textcolor[rgb]{1,0,0}{1}& gcf(9,1) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{2}& gcf(9,2) = \textcolor[rgb]{1,0,0}{1}\\
3& gcf(9,3) = 3\\
\textcolor[rgb]{1,0,0}{4}& gcf(9,4) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{5}& gcf(9,5) = \textcolor[rgb]{1,0,0}{1}\\
6& gcf(9,6) = 3\\
\textcolor[rgb]{1,0,0}{7}& gcf(9,7) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{8}& gcf(9,8) = \textcolor[rgb]{1,0,0}{1}\\
9& gcf(9,9) = 9\\
\end{array}
$
}}\\
9 \text{ has } \textcolor[rgb]{0,0,1}{6} \text{ relative primes } 1,2,4,5,7,8 \text { see the red color. So } \phi(9) = \textcolor[rgb]{0,0,1}{6}$$

$$\\\mathbf{Example\ 3: n = 7 }
$ and n is a prime number $\\
\samll{\text{
$
\begin{array}{rr}
\textcolor[rgb]{1,0,0}{1}& gcf(7,1) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{2}& gcf(7,2) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{3}& gcf(7,3) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{4}& gcf(7,4) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{5}& gcf(7,5) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{6}& gcf(7,6) = \textcolor[rgb]{1,0,0}{1}\\
7& gcf(7,7) = 7\\
\end{array}
$
}}\\
\small{
7 \text{ has } 7-1=\textcolor[rgb]{0,0,1}{6} \text{ relative primes } 1 \text{ until } 6 \text { see the red color. So } \phi(7) = \textcolor[rgb]{0,0,1}{6}.}\\
\text{ So } \phi \text{ of a prime number } \phi(p) = p-1$$

Thank you Heureka,  that explains it beautifully 

I have a further question Heureka. 

$$\!\ \varphi(6) = 2. \qquad 1 \text{ and } 5 \\\\
6 = 2 \cdot 3\\
\!\ \varphi(6)}=6\cdot (1-\frac{1}{2})( 1-\frac{1}{3} )}=2$$

Now I do get why phi(6)=2

But what is it about from 6=2.3

What is the relvance of the brackets.  How can this be used on other examples?

Thank you Heueka, you are a gem.