Hello I have a question about the following theorem proof. There I am particularly interested in the part that I have highlighted in thick.
Theorem: If $a$ and $b$ are two mutually prime integers, $\phi(a \cdot b) = \phi(a)\phi(b)$.
Proof: The theorem is proved by considering all integers $u$ smaller than $a \cdot b$ and define $u = aq+r$, $r=0,1,...,a-1$ and $q=0,1,...,b-1$. It is seen that $u$ is relatively prime to $a$ if $r$ is one of the $\phi(a)$ integers $r_i$ relatively prime to $a$. Thus, the $b\phi(a)$ integers $u_i$ given by $(r_i), (a+r_i),(2a+r_i),...,((b-1)a + r_i)$ are prime to $a$. If $q$ is chosen among the $\phi(b)$ integers smaller than $b$ and mutually prime with $b$, the corresponding integers $u_i$ are relatively prime to $b$, since no factor of $b$ can divide $a$ or $q$. Thus, there are $\phi(a)\phi(b)$ integers relatively prime to $a$ and $b$ and therefore relatively prime to $a \cdot b$.
Because I wanted to understand a little more precisely, I used an example to illustrate the theorem proof. For this I have considered the case $a=3$ and $b=5$, so $\phi(3 \cdot 5)$.
I now list once all possible $u_i$, we have $u = aq + r$ with $a=3$, $q=0,1,2,3,4$ and $r = 0,1,2$ according to the proof. I highlighted in red the terms $u_i$ that are coprime to $a$.
$$ \begin{array}{ccc} 3 \cdot 0 + 0 = 0 & 3 \cdot 1 + 0 = 3 & 3 \cdot 2 + 0 = 6, \\ 3 \cdot 0 + 1 = \color{red}{1} & 3 \cdot 1 + 1 = \color{red}{4} & 3 \cdot 2 + 1 = \color{red}{7}, \\ 3 \cdot 0 + 2 = \color{red}{2} & 3 \cdot 1 + 2 = \color{red}{5} & 3 \cdot 2 + 2 = \color{red}{8}, \\ 3 \cdot 3 + 0 = 9 & 3 \cdot 4 + 0 = 12 & ,\\ 3 \cdot 3 + 1 = \color{red}{10} & 3 \cdot 4 + 1 = \color{red}{13} & ,\\ 3 \cdot 3 + 2 = \color{red}{11} & 3 \cdot 4 + 2 = \color{red}{14} & , \end{array} $$
Now the proof states: If $q$ is chosen among the $\phi(b)$ integers smaller than $b$ and mutually prime with $b$, the corresponding integers $u_i$ are relatively prime to $b$, since no factor of $b$ can divide $a$ or $q$.
This means $q = 1,2,3,4$ since this are the values smaller than $b=5$ and they are coprime to $b$. I higlight this fact with blue in the following.
The text says also that the corresponding integers $u_i$ are relatively prime to $b$, I highlight this with green in the following.
Thus we have:
$$ \begin{array}{ccc} 3 \cdot 0 + 0 = 0 & 3 \cdot \color{blue}{1} + 0 = \color{green}{3} & 3 \cdot \color{blue}{2} + 0 = \color{green}{6} , \\ 3 \cdot 0 + 1 = \color{green}{1} & 3 \cdot \color{blue}{1} + 1 = \color{green}{4} & 3 \cdot \color{blue}{2} + 1 = \color{green}{7} ,\\ 3 \cdot 0 + 2 = \color{green}{2} & 3 \cdot \color{blue}{1} + 2 = 5 & 3 \cdot \color{blue}{2} + 2 = \color{green}{8} ,\\ 3 \cdot \color{blue}{3} + 0 = \color{green}{9} & 3 \cdot \color{blue}{4} + 0 = \color{green}{12} & ,\\ 3 \cdot \color{blue}{3} + 1 = 10 & 3 \cdot \color{blue}{4} + 1 = \color{green}{13} &, \\ 3 \cdot \color{blue}{3} + 2 = \color{green}{11} & 3 \cdot \color{blue}{4} + 2 = \color{green}{14} & , \end{array} $$
And the point leads me to my question. We see for example that $3 \cdot \color{blue}{3} + 1 = 10$ form that what the theorem proof states since $q$ is coprime to $b$ this $u$ term should be relatively prime to $b$ but $u_i = 10$ and $b = 5$ are not coprime.
So my question is in fact why states the text that If $q$ is chosen among the $\phi(b)$ integers smaller than $b$ and mutually prime with $b$, the corresponding integers $u_i$ are relatively prime to $b$? But we see from the previous consideration that there is a case wher this is not true.
So did I miss something here or is the text wrong?
I hope my thoughts are plausible.