I am new to proving math theorems and not sure that my proof is correct so can anyone please tell me if its correct or not and why ?
The problem : Let $m_1 , m_2 ,\ldots, m_n $ be pairwise relatively prime integers greater than or equal to 2. Show that if $a \equiv b \ \pmod{m_i}$ for $i = 1, 2,\ldots, n$, then $a \equiv b \pmod{m}$ where $m = m_1 m_2 \cdots m_n$
Note : I can't use CRT as this problem is to proof it .
My proof:
because $a \equiv b\pmod{m_i}$ therefore $m_i |\,a-b $
therefore $a-b = c_i \, m_i$, therefore $(a-b)^n = c\,m$
therefore $m \ |\ (a-b)^n$ therefore $m \ |\ ((a-b)^n + b^n) - b^n$
therefore $(a-b)^n + b^n \equiv b^n \pmod m$
therefore $(a-b)^n \equiv 0 \pmod m$
because $(a-b)$ is a possible solution to the congruence, therefore $(a-b)\equiv 0 \pmod m$
therefore $a \equiv b \pmod m$
Forget the powers. They make things messy and aren't relevant.
$a\equiv b \pmod {m_i}$ then $m_i| a-b$
so if all the $m_i|a-b$ and the $m_i$ are relatively prime then $m=\prod m_i|a-b$.
so $a\equiv b \pmod m$.
That's all there is to it.
I guess the only thing to worry about $d|k$ and $e|k$ then $\operatorname{lcm}(e,d)|k$. But I'm assuming you have or can prove that.