1
$\begingroup$

If $a\equiv b \pmod {m_i}$, $1\leq i\leq k$, there $m_1,m_2,\dots,m_k$ relatively prime, then $a\equiv b\pmod{m_1m_2\cdots m_k}$

My attempt: $$\frac{a-b}{m_i}=t_i, t_i\in Z$$ $$\frac{(a-b)^k}{m_1m_2\cdots m_k}=t_1t_2\cdots t_k$$ $$(a-b)^k\equiv 0 \pmod {m_1m_2\cdots m_k}$$ (not sure about this step) $$a-b\equiv 0 \pmod {m_1m_2\cdots m_k}$$ $$a\equiv b \pmod {m_1m_2\cdots m_k}$$ Is there an error in my proof? I didn't use the fact that $m_1,m_2,\dots,m_k$ are relatively prime and I guess it is given for a reason.

$\endgroup$
  • $\begingroup$ This is a part of the Chinese Reminder Theorem. $\endgroup$ – Lior B-S Mar 21 '13 at 9:55
1
$\begingroup$

This isn't sure right: $2^k = 0 (mod~4)$ for any $k \ge 2$, but $2\ne 0(mod~4)$.

You can prove your statement easily, note that $$\frac{a-b}{m_i}=t_i, t_i\in Z$$ [$m_i$ are coprime] $$\frac{(a-b)}{m_1m_2\cdots m_k} \in Z$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.