Prove that $a \equiv b \pmod{m_1m_2}\implies a \equiv b \pmod {m_1}$

Question

So, I have this problem:

if $$a \equiv b \mod(m_1m_2)$$ then (show) $$a \equiv b \mod(m_1)$$.

I have to do a proof, but I have no idea where to begin the proof. Can someone help?

---

Proof (Edit):

We know $a\equiv b \mod(m_1m_2)$, and $m_1$ and $m_2$ are positive integers and a and b are integers. We want to show $a≡b \mod(m_1)$

Consider a positive integer $n$, and integers $x$,$y$. By the definition of congruence, $x \equiv y \mod(n)$ means $n \mid x-y$. Since $n$, $x$, and $y$ $\in \mathbb{Z}$, then by the definition of divides $n \mid x-y$ becomes $\exists_c(nc=x-y)$ for some integer $c$.

Applying this definition:

Suppose $a \equiv b \mod(m_1m_2)$; this expression becomes $m_1m_2|a-b$ by the definition of congruence. Then, since $m_1, m_2, a, b \in \mathbb{Z}$ then: $\exists_c(m_1m_2c = a-b)$ for some integer $c$.

Now, let $c$ be $m_2c$. Thus, $m_1(m_2c) = a-b$, and since $(m_2c)$ is an integer, this proves that $m_1 \mid a-b$ that is, $a \equiv b \mod(m_1)$

Q.E.D.

Answer 1

Very simply: $a \equiv b \pmod{m_1 m_2} \implies a=b+m_1m_2k$ for some $k \in \mathbb{Z}$.

And we see $a \equiv b \pmod{m_1} \implies a=b+m_1l$ for some $l \in \mathbb{Z}$.

$l=m_2k$, so $a$ is of the form $b+m_1l$ and our condition for $a \equiv b \pmod{m_1}$ is satisfied.

Answer 2

Put it this way: $a\equiv b \mod(m_1m_2) \implies a = b+q \cdot m_1m_2 = b+q*m_2m_1= b+(qm_2)m_1=b+pm_1\implies a\equiv b \mod(m_1)$. Of course, this is assuming you are in real space, that is the $m_1m_2=m_2m_1$.

Answer 3

We are given $a-b \equiv 0 \pmod {m_1 m_2}$. This means that $(a-b) = km_1 m_2$, for a constant $k$

We want to prove that $(a - b) = j m_1$, for some other constant $j$. It should be clear from here.

Set $j = km_2$. The desired conclusion follows immediately.

Answer 4

$b \mod(m_1m_2)\equiv a $

It means that $\frac{b}{m_1m_2}=k+\frac{a}{m_1m_2}$ multiply $m_2$ to both sides of equation we get:

$\frac{b}{m_1}=km_2+\frac{a}{m_1}$ and it means that

$b \mod(m_1)\equiv a $