Question

if a=b(mod r) and a=b(mod s) then a=b(mod [r,s])

Answer 1

Answer on Question #82606 – Math – Abstract Algebra

Question

if $a = b (\bmod r)$ and $a = b (\bmod s)$ then $a = b (\bmod [r, s])$

Solution

We need to prove that if $a = b (\bmod r)$ and $a = b (\bmod s)$ then $a = b (\bmod [r, s])$.

Proof

$$a = b (\bmod r) \Leftrightarrow a = b + r \cdot k_1, \quad \forall k_1 \in \mathbb{Z}$$

and $\exists l \in \mathbb{Z}: [r, s] = r \cdot l$.

We need to prove that

$$a = b (\bmod [r, s]) \Leftrightarrow a = b + [r, s] \cdot k_2, \quad \forall k_2 \in \mathbb{Z}$$

Just let $k_1 = l \cdot k_2$. The first formula is correct for all $k_1$, and then for $k_1 = l \cdot k_2$. Thus, the second formula is correct for all $k_2$.

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