Calculus showing with mods

Question

I got this problem from my Calculus II teacher, I have no idea how to approach it...

Show that if $a,b,c$, and $d$ are integers such that $a\equiv b\pmod m$ and $c\equiv d\pmod m$, then $a+c\equiv b+d\pmod m$ and $ac\equiv bd\pmod m$.

If anyone can help, thanks!

Rewrite $b$ and $d$ as $a + mk_1$ and $c + mk_2$ by the definition of modular equivalence. Then calculate $b+d$ and $bd$. — Bruce Zheng, Oct 16 '14 at 01:45
So, would I literally do: $b+d=a+mk_1+c+mk_2$ and $bd=2ac+cmk_1+amk_2$? — KFC, Oct 16 '14 at 01:52
Just $ac$ on the second one, but yes. Can you use the definition of $x \equiv y \mod m$ to show the end result? — Bruce Zheng, Oct 16 '14 at 01:53

Aaron Maroja · Answer 1 · 2014-10-16T02:33:39.167

0

If $a \equiv b\ \ \text{mod}\ m$ and $c \equiv d\ \ \text{mod}\ m$ it follows* that $m |(a-b)$ and $m|(c-d)$ then there exist $r_1,r_2 \in \mathbb{Z}$ such that $$a-b = m\ r_1\ \ \text{and}\ \ c-d = m\ r_2$$ What can you say about $(a-b) + (c-d)$?

You take from here.

Same idea goes for $ac \equiv bd\ \ \text{mod}\ m$.

(*) Show that $a \equiv b\ \ \text{mod}\ \ m \Leftrightarrow m|(a-b)$ (It folllows directly from two facts : (i) Euclidean Division (ii) if $x|(y+z)$ and $x|y$ then $x|z$).

edited Oct 16 '14 at 02:33

answered Oct 16 '14 at 01:52

Aaron Maroja

17,571

I can say that it is equal to $m(r1+r2)$? – KFC Oct 16 '14 at 01:57
That's right! You got it. – Aaron Maroja Oct 16 '14 at 01:57
Wait, so the final answer is $m(r1+r2)$? – KFC Oct 16 '14 at 01:58
You want to get that $m|(a-b)+(c-d)$. If you let $r_3=r_1+r_2 \in \mathbb{Z}$ then you already have the result, because $(a-b) + (c-d)$ is a multiple of $m$. – Aaron Maroja Oct 16 '14 at 02:00
I'm not 100% sure as to what that means. – KFC Oct 16 '14 at 02:13
It means that $(a-b) + (c-d)\equiv 0\ \ \text{mod}\ \ m \Rightarrow a + c\equiv b+d\ \ \text{mod}\ \ m $. – Aaron Maroja Oct 16 '14 at 02:25

Calculus showing with mods

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