2

This is from Discrete Mathematics and its applications enter image description here

To do this proof, I used this mod property enter image description here

Here is my work enter image description here

What I did was basically expand both sides of (a-c) mod m and (b - d) mod m with that property. Then I saw that with what I was given, the two sides looked the same. I am not sure if you're allowed to use that property to do this though.

Community
  • 1
NotAsCommitedAndroider
  • 89
  • The exercise is about congruences (mod as a relation), but the corollary and your work is about mod as an operator. Do you understand the difference between the two? To get started, see here for proofs of similar congruence rules. – Bill Dubuque Feb 07 '15 at 04:08
  • is congruence just saying a≡b(modm) or that m divides a - b? – NotAsCommitedAndroider Feb 07 '15 at 04:11
  • Yes, by definition, $,a\equiv b\pmod m,$ is true iff $\ m\mid a-b\ \ $ – Bill Dubuque Feb 07 '15 at 04:13
  • @NotAsCommitedAndroider What does $a\equiv b\pmod{m}$ mean verbally? It means, "$a$ is congruent to be $b$ modulo $m$." Notationally, we see that $a\equiv b\pmod{m} \Longleftrightarrow a-b=\ell\cdot m$ for some $\ell\in\mathbb{Z}$; that is, just as Bill said, $m\mid (a-b)$. – Daniel W. Farlow Feb 07 '15 at 04:19

1 Answers1

2

From $a\equiv b\pmod{m}$ we know that $b=a+sm$ for some integer $s$. Similarly, $d=c+tm$. Substracting, we have $b-d=(a-c)+(s-t)m$, which means that $a-c\equiv b-d\pmod{m}$. $\Box$

Daniel W. Farlow
  • 22,531