Question

difference between a ≡ b (mod n) and a mod b?

Answer 1

i know a ≡ b (mod n) means n divides (a-b)

Answer 2

From my understanding the whole (mod n) part is sort of just an entire perspective on an equation, not really an operator. Except instead of an equation, it's considered a congruence because they're trying to be fancy, but I think the extra notation is superfluous. So as long as you see (mod n) next to an equation I interpret that to mean I can subtract n from the equation as many times as I want or add it as many times as I want without changing the meaning of the equation.

Answer 3

\(a \pmod{b}\) is "THE remainder" when you divide \(a\) by \(b\)

so the value will always be between \(0\) and \(b\)

Answer 4

\(a\equiv b \pmod{n}\) is a congruence

it is equivalent to saying \(n | (a-b)\)

which just means \(a-b = nk\) for some integer \(k\)

Answer 5

The way I understand it the 'mod b' is not an operator. it tells that your operation is done with 'modular arithmetic system':
http://en.wikipedia.org/wiki/Modular_arithmetic

In such a system, values are 'replaced' by their 'modulus' value.
For example in modular system of 8, then 1 is same as 9 which is same as 17, because their remainder in division in 8 is 1. They are all said to be 'equivalent' and therefore the notation:
$$ a\equiv b \mod 8$$
Notice the sign \(\equiv\) (equivalent) is used to say they are 'congruent modulo' as described by wiki.

Answer 6

for example :

\(\large 12 \pmod {5} = 2 \)

\(\large 13\equiv 2 \pmod {11} \)

Notice the strict differencec between an "identically equal to" and the "congruence" when using mod in both situations

Answer 7

Thank you everyone :)

@ganeshie8 , so is there a *formal* definition for a mod b? I can see that it's number, particularly the remainder of a/b.

So if I were to say, c = a mod b, then what does this mean?

Answer 8

btw, that's an *equal sign". No typo here

Answer 9

\(c = a\pmod{b}\) means \(c\) is the remainder when you divide \(a\) by \(b\) and \(c\) must be between \(0\) and\( b\)

\(c \equiv a\pmod{b}\) means \(b|(c-a)\).
for example :
\(2\equiv 7 \equiv 12\equiv 17\pmod{5}\)

Answer 10

:o no no no, that really *is* an equal sign.

Answer 11

however \(a\pmod{b}\) when used standalone evaluates to a single number

Answer 12

@ganeshie8

I believe there is a misunderstanding.

I know the definition a ≡ b (mod n).

However, as you said, a mod b (notice there is no congruent symbol) is the remainder of a/b. Which means. a mod b is a *number*.

So if i were to let c be the remainder of a/b. I.e
c = a mod b, (yes, it's an =, the equal sign) then is there a definition for this equation?

Answer 13

c = a mod b

c is the remainder when you divide a by b

thats the definition

Answer 14

but that's in words. Is there like an equation?

Answer 15

you almost never use the equality anywhere so im still not sure what you're lookign for here

Answer 16

Is mod in \(a\text{ mod }b\) a binary operator like + in \(a+b\)?

Answer 17

just like a ≡ b (mod n) means, in words, "n divides the difference of a and b". But you can also say in terms of equation,

nk = a - b, for some integer k <--- this is what i was preferring to but for c = a mod b instead. Does this make sense?

Answer 18

\[\Large \frac{5}{2} = 2 + \frac{1}{2}\] So 1 is the remainder since it can't be divided, there is no equation for the remainder.

Answer 19

below is the link between "congruence" and "equality" :

\(c\equiv a \pmod{b}\) \(\iff\) \(c\pmod{b} = a\pmod{b}\)

Answer 20

Well except I guess you could say this is the equation: \[1 = 5 (mod \ 2)\]

Answer 21

that relation might help in appreciating the difference between using mod in a congruence and standalone

Answer 22

\(1 = 5\pmod 2\) is a true statement because \(1\) is the remainder when 5 is divided by 2 and \(0\le 1\lt 2\)

Answer 23

below is NOT a true statement :
\(3 = 5\pmod{2}\)

Answer 24

Again below is a true statement :
\(3 \equiv 5\pmod{2}\) because \(2|(3-5)\)

Answer 25

huhm... maybe i was over thinking. I guess a mod b means the remainder of a/b is good enough

Answer 26

Yep. below is legit too :

\( \sqrt{2} = \pi^2 + \sqrt{2} \pmod {\pi} \)

Answer 27

convince yourself by saying "a (mod b) when used standalone just means remainder"

I bet you will almose never get to use a(mod b) standalone anywhere

Answer 28

Ok. Thanks everyone for your time ^^

Answer 29

Here are some properties if you want to look at congruences using remainders

http://gyazo.com/f3b22e981de6c7309737373e738743e5

but i feel this is useless because congruences are more neat and easier to make sense of directly