Please explain congruence modulo to a retard. I read lots of

>>7672770
Two numbers (a and b) are congruent by modulo n iff the remainder from dividing a by n is the same as the remainder from dividing b by n.
I find this way of thinking about it to be simpler, as then you can think about them as being the same distance from a multiple of n.
The other way of stating it is that two numbers are congruent by mod n if the distance between them (a-b) is divisible by n.

>>7672779
To give an example with this guys story:
a = 9
b= 18
Then a and b are congruent in modulo 9,
Because if you divide 9 by 9, your remainder is 0.
If you divide 18 by 9 your remainder is also 0.
I've never heard of the concept, but I just understood it by the name. Why is this so hard for you?

>>7672781
>Why is this so hard for you?

I get the definition, but I don't get:
>The other way of stating it is that two numbers are congruent by mod n if the distance between them (a-b) is divisible by n.

Think about a clock. Suppose it's 1:00. As time passes, the time on the clock keeps "increasing", in the sense that the number on the clock increases. In 11 hours, it'll be 12:00. One hour after that, though, it won't be 13:00 (unless you live somewhere on 24 hour time); it'll loop back to 1:00.

The reading on a clock represents a number modulo 12. Even though 1 and 13 (and 25, 37, etc.) are different numbers, they are equivalent modulo 12 - as far as the clock is concerned, 1 and 13 are the same.

>>7672791
I get this, but I don't get:
>The other way of stating it is that two numbers are congruent by mod n if the distance between them (a-b) is divisible by n

>>7672789
Think about how the definitions match.
Take 9 and 24 modulo 5.
9/5 gives a remainder 4, as does 24/5.
So you can write them as 9=5+4, 24=4*5+4
Then when you subtract the two 24-9=4*5+4-5-4=3*5 a multiple of 5.
This works generally with any modulus as the remainder terms will always cancel in the subtraction, just leaving multiples of the modulo number.

>>7672802
Alright, observe that the (positive) integers equivalent to 1 modulo 12 are:

1, 13, 25, 37, 49, 61, 73, ...

To get the next integer equivalent to 1 modulo 12, you just keep adding 12. To get from 1 to 37, you have to add 12 three times; thus, the difference between 37 and 1 is 12*3, which is a multiple of 12.

Slightly more generally, to get from any of these integers to any other integer in the list, you just add or subtract 12 an appropriate number of times. This is why the difference is a multiple of 12.

Consider the ring of integers [math] \mathbb{Z} [/math]. We can define an equivalence relation [math]\sim_n[/math] on it such that
[math]x \sim_n y [/math] iff [math]x-y = k n[/math] for a [math] k \in \mathbb{Z} [/math]. We also define equivalence classes [math] [x]_{\sim_n} := \{ y \in \mathbb{Z} | x \sim_n y \}[/math].

Now we just say that [math]x \equiv y \; (\mathrm{mod} \; n)[/math] if [math] [x]_{\sim_n} = [y]_{\sim_n} [/math].

>>7672789
a = 9
b = 18
their distance is 9, which is the base of which you are taking the modulo.
>also on a semi related topic
did you know that
a mod 9 = the sum of the numbers of A,
for example 1234567 mod 9 = 1 + 2 + 3...
= 28 --> 2+ 8 = 10 --> 1 + 0 = 1.
which is equal to 1234567 mod 9.
neato huh?

>>7672865
>general case
let X be a number in base N
then X mod (N-1) = the sum of the numbers of X.
I find this fact fappable

If [math]a \equiv b \ (mod \ n)[/math], then [math]a=kn+r[/math] and [math]b=mn+r[/math], where [math]k, m \in \mathbb{Z}[/math] and [math]0 \le r < n[/math]. Now, [math]a-b=kn+r-mn-r=(k-m)n \equiv 0 \ (mod \ n)[/math].