Question

Teach me Modular arithmetic
@ParthKohli

Answer 1

please

Answer 2

Thank you for qualifying me :P

Answer 3

mod(x,y) = The remainder you get by dividing x and y.

Answer 4

so mod(1,x)=1 always?

Answer 5

x/y or y/x

Answer 6

Oops sorry

Answer 7

y/x

Answer 8

I'm new to modular mathematics too.

Answer 9

the principle is like a circular ruler ?

Answer 10

Aw snap

Answer 11

\[a\mod b=x\forall x<b,\frac{a-x}b\in\mathbb Z\]

Answer 12

\( \color{Black}{\Rightarrow x = y(modz) }\)
This means that when you divide x and z, then you get y as the remainder.

Answer 13

I'm trying to think if a more rigorous definition is required.

Answer 14

ok new notation for me can you read that out in words?

Answer 15

\( \color{Black}{\Rightarrow x = y(modz) \Longleftrightarrow mod(z,x) = y }\)

Answer 16

so i would say that like this?
if x is y mod z then mod of z and x is equal to y

Answer 17

Yes. Modulus.

Answer 18

mod is short for modulus.

Answer 19

it is modulus oh good.

Answer 20

So did you understand now?

Answer 21

so

if x is y the modulus of z then the modulus of z and x is equal to y

Answer 22

*if x is y times the modulus of z ....

Answer 23

\( \color{Black}{\Rightarrow \mod{(1,\mathbb{Z}) = 0} }\)

Answer 24

No not times.

Answer 25

http://en.wikipedia.org/wiki/Modular_arithmetic look at this.

Answer 26

You should Google for the notation.

Answer 27

oh,

ah i see the modulus of one and an integer is equal to zero,
makes sense their would be no remainder

Answer 28

Exactly.

Answer 29

is \[\mod(x,y)=\mod(y,x)\] ?

Answer 30

Try it with two numbers.

Answer 31

It'd surely work if x and y are 1.

Answer 32

It'd certainly work if \(x = y\).

Answer 33

\[\mod(5,11)=6\]
\[\mod(11,5)=5\],
or did i write that the wrong way arround

Answer 34

You are correct here :)

Answer 35

mod(5,11)=1
?

Answer 36

so the order is important

Answer 37

mod(5,11) = 5

Answer 38

Yes it is very very very important.

Answer 39

Do you want to know a use of modular arithmetic?

Answer 40

donuts ?

Answer 41

lol

Answer 42

yeah i guess another example wouldn't hurt

Answer 43

If you want to add hours according to a clock, for example you want to add x hours to y'o clock, then you'll use the clock 7 notation.

Answer 44

\(\mod{(x + y,12)}\)

Answer 45

yeh i saw you asking a question about that earlier

Answer 46

3 hours added to 11'o clock is not 14 'o clock lol

Answer 47

Yeah I did xD

Answer 48

clock 7 notation
?
why the seven?

Answer 49

12*

Answer 50

ah , that makes sense,
now

Answer 51

Hmm, I'm glad it does :D

Answer 52

then should we say
\(\mod(x,y)\) is the modulus of x with respect to y? or to base y or something like that

Answer 53

Hmm? I don't get the question. I need examples :P

Answer 54

well can the time '3 hours after 11' being displayed on a 12-hr clock
be said
the time modulus of (3) to base twelve is two
\[\mod(3+11,12)=2\]

Answer 55

Yes exactly

Answer 56

you would say to base twelve?

Answer 57

No

Answer 58

14 modulus 12

Answer 59

and you say somthing about your fourmula @badreferences

\[a\mod b=x\forall x<b,\frac{a-x}b\in\mathbb Z\]
how do you say it?

Answer 60

\(\forall\) means "for all", or "these are the conditions for which the statement is true".

Answer 61

the modulus of b is
for all x less that b
and withan integer remainder of (a minus x) over b
is equal to y .

Answer 62

is that right?

Answer 63

?<

Answer 64

Well, actually, Parth is teaching you I think another notation of modulo arithmetic. I'm more familiar with the modulo as an operator.

And right now I have a bit of a headache, so you can ignore me. ;) It's not that important. Parth was pretty dead on.

Answer 65

Bwaha I'm new too so I'm not the perfect teacher