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mod?

yes mod

modulus

mod is 2

idk what div is.

By div, do you mean quotient?

hmm

...why are you asking me?

Hmm... mod is 1.

becoz i m not getting ur question :(

\[\large \rm 3 \cdot (-4) + \underbrace{1}_{\large modulus} = -11\]

and what does -11/3, -11|3| mean @jiteshmeghwal9 ?

and why is @bahrom7893 wrong @ParthKohli ?

Depends on your question...\[\large \rm a ~div~b=a\div b?\]

yes

@lgbasallote How is the modulus 2? Can you show so?

again...why are you asking me?

Then,\[\rm \large -11~div~3 ={-11 \over 3} = -3{2 \over 3}\]

mmhmm

it is...

Yes, that's what...

oh it's -11.. shoot i got my mod wrong

so then...?

A mod B is A%B, or the remainder when A is divided by B

-11/3*

mod is modulo @jiteshmeghwal9

@jiteshmeghwal9 a mod b is the remainder when a is divided by b. Yep.

ohh ok !

|dw:1350970551590:dw|

Mod is -2

i agree..

|dw:1350970548817:dw|

|dw:1350970607290:dw|

question is... why is @ParthKohli saying something different...

@bahrom7893 now prove that
-2 mod 3 is equivalent to 1 mod 3.

^?

lol parth i have no clue how to prove that.

but it must be hahah

Dude
-2 mod 3 = 1
1 mod 3 = 1
So they are equivalent.

-2 mod 3 is -2...

\[\Huge{-1 \times \frac{11}{3} \approx -3.67}\]

^?

\[\rm \large -3 \cdot 1+\underbrace1_{modulus} =-2\]

According to definition, actually, modulus is 'c' in the following:
n = a x b + c

since when did -2 divided by 3 become 1...you have to teach me some arithmetic master

Modulus is NOT NECESSARILY THE REMAINDER, SIRE.

....

Do you know the Euclidean Algorithm?

of course

do you?

Yes.

Use the Euclidean Algorithm then.

a variation of Euclidean Algorithm actually...

euclidian algorithm is used for finding gcf....

The above is the actual definition of modulus.

yes

b is n/a

how?

And\[\rm \large n = -2\]

how is b = -1?

\[\rm \large 3\cdot(-1) + 1 = -2\]\[\rm \large a\cdot b +c=n \]

you're not really telling me anything...you're just circling around

lol I am... I just proved you that \(\rm \large (-2) mod3 = 1\)

you just have -2 mod 3 = ?
show me how you get that to be 1

I just did...

you didn't really prove...because you just assumed a value of b and c and equated them

That's the definition of modulus, dude.

show me a solution that tells me -2 mod 3 is 1

you have two unknows..you can't use the definition

I'd show you all of it in one post.

sure.

Would you like it with hearts and smileys?
Oh, also see that b is an integer.

but the question you refuse to answer is... how do you get those values???

What values?

1and -1..you can't pull them out of nowhere

i'm not asking about the definition! i'm asking about your numbers

Oh, and you have to try and test in the Euclidean Algorithm too.

i can also say 3*(-2) + 4 = -2
so i can say -2 mod 3 is 4

so you're wrong by *my* answer

-2 mod 3 should be 4 not 1

Yes, it's also 4.

...........

no it isn't

-2 mod 3 is 1

That's the fun of modular arithmetic: you have -2 mod 3 = 1,4,7,10...

nope

Yes. Would you like a machine to answer it for you?

try wolfram

Sure.

you'll see it's 1

or...you can just watch more khanacademy videos....

XD

http://www.wolframalpha.com/input/?i=-2+mod+3 You may have a look.

And remember, there's no equality in modulus... there's congruence.\[\rm \large a \mod b \equiv c\]

you don't really get what my question was...do you....

Though I have to admit, I forgot the above too...

-11 mod 3

\[\rm \large 3\cdot(-4)+1 = -11 \]Right?

\[-11\mod3\equiv 1\]But also...

\[\rm \large 3\cdot(-5) + 4 = -11\]

So we can say that\[\rm \Huge -11\mod3 \equiv1+3k\]where k is an integer.

You're welcome. @lgbasallote

that was *not* what i was asking

You're welcome anyway.
I'm glad I could be of help.

Sorry, if it was modulo, then it's just the remainder.