if a = -11 and b = 3
then what is
a div b
and
a mod b

- lgbasallote

if a = -11 and b = 3
then what is
a div b
and
a mod b

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- anonymous

mod?

- lgbasallote

yes mod

- bahrom7893

modulus

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## More answers

- bahrom7893

mod is 2

- bahrom7893

idk what div is.

- ParthKohli

By div, do you mean quotient?

- lgbasallote

hmm

- lgbasallote

...why are you asking me?

- ParthKohli

Hmm... mod is 1.

- jiteshmeghwal9

becoz i m not getting ur question :(

- ParthKohli

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

- lgbasallote

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

- lgbasallote

and why is @bahrom7893 wrong @ParthKohli ?

- ParthKohli

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

- lgbasallote

yes

- ParthKohli

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

- lgbasallote

again...why are you asking me?

- ParthKohli

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

- lgbasallote

mmhmm

- bahrom7893

Parth... i thought modulus was a remainder when a is divided by b. That's what it means in programming.

- lgbasallote

it is...

- ParthKohli

Yes, that's what...

- jiteshmeghwal9

\[\Huge{\frac{-11}{2}=a \div b}\] accrding to ur question .
but i'm not getting what do u mean by \(\Large{a\space mod\space b}\).
@lgbasallote

- bahrom7893

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

- lgbasallote

so then...?

- bahrom7893

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

- jiteshmeghwal9

-11/3*

- lgbasallote

mod is modulo @jiteshmeghwal9

- ParthKohli

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

- jiteshmeghwal9

ohh ok !

- bahrom7893

|dw:1350970551590:dw|

- bahrom7893

Mod is -2

- lgbasallote

i agree..

- jiteshmeghwal9

|dw:1350970548817:dw|

- bahrom7893

|dw:1350970607290:dw|

- lgbasallote

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

- ParthKohli

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

- lgbasallote

^?

- bahrom7893

lol parth i have no clue how to prove that.

- bahrom7893

but it must be hahah

- ParthKohli

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

- lgbasallote

-2 mod 3 is -2...

- jiteshmeghwal9

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

- lgbasallote

^?

- ParthKohli

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

- ParthKohli

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

- lgbasallote

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

- ParthKohli

Modulus is NOT NECESSARILY THE REMAINDER, SIRE.

- lgbasallote

....

- ParthKohli

Do you know the Euclidean Algorithm?

- lgbasallote

of course

- lgbasallote

do you?

- ParthKohli

Yes.

- ParthKohli

Use the Euclidean Algorithm then.

- ParthKohli

a variation of Euclidean Algorithm actually...

- lgbasallote

euclidian algorithm is used for finding gcf....

- ParthKohli

If this is true:\[\large \rm a\cdot b +c=n\]Then this is true:\[\rm \large n~mod~a=c\]Where \(\rm b\) is an integer.

- ParthKohli

The above is the actual definition of modulus.

- lgbasallote

yes

- lgbasallote

b is n/a

- ParthKohli

And that definition does satisfy the following:\[\rm \large 3\cdot(-1)+1 = -2\]Therefore\[-2\mod 3=1\]

- lgbasallote

how?

- ParthKohli

compare the above with the definition:\[\rm \large a = 3\]\[\large \rm b = -1\]\[\rm \large c = 1 = -2mod3\]

- ParthKohli

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

- lgbasallote

how is b = -1?

- ParthKohli

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

- lgbasallote

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

- ParthKohli

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

- lgbasallote

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

- ParthKohli

I just did...

- lgbasallote

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

- ParthKohli

That's the definition of modulus, dude.

- lgbasallote

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

- lgbasallote

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

- ParthKohli

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

- lgbasallote

sure.

- ParthKohli

The definition is as follows:-
If this is true,
a * b + c = n
Then this is also true,
n mod a = c
Now, this is true:
3 * (-1) + 1 = -2
Then, this is ought to be true:
-2 mod 3 = 1

- ParthKohli

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

- lgbasallote

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

- ParthKohli

What values?

- lgbasallote

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

- ParthKohli

Do you think that my answer works according to the definition?
Oh, and you have to try and test in the Euclidean Algorithm too.

- lgbasallote

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

- ParthKohli

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

- lgbasallote

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

- lgbasallote

so you're wrong by *my* answer

- lgbasallote

-2 mod 3 should be 4 not 1

- ParthKohli

Yes, it's also 4.

- lgbasallote

...........

- lgbasallote

no it isn't

- lgbasallote

-2 mod 3 is 1

- ParthKohli

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

- lgbasallote

nope

- ParthKohli

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

- lgbasallote

try wolfram

- ParthKohli

Sure.

- lgbasallote

you'll see it's 1

- lgbasallote

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

- ParthKohli

XD

##### 1 Attachment

- ParthKohli

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

- ParthKohli

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

- lgbasallote

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

- ParthKohli

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

- ParthKohli

-11 mod 3

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

You're welcome. @lgbasallote

- lgbasallote

that was *not* what i was asking

- ParthKohli

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

- lgbasallote

i already know everything you were saying (you just overcomplicated everything by circling around)....no one was able to answer my question actually...
i'll just wait for the big boys to go online...then i can ask them...

- ParthKohli

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

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