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

Discrete Math
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mod?
yes mod
modulus

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Other answers:

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
Parth... i thought modulus was a remainder when a is divided by b. That's what it means in programming.
it is...
Yes, that's what...
\[\Huge{\frac{-11}{2}=a \div b}\] accrding to ur question . but i'm not getting what do u mean by \(\Large{a\space mod\space b}\). @lgbasallote
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....
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.
The above is the actual definition of modulus.
yes
b is n/a
And that definition does satisfy the following:\[\rm \large 3\cdot(-1)+1 = -2\]Therefore\[-2\mod 3=1\]
how?
compare the above with the definition:\[\rm \large a = 3\]\[\large \rm b = -1\]\[\rm \large c = 1 = -2mod3\]
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.
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
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
Do you think that my answer works according to the definition? Oh, and you have to try and test in the Euclidean Algorithm too.
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
1 Attachment
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.
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...
Sorry, if it was modulo, then it's just the remainder.

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