lgbasallote
  • lgbasallote
if a = -11 and b = 3 then what is a div b and a mod b
Discrete Math
chestercat
  • chestercat
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anonymous
  • anonymous
mod?
lgbasallote
  • lgbasallote
yes mod
bahrom7893
  • bahrom7893
modulus

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bahrom7893
  • bahrom7893
mod is 2
bahrom7893
  • bahrom7893
idk what div is.
ParthKohli
  • ParthKohli
By div, do you mean quotient?
lgbasallote
  • lgbasallote
hmm
lgbasallote
  • lgbasallote
...why are you asking me?
ParthKohli
  • ParthKohli
Hmm... mod is 1.
jiteshmeghwal9
  • jiteshmeghwal9
becoz i m not getting ur question :(
ParthKohli
  • ParthKohli
\[\large \rm 3 \cdot (-4) + \underbrace{1}_{\large modulus} = -11\]
lgbasallote
  • lgbasallote
and what does -11/3, -11|3| mean @jiteshmeghwal9 ?
lgbasallote
  • lgbasallote
and why is @bahrom7893 wrong @ParthKohli ?
ParthKohli
  • ParthKohli
Depends on your question...\[\large \rm a ~div~b=a\div b?\]
lgbasallote
  • lgbasallote
yes
ParthKohli
  • ParthKohli
@lgbasallote How is the modulus 2? Can you show so?
lgbasallote
  • lgbasallote
again...why are you asking me?
ParthKohli
  • ParthKohli
Then,\[\rm \large -11~div~3 ={-11 \over 3} = -3{2 \over 3}\]
lgbasallote
  • lgbasallote
mmhmm
bahrom7893
  • bahrom7893
Parth... i thought modulus was a remainder when a is divided by b. That's what it means in programming.
lgbasallote
  • lgbasallote
it is...
ParthKohli
  • ParthKohli
Yes, that's what...
jiteshmeghwal9
  • 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
  • bahrom7893
oh it's -11.. shoot i got my mod wrong
lgbasallote
  • lgbasallote
so then...?
bahrom7893
  • bahrom7893
A mod B is A%B, or the remainder when A is divided by B
jiteshmeghwal9
  • jiteshmeghwal9
-11/3*
lgbasallote
  • lgbasallote
mod is modulo @jiteshmeghwal9
ParthKohli
  • ParthKohli
@jiteshmeghwal9 a mod b is the remainder when a is divided by b. Yep.
jiteshmeghwal9
  • jiteshmeghwal9
ohh ok !
bahrom7893
  • bahrom7893
|dw:1350970551590:dw|
bahrom7893
  • bahrom7893
Mod is -2
lgbasallote
  • lgbasallote
i agree..
jiteshmeghwal9
  • jiteshmeghwal9
|dw:1350970548817:dw|
bahrom7893
  • bahrom7893
|dw:1350970607290:dw|
lgbasallote
  • lgbasallote
question is... why is @ParthKohli saying something different...
ParthKohli
  • ParthKohli
@bahrom7893 now prove that -2 mod 3 is equivalent to 1 mod 3.
lgbasallote
  • lgbasallote
^?
bahrom7893
  • bahrom7893
lol parth i have no clue how to prove that.
bahrom7893
  • bahrom7893
but it must be hahah
ParthKohli
  • ParthKohli
Dude -2 mod 3 = 1 1 mod 3 = 1 So they are equivalent.
lgbasallote
  • lgbasallote
-2 mod 3 is -2...
jiteshmeghwal9
  • jiteshmeghwal9
\[\Huge{-1 \times \frac{11}{3} \approx -3.67}\]
lgbasallote
  • lgbasallote
^?
ParthKohli
  • ParthKohli
\[\rm \large -3 \cdot 1+\underbrace1_{modulus} =-2\]
ParthKohli
  • ParthKohli
According to definition, actually, modulus is 'c' in the following: n = a x b + c
lgbasallote
  • lgbasallote
since when did -2 divided by 3 become 1...you have to teach me some arithmetic master
ParthKohli
  • ParthKohli
Modulus is NOT NECESSARILY THE REMAINDER, SIRE.
lgbasallote
  • lgbasallote
....
ParthKohli
  • ParthKohli
Do you know the Euclidean Algorithm?
lgbasallote
  • lgbasallote
of course
lgbasallote
  • lgbasallote
do you?
ParthKohli
  • ParthKohli
Yes.
ParthKohli
  • ParthKohli
Use the Euclidean Algorithm then.
ParthKohli
  • ParthKohli
a variation of Euclidean Algorithm actually...
lgbasallote
  • lgbasallote
euclidian algorithm is used for finding gcf....
ParthKohli
  • 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
  • ParthKohli
The above is the actual definition of modulus.
lgbasallote
  • lgbasallote
yes
lgbasallote
  • lgbasallote
b is n/a
ParthKohli
  • ParthKohli
And that definition does satisfy the following:\[\rm \large 3\cdot(-1)+1 = -2\]Therefore\[-2\mod 3=1\]
lgbasallote
  • lgbasallote
how?
ParthKohli
  • ParthKohli
compare the above with the definition:\[\rm \large a = 3\]\[\large \rm b = -1\]\[\rm \large c = 1 = -2mod3\]
ParthKohli
  • ParthKohli
And\[\rm \large n = -2\]
lgbasallote
  • lgbasallote
how is b = -1?
ParthKohli
  • ParthKohli
\[\rm \large 3\cdot(-1) + 1 = -2\]\[\rm \large a\cdot b +c=n \]
lgbasallote
  • lgbasallote
you're not really telling me anything...you're just circling around
ParthKohli
  • ParthKohli
lol I am... I just proved you that \(\rm \large (-2) mod3 = 1\)
lgbasallote
  • lgbasallote
you just have -2 mod 3 = ? show me how you get that to be 1
ParthKohli
  • ParthKohli
I just did...
lgbasallote
  • lgbasallote
you didn't really prove...because you just assumed a value of b and c and equated them
ParthKohli
  • ParthKohli
That's the definition of modulus, dude.
lgbasallote
  • lgbasallote
show me a solution that tells me -2 mod 3 is 1
lgbasallote
  • lgbasallote
you have two unknows..you can't use the definition
ParthKohli
  • ParthKohli
I'd show you all of it in one post.
lgbasallote
  • lgbasallote
sure.
ParthKohli
  • 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
  • ParthKohli
Would you like it with hearts and smileys? Oh, also see that b is an integer.
lgbasallote
  • lgbasallote
but the question you refuse to answer is... how do you get those values???
ParthKohli
  • ParthKohli
What values?
lgbasallote
  • lgbasallote
1and -1..you can't pull them out of nowhere
ParthKohli
  • 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
  • lgbasallote
i'm not asking about the definition! i'm asking about your numbers
ParthKohli
  • ParthKohli
Oh, and you have to try and test in the Euclidean Algorithm too.
lgbasallote
  • lgbasallote
i can also say 3*(-2) + 4 = -2 so i can say -2 mod 3 is 4
lgbasallote
  • lgbasallote
so you're wrong by *my* answer
lgbasallote
  • lgbasallote
-2 mod 3 should be 4 not 1
ParthKohli
  • ParthKohli
Yes, it's also 4.
lgbasallote
  • lgbasallote
...........
lgbasallote
  • lgbasallote
no it isn't
lgbasallote
  • lgbasallote
-2 mod 3 is 1
ParthKohli
  • ParthKohli
That's the fun of modular arithmetic: you have -2 mod 3 = 1,4,7,10...
lgbasallote
  • lgbasallote
nope
ParthKohli
  • ParthKohli
Yes. Would you like a machine to answer it for you?
lgbasallote
  • lgbasallote
try wolfram
ParthKohli
  • ParthKohli
Sure.
lgbasallote
  • lgbasallote
you'll see it's 1
lgbasallote
  • lgbasallote
or...you can just watch more khanacademy videos....
ParthKohli
  • ParthKohli
XD
1 Attachment
ParthKohli
  • ParthKohli
http://www.wolframalpha.com/input/?i=-2+mod+3 You may have a look.
ParthKohli
  • ParthKohli
And remember, there's no equality in modulus... there's congruence.\[\rm \large a \mod b \equiv c\]
lgbasallote
  • lgbasallote
you don't really get what my question was...do you....
ParthKohli
  • ParthKohli
Though I have to admit, I forgot the above too...
ParthKohli
  • ParthKohli
-11 mod 3
ParthKohli
  • ParthKohli
\[\rm \large 3\cdot(-4)+1 = -11 \]Right?
ParthKohli
  • ParthKohli
\[-11\mod3\equiv 1\]But also...
ParthKohli
  • ParthKohli
\[\rm \large 3\cdot(-5) + 4 = -11\]
ParthKohli
  • ParthKohli
So we can say that\[\rm \Huge -11\mod3 \equiv1+3k\]where k is an integer.
ParthKohli
  • ParthKohli
You're welcome. @lgbasallote
lgbasallote
  • lgbasallote
that was *not* what i was asking
ParthKohli
  • ParthKohli
You're welcome anyway. I'm glad I could be of help.
lgbasallote
  • 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
  • ParthKohli
Sorry, if it was modulo, then it's just the remainder.

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