lgbasallote
if a = -11 and b = 3
then what is
a div b
and
a mod b
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Dwade03
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mod?
lgbasallote
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yes mod
bahrom7893
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modulus
bahrom7893
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mod is 2
bahrom7893
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idk what div is.
ParthKohli
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By div, do you mean quotient?
lgbasallote
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hmm
lgbasallote
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...why are you asking me?
ParthKohli
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Hmm... mod is 1.
jiteshmeghwal9
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becoz i m not getting ur question :(
ParthKohli
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\[\large \rm 3 \cdot (-4) + \underbrace{1}_{\large modulus} = -11\]
lgbasallote
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and what does -11/3, -11|3| mean @jiteshmeghwal9 ?
lgbasallote
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and why is @bahrom7893 wrong @ParthKohli ?
ParthKohli
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Depends on your question...\[\large \rm a ~div~b=a\div b?\]
lgbasallote
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yes
ParthKohli
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@lgbasallote How is the modulus 2? Can you show so?
lgbasallote
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again...why are you asking me?
ParthKohli
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Then,\[\rm \large -11~div~3 ={-11 \over 3} = -3{2 \over 3}\]
lgbasallote
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mmhmm
bahrom7893
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Parth... i thought modulus was a remainder when a is divided by b. That's what it means in programming.
lgbasallote
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it is...
ParthKohli
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Yes, that's what...
jiteshmeghwal9
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\[\Huge{\frac{-11}{2}=a \div b}\] accrding to ur question .
but i'm not getting what do u mean by \(\Large{a\space mod\space b}\).
@lgbasallote
bahrom7893
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oh it's -11.. shoot i got my mod wrong
lgbasallote
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so then...?
bahrom7893
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A mod B is A%B, or the remainder when A is divided by B
jiteshmeghwal9
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-11/3*
lgbasallote
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mod is modulo @jiteshmeghwal9
ParthKohli
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@jiteshmeghwal9 a mod b is the remainder when a is divided by b. Yep.
jiteshmeghwal9
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ohh ok !
bahrom7893
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|dw:1350970551590:dw|
bahrom7893
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Mod is -2
lgbasallote
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i agree..
jiteshmeghwal9
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|dw:1350970548817:dw|
bahrom7893
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|dw:1350970607290:dw|
lgbasallote
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question is... why is @ParthKohli saying something different...
ParthKohli
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@bahrom7893 now prove that
-2 mod 3 is equivalent to 1 mod 3.
lgbasallote
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^?
bahrom7893
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lol parth i have no clue how to prove that.
bahrom7893
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but it must be hahah
ParthKohli
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Dude
-2 mod 3 = 1
1 mod 3 = 1
So they are equivalent.
lgbasallote
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-2 mod 3 is -2...
jiteshmeghwal9
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\[\Huge{-1 \times \frac{11}{3} \approx -3.67}\]
lgbasallote
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^?
ParthKohli
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\[\rm \large -3 \cdot 1+\underbrace1_{modulus} =-2\]
ParthKohli
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According to definition, actually, modulus is 'c' in the following:
n = a x b + c
lgbasallote
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since when did -2 divided by 3 become 1...you have to teach me some arithmetic master
ParthKohli
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Modulus is NOT NECESSARILY THE REMAINDER, SIRE.
lgbasallote
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....
ParthKohli
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Do you know the Euclidean Algorithm?
lgbasallote
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of course
lgbasallote
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do you?
ParthKohli
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Yes.
ParthKohli
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Use the Euclidean Algorithm then.
ParthKohli
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a variation of Euclidean Algorithm actually...
lgbasallote
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euclidian algorithm is used for finding gcf....
ParthKohli
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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
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The above is the actual definition of modulus.
lgbasallote
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yes
lgbasallote
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b is n/a
ParthKohli
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And that definition does satisfy the following:\[\rm \large 3\cdot(-1)+1 = -2\]Therefore\[-2\mod 3=1\]
lgbasallote
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how?
ParthKohli
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compare the above with the definition:\[\rm \large a = 3\]\[\large \rm b = -1\]\[\rm \large c = 1 = -2mod3\]
ParthKohli
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And\[\rm \large n = -2\]
lgbasallote
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how is b = -1?
ParthKohli
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\[\rm \large 3\cdot(-1) + 1 = -2\]\[\rm \large a\cdot b +c=n \]
lgbasallote
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you're not really telling me anything...you're just circling around
ParthKohli
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lol I am... I just proved you that \(\rm \large (-2) mod3 = 1\)
lgbasallote
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you just have -2 mod 3 = ?
show me how you get that to be 1
ParthKohli
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I just did...
lgbasallote
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you didn't really prove...because you just assumed a value of b and c and equated them
ParthKohli
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That's the definition of modulus, dude.
lgbasallote
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show me a solution that tells me -2 mod 3 is 1
lgbasallote
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you have two unknows..you can't use the definition
ParthKohli
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I'd show you all of it in one post.
lgbasallote
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sure.
ParthKohli
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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
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Would you like it with hearts and smileys?
Oh, also see that b is an integer.
lgbasallote
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but the question you refuse to answer is... how do you get those values???
ParthKohli
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What values?
lgbasallote
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1and -1..you can't pull them out of nowhere
ParthKohli
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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
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i'm not asking about the definition! i'm asking about your numbers
ParthKohli
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Oh, and you have to try and test in the Euclidean Algorithm too.
lgbasallote
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i can also say 3*(-2) + 4 = -2
so i can say -2 mod 3 is 4
lgbasallote
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so you're wrong by *my* answer
lgbasallote
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-2 mod 3 should be 4 not 1
ParthKohli
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Yes, it's also 4.
lgbasallote
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...........
lgbasallote
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no it isn't
lgbasallote
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-2 mod 3 is 1
ParthKohli
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That's the fun of modular arithmetic: you have -2 mod 3 = 1,4,7,10...
lgbasallote
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nope
ParthKohli
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Yes. Would you like a machine to answer it for you?
lgbasallote
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try wolfram
ParthKohli
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Sure.
lgbasallote
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you'll see it's 1
lgbasallote
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or...you can just watch more khanacademy videos....
ParthKohli
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XD
ParthKohli
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And remember, there's no equality in modulus... there's congruence.\[\rm \large a \mod b \equiv c\]
lgbasallote
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you don't really get what my question was...do you....
ParthKohli
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Though I have to admit, I forgot the above too...
ParthKohli
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-11 mod 3
ParthKohli
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\[\rm \large 3\cdot(-4)+1 = -11 \]Right?
ParthKohli
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\[-11\mod3\equiv 1\]But also...
ParthKohli
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\[\rm \large 3\cdot(-5) + 4 = -11\]
ParthKohli
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So we can say that\[\rm \Huge -11\mod3 \equiv1+3k\]where k is an integer.
ParthKohli
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You're welcome. @lgbasallote
lgbasallote
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that was *not* what i was asking
ParthKohli
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You're welcome anyway.
I'm glad I could be of help.
lgbasallote
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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
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Sorry, if it was modulo, then it's just the remainder.