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lgbasallote Group Title

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

  • one year ago
  • one year ago

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  1. Dwade03 Group Title
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    mod?

    • one year ago
  2. lgbasallote Group Title
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    yes mod

    • one year ago
  3. bahrom7893 Group Title
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    modulus

    • one year ago
  4. bahrom7893 Group Title
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    mod is 2

    • one year ago
  5. bahrom7893 Group Title
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    idk what div is.

    • one year ago
  6. ParthKohli Group Title
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    By div, do you mean quotient?

    • one year ago
  7. lgbasallote Group Title
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    hmm

    • one year ago
  8. lgbasallote Group Title
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    ...why are you asking me?

    • one year ago
  9. ParthKohli Group Title
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    Hmm... mod is 1.

    • one year ago
  10. jiteshmeghwal9 Group Title
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    becoz i m not getting ur question :(

    • one year ago
  11. ParthKohli Group Title
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    \[\large \rm 3 \cdot (-4) + \underbrace{1}_{\large modulus} = -11\]

    • one year ago
  12. lgbasallote Group Title
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    and what does -11/3, -11|3| mean @jiteshmeghwal9 ?

    • one year ago
  13. lgbasallote Group Title
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    and why is @bahrom7893 wrong @ParthKohli ?

    • one year ago
  14. ParthKohli Group Title
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    Depends on your question...\[\large \rm a ~div~b=a\div b?\]

    • one year ago
  15. lgbasallote Group Title
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    yes

    • one year ago
  16. ParthKohli Group Title
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    @lgbasallote How is the modulus 2? Can you show so?

    • one year ago
  17. lgbasallote Group Title
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    again...why are you asking me?

    • one year ago
  18. ParthKohli Group Title
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    Then,\[\rm \large -11~div~3 ={-11 \over 3} = -3{2 \over 3}\]

    • one year ago
  19. lgbasallote Group Title
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    mmhmm

    • one year ago
  20. bahrom7893 Group Title
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    Parth... i thought modulus was a remainder when a is divided by b. That's what it means in programming.

    • one year ago
  21. lgbasallote Group Title
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    it is...

    • one year ago
  22. ParthKohli Group Title
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    Yes, that's what...

    • one year ago
  23. jiteshmeghwal9 Group Title
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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

    • one year ago
  24. bahrom7893 Group Title
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    oh it's -11.. shoot i got my mod wrong

    • one year ago
  25. lgbasallote Group Title
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    so then...?

    • one year ago
  26. bahrom7893 Group Title
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    A mod B is A%B, or the remainder when A is divided by B

    • one year ago
  27. jiteshmeghwal9 Group Title
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    -11/3*

    • one year ago
  28. lgbasallote Group Title
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    mod is modulo @jiteshmeghwal9

    • one year ago
  29. ParthKohli Group Title
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    @jiteshmeghwal9 a mod b is the remainder when a is divided by b. Yep.

    • one year ago
  30. jiteshmeghwal9 Group Title
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    ohh ok !

    • one year ago
  31. bahrom7893 Group Title
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    |dw:1350970551590:dw|

    • one year ago
  32. bahrom7893 Group Title
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    Mod is -2

    • one year ago
  33. lgbasallote Group Title
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    i agree..

    • one year ago
  34. jiteshmeghwal9 Group Title
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    |dw:1350970548817:dw|

    • one year ago
  35. bahrom7893 Group Title
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    |dw:1350970607290:dw|

    • one year ago
  36. lgbasallote Group Title
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    question is... why is @ParthKohli saying something different...

    • one year ago
  37. ParthKohli Group Title
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    @bahrom7893 now prove that -2 mod 3 is equivalent to 1 mod 3.

    • one year ago
  38. lgbasallote Group Title
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    ^?

    • one year ago
  39. bahrom7893 Group Title
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    lol parth i have no clue how to prove that.

    • one year ago
  40. bahrom7893 Group Title
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    but it must be hahah

    • one year ago
  41. ParthKohli Group Title
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    Dude -2 mod 3 = 1 1 mod 3 = 1 So they are equivalent.

    • one year ago
  42. lgbasallote Group Title
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    -2 mod 3 is -2...

    • one year ago
  43. jiteshmeghwal9 Group Title
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    \[\Huge{-1 \times \frac{11}{3} \approx -3.67}\]

    • one year ago
  44. lgbasallote Group Title
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    ^?

    • one year ago
  45. ParthKohli Group Title
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    \[\rm \large -3 \cdot 1+\underbrace1_{modulus} =-2\]

    • one year ago
  46. ParthKohli Group Title
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    According to definition, actually, modulus is 'c' in the following: n = a x b + c

    • one year ago
  47. lgbasallote Group Title
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    since when did -2 divided by 3 become 1...you have to teach me some arithmetic master

    • one year ago
  48. ParthKohli Group Title
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    Modulus is NOT NECESSARILY THE REMAINDER, SIRE.

    • one year ago
  49. lgbasallote Group Title
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    ....

    • one year ago
  50. ParthKohli Group Title
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    Do you know the Euclidean Algorithm?

    • one year ago
  51. lgbasallote Group Title
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    of course

    • one year ago
  52. lgbasallote Group Title
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    do you?

    • one year ago
  53. ParthKohli Group Title
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    Yes.

    • one year ago
  54. ParthKohli Group Title
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    Use the Euclidean Algorithm then.

    • one year ago
  55. ParthKohli Group Title
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    a variation of Euclidean Algorithm actually...

    • one year ago
  56. lgbasallote Group Title
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    euclidian algorithm is used for finding gcf....

    • one year ago
  57. ParthKohli Group Title
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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.

    • one year ago
  58. ParthKohli Group Title
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    The above is the actual definition of modulus.

    • one year ago
  59. lgbasallote Group Title
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    yes

    • one year ago
  60. lgbasallote Group Title
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    b is n/a

    • one year ago
  61. ParthKohli Group Title
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    And that definition does satisfy the following:\[\rm \large 3\cdot(-1)+1 = -2\]Therefore\[-2\mod 3=1\]

    • one year ago
  62. lgbasallote Group Title
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    how?

    • one year ago
  63. ParthKohli Group Title
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    compare the above with the definition:\[\rm \large a = 3\]\[\large \rm b = -1\]\[\rm \large c = 1 = -2mod3\]

    • one year ago
  64. ParthKohli Group Title
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    And\[\rm \large n = -2\]

    • one year ago
  65. lgbasallote Group Title
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    how is b = -1?

    • one year ago
  66. ParthKohli Group Title
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    \[\rm \large 3\cdot(-1) + 1 = -2\]\[\rm \large a\cdot b +c=n \]

    • one year ago
  67. lgbasallote Group Title
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    you're not really telling me anything...you're just circling around

    • one year ago
  68. ParthKohli Group Title
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    lol I am... I just proved you that \(\rm \large (-2) mod3 = 1\)

    • one year ago
  69. lgbasallote Group Title
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    you just have -2 mod 3 = ? show me how you get that to be 1

    • one year ago
  70. ParthKohli Group Title
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    I just did...

    • one year ago
  71. lgbasallote Group Title
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    you didn't really prove...because you just assumed a value of b and c and equated them

    • one year ago
  72. ParthKohli Group Title
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    That's the definition of modulus, dude.

    • one year ago
  73. lgbasallote Group Title
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    show me a solution that tells me -2 mod 3 is 1

    • one year ago
  74. lgbasallote Group Title
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    you have two unknows..you can't use the definition

    • one year ago
  75. ParthKohli Group Title
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    I'd show you all of it in one post.

    • one year ago
  76. lgbasallote Group Title
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    sure.

    • one year ago
  77. ParthKohli Group Title
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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

    • one year ago
  78. ParthKohli Group Title
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    Would you like it with hearts and smileys? Oh, also see that b is an integer.

    • one year ago
  79. lgbasallote Group Title
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    but the question you refuse to answer is... how do you get those values???

    • one year ago
  80. ParthKohli Group Title
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    What values?

    • one year ago
  81. lgbasallote Group Title
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    1and -1..you can't pull them out of nowhere

    • one year ago
  82. ParthKohli Group Title
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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.

    • one year ago
  83. lgbasallote Group Title
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    i'm not asking about the definition! i'm asking about your numbers

    • one year ago
  84. ParthKohli Group Title
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    Oh, and you have to try and test in the Euclidean Algorithm too.

    • one year ago
  85. lgbasallote Group Title
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    i can also say 3*(-2) + 4 = -2 so i can say -2 mod 3 is 4

    • one year ago
  86. lgbasallote Group Title
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    so you're wrong by *my* answer

    • one year ago
  87. lgbasallote Group Title
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    -2 mod 3 should be 4 not 1

    • one year ago
  88. ParthKohli Group Title
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    Yes, it's also 4.

    • one year ago
  89. lgbasallote Group Title
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    ...........

    • one year ago
  90. lgbasallote Group Title
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    no it isn't

    • one year ago
  91. lgbasallote Group Title
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    -2 mod 3 is 1

    • one year ago
  92. ParthKohli Group Title
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    That's the fun of modular arithmetic: you have -2 mod 3 = 1,4,7,10...

    • one year ago
  93. lgbasallote Group Title
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    nope

    • one year ago
  94. ParthKohli Group Title
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    Yes. Would you like a machine to answer it for you?

    • one year ago
  95. lgbasallote Group Title
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    try wolfram

    • one year ago
  96. ParthKohli Group Title
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    Sure.

    • one year ago
  97. lgbasallote Group Title
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    you'll see it's 1

    • one year ago
  98. lgbasallote Group Title
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    or...you can just watch more khanacademy videos....

    • one year ago
  99. ParthKohli Group Title
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    XD

    • one year ago
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  100. ParthKohli Group Title
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    http://www.wolframalpha.com/input/?i=-2+mod+3 You may have a look.

    • one year ago
  101. ParthKohli Group Title
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    And remember, there's no equality in modulus... there's congruence.\[\rm \large a \mod b \equiv c\]

    • one year ago
  102. lgbasallote Group Title
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    you don't really get what my question was...do you....

    • one year ago
  103. ParthKohli Group Title
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    Though I have to admit, I forgot the above too...

    • one year ago
  104. ParthKohli Group Title
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    -11 mod 3

    • one year ago
  105. ParthKohli Group Title
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    \[\rm \large 3\cdot(-4)+1 = -11 \]Right?

    • one year ago
  106. ParthKohli Group Title
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    \[-11\mod3\equiv 1\]But also...

    • one year ago
  107. ParthKohli Group Title
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    \[\rm \large 3\cdot(-5) + 4 = -11\]

    • one year ago
  108. ParthKohli Group Title
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    So we can say that\[\rm \Huge -11\mod3 \equiv1+3k\]where k is an integer.

    • one year ago
  109. ParthKohli Group Title
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    You're welcome. @lgbasallote

    • one year ago
  110. lgbasallote Group Title
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    that was *not* what i was asking

    • one year ago
  111. ParthKohli Group Title
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    You're welcome anyway. I'm glad I could be of help.

    • one year ago
  112. lgbasallote Group Title
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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...

    • one year ago
  113. ParthKohli Group Title
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    Sorry, if it was modulo, then it's just the remainder.

    • one year ago
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