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lgbasallote

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
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    mod?

    • one year ago
  2. lgbasallote
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    yes mod

    • one year ago
  3. bahrom7893
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    modulus

    • one year ago
  4. bahrom7893
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    mod is 2

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

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

    • one year ago
  7. lgbasallote
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    hmm

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

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

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

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

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

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

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

    • one year ago
  15. lgbasallote
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    yes

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

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

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

    • one year ago
  19. lgbasallote
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    mmhmm

    • one year ago
  20. 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.

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

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

    • one year ago
  23. 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

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

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

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

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

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

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

    • one year ago
  30. jiteshmeghwal9
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    ohh ok !

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

    • one year ago
  32. bahrom7893
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    Mod is -2

    • one year ago
  33. lgbasallote
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    i agree..

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

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

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

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

    • one year ago
  38. lgbasallote
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    ^?

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

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

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

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

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

    • one year ago
  44. lgbasallote
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    ^?

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

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

    • one year ago
  47. lgbasallote
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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
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    Modulus is NOT NECESSARILY THE REMAINDER, SIRE.

    • one year ago
  49. lgbasallote
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    ....

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

    • one year ago
  51. lgbasallote
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    of course

    • one year ago
  52. lgbasallote
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    do you?

    • one year ago
  53. ParthKohli
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    Yes.

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

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

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

    • one year ago
  57. 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.

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

    • one year ago
  59. lgbasallote
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    yes

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

    • one year ago
  61. ParthKohli
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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
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    how?

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

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

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

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

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

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

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

    • one year ago
  71. lgbasallote
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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
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    That's the definition of modulus, dude.

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

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

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

    • one year ago
  76. lgbasallote
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    sure.

    • one year ago
  77. 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

    • one year ago
  78. ParthKohli
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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
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    but the question you refuse to answer is... how do you get those values???

    • one year ago
  80. ParthKohli
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    What values?

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

    • one year ago
  82. 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.

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

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

    • one year ago
  85. lgbasallote
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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
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    so you're wrong by *my* answer

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

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

    • one year ago
  89. lgbasallote
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    ...........

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

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

    • one year ago
  92. ParthKohli
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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
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    nope

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

    • one year ago
  95. lgbasallote
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    try wolfram

    • one year ago
  96. ParthKohli
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    Sure.

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

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

    • one year ago
  99. ParthKohli
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    XD

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

    • one year ago
  101. ParthKohli
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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
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    you don't really get what my question was...do you....

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

    • one year ago
  104. ParthKohli
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    -11 mod 3

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

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

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

    • one year ago
  108. ParthKohli
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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
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    You're welcome. @lgbasallote

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

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

    • one year ago
  112. 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...

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

    • one year ago
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