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lgbasallote

  • 2 years ago

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

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  1. Dwade03
    • 2 years ago
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    mod?

  2. lgbasallote
    • 2 years ago
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    yes mod

  3. bahrom7893
    • 2 years ago
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    modulus

  4. bahrom7893
    • 2 years ago
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    mod is 2

  5. bahrom7893
    • 2 years ago
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    idk what div is.

  6. ParthKohli
    • 2 years ago
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    By div, do you mean quotient?

  7. lgbasallote
    • 2 years ago
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    hmm

  8. lgbasallote
    • 2 years ago
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    ...why are you asking me?

  9. ParthKohli
    • 2 years ago
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    Hmm... mod is 1.

  10. jiteshmeghwal9
    • 2 years ago
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    becoz i m not getting ur question :(

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

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

  13. lgbasallote
    • 2 years ago
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    and why is @bahrom7893 wrong @ParthKohli ?

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

  15. lgbasallote
    • 2 years ago
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    yes

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

  17. lgbasallote
    • 2 years ago
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    again...why are you asking me?

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

  19. lgbasallote
    • 2 years ago
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    mmhmm

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

  21. lgbasallote
    • 2 years ago
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    it is...

  22. ParthKohli
    • 2 years ago
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    Yes, that's what...

  23. jiteshmeghwal9
    • 2 years ago
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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

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

  25. lgbasallote
    • 2 years ago
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    so then...?

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

  27. jiteshmeghwal9
    • 2 years ago
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    -11/3*

  28. lgbasallote
    • 2 years ago
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    mod is modulo @jiteshmeghwal9

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

  30. jiteshmeghwal9
    • 2 years ago
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    ohh ok !

  31. bahrom7893
    • 2 years ago
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    |dw:1350970551590:dw|

  32. bahrom7893
    • 2 years ago
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    Mod is -2

  33. lgbasallote
    • 2 years ago
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    i agree..

  34. jiteshmeghwal9
    • 2 years ago
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    |dw:1350970548817:dw|

  35. bahrom7893
    • 2 years ago
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    |dw:1350970607290:dw|

  36. lgbasallote
    • 2 years ago
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    question is... why is @ParthKohli saying something different...

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

  38. lgbasallote
    • 2 years ago
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    ^?

  39. bahrom7893
    • 2 years ago
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    lol parth i have no clue how to prove that.

  40. bahrom7893
    • 2 years ago
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    but it must be hahah

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

  42. lgbasallote
    • 2 years ago
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    -2 mod 3 is -2...

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

  44. lgbasallote
    • 2 years ago
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    ^?

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

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

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

  48. ParthKohli
    • 2 years ago
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    Modulus is NOT NECESSARILY THE REMAINDER, SIRE.

  49. lgbasallote
    • 2 years ago
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    ....

  50. ParthKohli
    • 2 years ago
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    Do you know the Euclidean Algorithm?

  51. lgbasallote
    • 2 years ago
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    of course

  52. lgbasallote
    • 2 years ago
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    do you?

  53. ParthKohli
    • 2 years ago
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    Yes.

  54. ParthKohli
    • 2 years ago
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    Use the Euclidean Algorithm then.

  55. ParthKohli
    • 2 years ago
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    a variation of Euclidean Algorithm actually...

  56. lgbasallote
    • 2 years ago
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    euclidian algorithm is used for finding gcf....

  57. ParthKohli
    • 2 years ago
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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.

  58. ParthKohli
    • 2 years ago
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    The above is the actual definition of modulus.

  59. lgbasallote
    • 2 years ago
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    yes

  60. lgbasallote
    • 2 years ago
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    b is n/a

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

  62. lgbasallote
    • 2 years ago
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    how?

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

  64. ParthKohli
    • 2 years ago
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    And\[\rm \large n = -2\]

  65. lgbasallote
    • 2 years ago
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    how is b = -1?

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

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

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

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

  70. ParthKohli
    • 2 years ago
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    I just did...

  71. lgbasallote
    • 2 years ago
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    you didn't really prove...because you just assumed a value of b and c and equated them

  72. ParthKohli
    • 2 years ago
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    That's the definition of modulus, dude.

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

  74. lgbasallote
    • 2 years ago
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    you have two unknows..you can't use the definition

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

  76. lgbasallote
    • 2 years ago
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    sure.

  77. ParthKohli
    • 2 years ago
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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

  78. ParthKohli
    • 2 years ago
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    Would you like it with hearts and smileys? Oh, also see that b is an integer.

  79. lgbasallote
    • 2 years ago
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    but the question you refuse to answer is... how do you get those values???

  80. ParthKohli
    • 2 years ago
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    What values?

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

  82. ParthKohli
    • 2 years ago
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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.

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

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

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

  86. lgbasallote
    • 2 years ago
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    so you're wrong by *my* answer

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

  88. ParthKohli
    • 2 years ago
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    Yes, it's also 4.

  89. lgbasallote
    • 2 years ago
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    ...........

  90. lgbasallote
    • 2 years ago
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    no it isn't

  91. lgbasallote
    • 2 years ago
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    -2 mod 3 is 1

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

  93. lgbasallote
    • 2 years ago
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    nope

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

  95. lgbasallote
    • 2 years ago
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    try wolfram

  96. ParthKohli
    • 2 years ago
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    Sure.

  97. lgbasallote
    • 2 years ago
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    you'll see it's 1

  98. lgbasallote
    • 2 years ago
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    or...you can just watch more khanacademy videos....

  99. ParthKohli
    • 2 years ago
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    XD

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

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

  102. lgbasallote
    • 2 years ago
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    you don't really get what my question was...do you....

  103. ParthKohli
    • 2 years ago
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    Though I have to admit, I forgot the above too...

  104. ParthKohli
    • 2 years ago
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    -11 mod 3

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

  106. ParthKohli
    • 2 years ago
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    \[-11\mod3\equiv 1\]But also...

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

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

  109. ParthKohli
    • 2 years ago
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    You're welcome. @lgbasallote

  110. lgbasallote
    • 2 years ago
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    that was *not* what i was asking

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

  112. lgbasallote
    • 2 years ago
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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...

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

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