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aman07

Let * be a binary operation on Real numbers defined by a*b=ab/4 Given a*b=9 Find a and b

  • one year ago
  • one year ago

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  1. aman07
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    help ! @TuringTest , @Mani_Jha , @dumbcow ,

    • one year ago
  2. mathslover
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    ok so we are given with the following information : a * b = ab/4 a*b = 9 this means \(\large{\frac{ab}{4} = 9}\) , right ?

    • one year ago
  3. Hoa
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    yes

    • one year ago
  4. aman07
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    yes

    • one year ago
  5. Hoa
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    ---> ab =36

    • one year ago
  6. mathslover
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    Now , can you find what will be the value of "ab" ?

    • one year ago
  7. aman07
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    ab=36 ?

    • one year ago
  8. Hoa
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    100010 =36

    • one year ago
  9. aman07
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    but how i find value of a and b ?

    • one year ago
  10. mathslover
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    Yes right @aman07

    • one year ago
  11. mathslover
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    OK let me think for what to do after this, wait!

    • one year ago
  12. aman07
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    ok

    • one year ago
  13. mathslover
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    What we have yet as a first way to get rid of this is "hit and trial" but really I don't think that hit and trial will be a satisfactory method ....

    • one year ago
  14. aman07
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    well i knw to do that ..but there is some method using identity element and stuff

    • one year ago
  15. mathslover
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    http://www.math.csusb.edu/notes/binop/node2.html I saw something here

    • one year ago
  16. aman07
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    :( i have that already in text.. i am stuck in this particular qn :(

    • one year ago
  17. Hoa
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    i got it now. you have ab =36 that means a=3 and b =6. in binary system, 3= 10 (one zero) and 6 = 1001. make multiplication between them , you got exactly 36 in both code (binary and 10 system)

    • one year ago
  18. aman07
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    wth ??

    • one year ago
  19. goformit100
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    ok

    • one year ago
  20. mathslover
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    I tried to search some relative problems on binary operations on internet and I found this question : if : \(\large{a\alpha b = |a-b| }\) then what will be \(\large{6\alpha 8 }\) . Of course that will be 2 but the question given is not exactly what we have in m

    • one year ago
  21. mathslover
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    In my question (contd. from the previous post.) Ok, I have never learnt it but I thought it is a type of mental ability question... Is it also related to binary numbers ?

    • one year ago
  22. Hoa
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    |dw:1361628824557:dw|

    • one year ago
  23. aman07
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    nope ..i think no one understands the type of qn i am referring to ..leave it ..i am leaving thanks for trying to help

    • one year ago
  24. ParthKohli
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    @hoa "binary" doesn't mean "binary numbers"; it is just an operation you are doing with two numbers. Remember that "binary" stands for the numeral two.

    • one year ago
  25. mathslover
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    Don't get depressed from the help here given by me @aman07 ... we have at present many users seeing this question and I am sure that some people are still here who are aware of these type of problems. Just have patience aman... wait for them to respond

    • one year ago
  26. aman07
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    ok

    • one year ago
  27. ParthKohli
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    We don't have sufficient information in this question.

    • one year ago
  28. ParthKohli
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    Do you have the answer?

    • one year ago
  29. aman07
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    no i do not have

    • one year ago
  30. ParthKohli
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    there are infinite ordered pairs of \((a,b)\).

    • one year ago
  31. mathslover
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    Yes right @ParthKohli . There is no sufficient information to solve for a and b.. though i can say : a = 36/b and b = 36/a :)

    • one year ago
  32. mathslover
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    but that values is where I am stuck ...

    • one year ago
  33. Hoa
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    lol, interesting question and interesting asker

    • one year ago
  34. ParthKohli
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    There are infinite values for \((a,b)\). We could have considered a particular pair if we had a much more restricted set/interval or the value of either of them.

    • one year ago
  35. ParthKohli
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    If you want to find \(ab\), then it's 36 :-)

    • one year ago
  36. walters
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    let \[x\]element R be the identity with respect to * ,then x*a=a*x=a for all a element R now|dw:1361630564587:dw| |dw:1361630761828:dw|

    • one year ago
  37. mathteacher1729
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    ab/4 = 9 means... Step 1: multiply a times b Step 2: divide by 4 the result is 9. That means 9*4 = a*b = 36 So a and b could be 9 and 4 or 18 and 2 or 3 and 12 or any two digits that multiply to 36. We'd need more information to narrow down the possibilities of what a and b could be. Are they both positive? Are they both positive integers? Are they both positive integers within a certain range? etc.

    • one year ago
  38. aman07
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    well i am speAKing about the working given above by @walters using identity element but still i am having confusion

    • one year ago
  39. walters
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    where

    • one year ago
  40. walters
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    b*a does not neccesary means b multiply by a ,that's why i let x to identity knowing that i will have xa/4 and assign it to a but since a*b =9 this will also means that if a take x*a i will get the same value as a*b because of the binary operator

    • one year ago
  41. aman07
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    can u explain it step by step pleas e?

    • one year ago
  42. aman07
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    why doesnt b*a imply ba ?

    • one year ago
  43. aman07
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    @walters ?

    • one year ago
  44. aman07
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    i dont get the step ab=36 then how u write xa=36 ?

    • one year ago
  45. walters
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    because * (it is the star operator not multiplications) they are not equal

    • one year ago
  46. walters
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    |dw:1361638259187:dw|

    • one year ago
  47. aman07
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    |dw:1361638483005:dw|

    • one year ago
  48. aman07
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    b can be any number

    • one year ago
  49. walters
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    do u know the defination of the binary operator

    • one year ago
  50. aman07
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    yes

    • one year ago
  51. walters
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    are doing Groups (abstract algebra)

    • one year ago
  52. aman07
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    .

    • one year ago
  53. walters
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    ur questions means a*b if u take any 2 real numbers and map them the result will be in the form of multiplication of ur 2 real numbers divide by 4 and we know that tha mapping of those two numbers is equal to 9 tht's y i 've used 36 on my calculation because of the conditions i have

    • one year ago
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