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What's (±1)(±1) equal to?

  1. Sep 4, 2011 #1
    What does (±1)(±1) equal to, is it just positive 1?
     
  2. jcsd
  3. Sep 4, 2011 #2
    wouldn't it be (1)(1) = 1
    (1)(-1)= -1
    (-1)(1) = -1
    (-1)(-1) = 1?
     
  4. Sep 4, 2011 #3
    so.. plus or minus 1?
     
  5. Sep 4, 2011 #4
    I thought it would have to be just positive one because I thought that (±1)(±1) = (1)(1) or (-1)(-1) which both equal 1, I thought they both had to be both either positive or negative at the same time, in which case (±1)(±1) = 1?
     
  6. Sep 4, 2011 #5
    I don't see why they should have to be positive or negative at the same time, unless the reality of the problem dictates it.
     
  7. Sep 4, 2011 #6
    I thought that if one wanted to distinguish them being either positive or negative at different times you would put (±1)(-+1)

    -+ is suppose to be ± rotated 180 degrees?
     
  8. Sep 4, 2011 #7

    gb7nash

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    That's the way I interpret it. Sometimes you'll see things like:

    [tex](5 \pm 5) \mp 10[/tex]

    where you get 0 or 10. The rule is to take the top operation to get one answer, then take the bottom operation to obtain the second answer. In the OP's problem, both answers would be 1.
     
  9. Sep 4, 2011 #8

    eumyang

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    You mean this?
    [itex](\pm 1)(\mp 1)[/itex]

    I've seen the "minus-plus" symbol before. The cosine of a sum and difference can be written in one formula like so:
    [itex]\cos (a \pm b) = \cos a \cos b \mp \sin a \sin b[/itex]
    ... indicating that the symbol on the RHS is different from the one on the LHS.


    EDIT: gb7nash beat me to it. ;)
     
  10. Sep 4, 2011 #9
    well you've lost me. forget I said anything :P
     
  11. Sep 4, 2011 #10

    uart

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    Id say both interpretations were possible, depending on the context.

    Say X was a two state random variable that could be either +1 or -1. Similarly Y is an independent two state random variable. The product XY is [itex](\pm 1)(\pm 1)[/itex], but it's certainly not always +1 in this case.
     
  12. Sep 4, 2011 #11

    Mark44

    Staff: Mentor

    This would be [itex]\pm 1[/itex]. The first factor could be either positive or negative, and so could the second factor. You can't assume (and shouldn't) that if the first factor is positive, so is the second.
     
  13. Sep 4, 2011 #12
    what about [tex]\mp 1[/tex]?
     
  14. Sep 5, 2011 #13

    uart

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    I could accept the idea of "coupled" plus or minuses as a shorthand notation in some specific circumstances, eg:

    [tex] \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b [/tex]

    [tex] \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b [/tex]

    In general however, without any specific context as in the OP, I would never consider all the ([itex]\pm[/itex])'s in an equation (or set of equations) to be coupled in this way.
     
  15. Sep 5, 2011 #14

    NascentOxygen

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    In what context did you encounter this?
     
  16. Sep 5, 2011 #15

    Mentallic

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    When I used to deal with equations that involved [itex]\pm[/itex] that were both dependent and independent of others, I would label them with numbers such as [itex]\pm_1, \pm_2[/itex] for example. I think later on when I saw them being used in formal writing, they were denoted by dashes, such as what you see when dealing with derivatives, [itex]\pm', \pm''[/itex] etc.
     
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