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Why is -1*-1=1

  1. Aug 4, 2010 #1

    hunt_mat

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    I have to teach an elementary mathematics course next term to people who don't have A-level maths and I am not too sure where to start from. One of the questions that popped into my head was why was [tex](-1)^{2}=1[/tex]? I can show it's postive quite easily but I don't know if it is 1 by definition or not.

    Mat
     
  2. jcsd
  3. Aug 4, 2010 #2

    Mark44

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    1 + (-1) = 0, because 1 and -1 are additive inverses.
    1(1 + (-1)) = 1(0) = 0, because 1 is the multiplicative inverse.
    It must also be true that -[1(1 + (-1))] = 0, because 0 is its own additive inverse.
    On the other hand, -1(1 + (-1)) = -1(1) + (-1)(-1), due to the distributive law.
    So we have -1(1) + (-1)(-1) = -1 + (-1)(-1) = 0, from which we see that (-1)(-1) = 1. The only number I can add to -1 to get 0 is its additive inverse, + 1.
    Therefore, (-1)(-1) = 1.
     
  4. Aug 4, 2010 #3

    D H

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    1(1 + (-1)) = 1(0) = 0, because 1 is the multiplicative identity.
     
  5. Aug 4, 2010 #4
    If you are doing an axiomatic treatment of the integers, you define 1 to be the multiplicative identity. -1 is defined to be the additive inverse of 1 (so 1 + -1 = 0).

    -1*-1 = -1*(-1 + 0) [Use that 0 is the additive identity]
    -1*(-1 + 0) = -1*(-1 + (1 + -1)) [Use that 0 = 1 + -1]
    -1*(-1 + (1 + -1)) = -1*-1 + -1*1 + -1*-1 [Distribution]

    This gives us that -1*-1 = -1*-1 + -1*1 + -1*-1

    If we assume that the additive identity is unique, then -1*-1 + -1*1 must be the additive identity.

    We have now that 0 = -1*-1 + -1 [Remember, -1*1 = -1]

    Then -1 is the additive inverse of -1*-1. If we assume additive inverses are unique, then since -1 is also the additive inverse of 1, it follows that -1*-1 = 1.

    I'm sure there's an easier way though!
     
  6. Aug 4, 2010 #5
    Ah, looking at Mark44's post, there is a nicer way.

    First we prove 0*0 = 0.

    This is true because 0*0 = 0*(0+0) = 0*0 + 0*0. By uniqueness of the additive identity, 0*0 = 0.

    Also, 1 + -1 = 0, by definition.

    Thus (1 + -1)*(1 + -1) = 0.

    Distribution gives us (1*1 + -1*1) + (1*-1 + -1*-1) = 0,
    or... (1 + -1) + (-1 + -1*-1) = 0.

    -1*-1 is then the additive inverse of -1, so -1*-1 = 1.


    This avoids the trickiness of saying -[1(1 + (-1))] = -1(1 + -1). But really there are a lot of ways to prove it. Technically, I don't even think uniqueness of inverses and identities is assumed!
     
  7. Aug 4, 2010 #6
    You guys are way overthinking.
    You have 1*1=1. No disputes there.
    Then -1*1=-1, since 1 is the multiplicative identity.
    Then -1*-1 can't be the same thing as 1*-1 otherwise -1=1,
    so -1*-1=1. This is the treatment given in Euler's Elements of Algebra, Art.33
    Just that simple.

    Aother treatment is to view -1 as an operator that flips quantities across the y-axis. So -1*2=-2 means -1 takes the point at 2 on the x axis and flips it to -2. Then -1*-1*2 flips 2 back to its original position. But I think the first way is the best.
     
  8. Aug 4, 2010 #7
    dimitri151, why couldn't -1*-1 be some number other than -1 or 1? Why not 0 or 2?
     
  9. Aug 4, 2010 #8
    There's no question about the absolute value of the product, only of the sign. You agree that for a and b positive a*b=ab, and that for a positive, b negative for example you have a*-b=-ab. So why would you wonder what -a*-b would be without regard to the sign?

    Look at it like this, say the product is an area, and you can visualize the area by taking a on the x axis and b on the y-axis, with positive a being on the positive x-axis, negative a on the negative x-axis, positive b on the positive y-axis, and so forth, and then erecting perpendiculars from each point so that they meet in their mutual quadrant. If you take -1 on the x axis, and -1 on the y-axis, noone seriously disputes that the area will be other than a unit area. Only the sign needs to be determined which is done as above.
     
  10. Aug 5, 2010 #9

    Hurkyl

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    There is a number system where -1 * -1 = 2. Lots of them, in fact. There's one where -1 * -1 = -1.

    I don't know of the utility of any of these number systems, though. They must have "bad" properties -- e.g. maybe the distributive law fails sometimes.


    The first, uncontroversial application of negative numbers was in finance. If you're tallying up entries in your ledger, what is the effect of striking out one debt of $1? This would be a product with both terms noted via sign and magnitude -- you gain -1 * -1 dollars.

    I suppose way back then, it might have been more mainstream to treat adding one debt of $1 and removing one debt of $1 as two totally different things, rather than the same kind of thing just with different signs, so I guess you wouldn't have seen sign and magnitude on both factors.
     
  11. Aug 5, 2010 #10

    Borek

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    Somehow I doubt everything what was told above won't be wasted on people who think squaring the number means drawing a square around.
     
  12. Aug 5, 2010 #11

    hunt_mat

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    Wow, I never expected this amout of response! Thanks. Borek, this is the first course I have been asked to teach, so I want to be prepared as possible.

    Cheers
     
  13. Aug 5, 2010 #12

    Mark44

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    My brain knows that, but apparently my fingers don't.
     
  14. Aug 5, 2010 #13
    How about this? In a system where -1*-1=-1, what is the multiplicative inverse of -1, that is, what is x so that -1*x=1? If -1 doesn't have an inverse then the system isn't even a group.
     
    Last edited: Aug 5, 2010
  15. Aug 5, 2010 #14

    Hurkyl

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    Theorem: If G is a magma with elements -1 and 1, then at least one of the following fails:
    • G is a group
    • -1 * -1 = -1
    • 1 * -1 = -1
    • 1 and -1 are unequal

    (A magma is simply a set with a binary operation)
     
  16. Aug 11, 2010 #15
    I've got an idea, if it for elementary maths I'd imagine they know what a function is.

    So why not draw [tex]f(x) = x^2[/tex] and show that [tex]f(-1) = (-1)^2 = 1[/tex] graphically :biggrin:
     
  17. Aug 11, 2010 #16
    The representation of the function as a graph assumes what is to be demonstrated. If you define -1*-1=-1 then f(x)=x2 will look something like what the graph of y=x3 looks lke.
     
  18. Aug 11, 2010 #17

    hunt_mat

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    I have to teach them what a function is.

    Is there a place on this site for people to upload their lecture notes for others? I have gone through all this effort to type up my notes in LaTeX I would want others to benefit from my hard work.

    Mat
     
  19. Sep 14, 2010 #18
    Refer to negative numbers and "the opposite" of positive numbers.
    That is, the "opposite" of 3 is (-3)
    Therefore, the "opposite" of (-3) is 3 (they are opposites of each other)

    So, (-1) * X = (-X) = "the opposite of X"

    For X = 5,
    (-1) * 5 = (-5)
    which can be shown using the definition of multiplication:
    (-1) * 5 = (-1) + (-1) + (-1) + (-1) + (-1) = (-5)
    this is the sum of five (-1)'s

    This can be shown to be true for all X>0 (there is no such thing as negative 0)



    Whatever mathematical rules hold true for positive numbers must also hold true for negative ones, so:

    for X = (-5), we have:
    (-1) * (-5) = -(-5) = 5 "the opposite of (-5)"

    and for X = (-1), we get
    (-1) * (-1) = -(-1) = "the opposite of (-1)" = 1
     
  20. Sep 23, 2010 #19
  21. Mar 31, 2011 #20
    a/a = 1, a <>0, -1/-1 = 1
     
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