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Multiplying negatives

  1. May 30, 2004 #1
    can anyone explain why for example
    -7 * -7 = +49

    after all isnt -7 * -7 = to saying - 7 + -7 seven times

    now if we take -7 + -7 sven times it would be -49
     
  2. jcsd
  3. May 30, 2004 #2
    here is a proof that -1*-1 = 1

    -1*0 = 0
    since 0 = 1-1
    -1*(1-1) = 0
    by distributivity we get
    -1*1 + -1*-1 = 0
    (-1) + (-1*-1) = 0
    since -1 + 1 =0
    then -1 *-1 = 1
     
  4. May 31, 2004 #3

    Gokul43201

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    No... look carefully again at what you claim. -7*-7 should be the same as adding -7 to itself "minus" seven times.

    What you are doing is adding it seven times. So that is the same as-7*7, which is why you get -49
     
    Last edited: May 31, 2004
  5. May 31, 2004 #4
    (-7) * (-7) = -[7 * (-7)] = -(-49) = 49.
    You can show this with complex numbers.
     
  6. Jun 1, 2004 #5

    It isn't -7 + -7 times. THis doesn't make sense, but its acutally -7 + -7 negative seven times...
    (really doesn't make sense!!!)
     
  7. Jun 7, 2004 #6
    this is how i would explain it; it's not intended to be a proof...

    let's work out why (-3)(-3)=9.

    (3)(3)=9
    (2)(3)=6
    (1)(3)=3
    (0)(3)=0

    the pattern is the answers go down by 3 when the first number is decreased by 1.

    (-1)(3)=-3
    (-2)(3)=-6
    (-3)(3)=-9

    you can do something similar to get the following results:

    (-1)(2)=-2
    (-2)(2)=-4
    (-3)(2)=-6

    (-1)(1)=-1
    (-2)(1)=-2
    (-3)(1)=-3

    (-1)(0)=0
    (-2)(0)=0
    (-3)(0)=0.

    look what happens at (-3)(x): when x goes down by 1, the answer goes UP by 3. therefore:

    (-3)(-1)=3 (0+3)
    (-3)(-2)=6 (3+3)
    (-3)(-3)=9 (6+3)
     
    Last edited: Jun 7, 2004
  8. Jun 8, 2004 #7
    Finally, there is a choo-choo train illustration.

    ------

    A train track runs from station A past station B onto station C, where A is to the west of B and C is to the east of B. Also, the track is perfectly straight and the distance AB equals distance BC. It's numerically 49 distance-units for this example.

    Positive velocity of a train is in the eastward direction and negative velocity is in the westward direction (assuming the train can run bidirectionally).

    Positive time durations are later and negative time durations are earlier.

    Positive distances are to the east and negative distances are to the west.

    OK.

    Suppose a train at B runs eastward at velocity +7 (distance-units/time-unit) and will run for +7 time-units. Then the distance will be (+7)(+7)=+49, or 49 distance-units to the east of the station B. That is where station C is.

    Suppose a train at B runs eastward at velocity +7 (distance-units/time-unit) and already ran for -7 time-units. Then the distance was (+7)(-7)=-49, or 49 distance-units to the west of the station B. That is where station A is.

    Suppose a train at B runs westward at velocity -7 (distance-units/time-unit) and will run for +7 time-units. Then the distance will be (-7)(+7)=-49, or 49 distance-units to the west of the station B. That is where station A is.

    Suppose a train at B runs westward at velocity -7 (distance-units/time-unit) and already ran for -7 time-units. Then the distance was (-7)(-7)=+49, or 49 distance-units to the east of the station B. That is where station C is.
     
  9. Jun 8, 2004 #8

    -7*-7 = 7(-1*-1)

    This is a PROOF
     
    Last edited: Jun 8, 2004
  10. Jun 8, 2004 #9
    somehow you set (-1*-1) = +1, even though that is what you were trying to prove!

    here is one way to get your head around it.

    -7 * -7 = (-1)(7)(-7) or -(7 * -7)

    seven -7s is -49, so

    -(7*-7) = -(-49)

    the negative of a negative number is always positive

    therefore -(-49) = 49
     
    Last edited: Jun 8, 2004
  11. Jun 8, 2004 #10
    Warr - You have accused hello3719 of something he did not do, and then you turn around and do the very thing yourself:

    This is what you are trying to show!

    To see that hello3719 did not commit this error, let me expand on the proof:

    Code (Text):
              -1*0 = 0       Anything times 0 = 0
          -1*(1-1) = 0       Substitution of 1-1 for 0
      -1*1 + -1*-1 = 0       Distributive Law
    (-1) + (-1*-1) = 0       Simplify the first term.
             -1*-1 = 1       Add 1 to both sides of the equation.
     
  12. Jun 8, 2004 #11
    no you didn't follow well
    (-1)(0) = 0
    (-1)(-1 + 1) = 0
    by distr
    (-1)(-1) + 1(-1) = 0
    1(-1) = -1
    so -1 + (-1)(-1) =0
    since -1 + 1 = 0 (eq 1)
    and that we have -1 + (-1)*(-1) = 0 eq 2)
    then if we substract eq 2 form eq 1 then we get (-1)*(-1)=1


    LOL .that IS what we have to prove, negative of a negative means
    (-1)(-1)
     
  13. Jun 9, 2004 #12
    sorry, I reread. My mistake
     
  14. Jun 12, 2004 #13
    Here is Hello's proof in more general terms:

    The proof will be in two steps. First, we will prove that x*(-y) = - k where x, y, and k are all integers

    Let a and b be any arbitrary integers

    Since the integers represent a number field, they are closed under multiplication
    Thus a*b = c where c is another integer.

    Since the integers are also closed under addition, we can find an integer d such that d-1 = b. (You could also cite Peano's Postulates for this.) We note that d must be greater than b.

    Substituting these into our original equation: a*(d-1) = c

    By the distributive property of the integers then, we can write:

    a*d + a*(-1) = c

    But notice: Since d > b, then a*d > c. Obviously then a*(-1) must be less than 0.

    Now, multiply a*d + a*(-1) = c through by -1. Based on our results from above we get:

    -a*d + (-a)*(-1) = -c

    Again since d > b, then -a*d must be less than c. Therefore, (-a)*(-1) must be greater than 0.
     
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