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Negatives and positives

  1. Jul 23, 2005 #1
    this is the dumbest question I've ever come up with, but I can't answer it, so that would make me horribly stupid.

    Why is:
    -1*-1=1
    1*1=1
    -1*1=-1
    1*-1=-1

    why can't:
    -1*-1=-1
    1*1=-1
    -1*1=1
    1*-1=1
    ?

    Is there any reason why we've defined things in the latter way rather than the former?

    p.s. sorry folks, I'm sure there is an obvious and simple reason, but I can't see it right now.
     
  2. jcsd
  3. Jul 23, 2005 #2

    honestrosewater

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    Well,
    1x = x​
    is an axiom. :smile: The general case probably has another proof, but I can't think of it just now. Right, here's a start.
    Prove: (-x)y = -(xy) = x(-y).

    xy + (-x)y = (x + -x)y = 0y = 0​
    Say you already know that if w + v = 0, then v = -w. That gives you the first equality, (-x)y = -(xy). The other, -(xy) = x(-y), is proven the same way.
    The next theorem you want is
    Prove: (-x)(-y) = xy.
    Use the previous theorem to prove this.
     
    Last edited: Jul 23, 2005
  4. Jul 23, 2005 #3

    HallsofIvy

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    In other words, if you define
    "-1*-1=-1
    1*1=-1
    -1*1=1
    1*-1=1 "

    Then the distributive law would give -1*(1+ -1)= -1*(1)+(-1)*(-1)= -1-1= -2

    But 1+ -1= 0 (by definition of -1) so it is really (-1)*0= 0. The definitions you give would mean that the distributive law (a very important algebraic law) would not be true.
     
  5. Jul 23, 2005 #4
    ahhh... I see! thanks, I was really only looking at commutivity i guess.

    well, are you sure?

    based on
    -1*-1=-1
    1*1=-1
    -1*1=1
    1*-1=1

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

    becasue -1*1=1, then the distributive law would still work.
    You just messed up in your example.

    hmm...

    "Well,

    1x = x

    is an axiom. The general case probably has another proof, but I can't think of it just now. Right, here's a start.

    Prove: (-x)y = -(xy) = x(-y).

    xy + (-x)y = (x + -x)y = 0y = 0

    Say you already know that if w + v = 0, then v = -w. That gives you the first equality, (-x)y = -(xy). The other, -(xy) = x(-y), is proven the same way.
    The next theorem you want is

    Prove: (-x)(-y) = xy.

    Use the previous theorem to prove this." - Honestrosewater

    I can't, based on this:
    -1*-1=-1
    1*1=-1
    -1*1=1
    1*-1=1
    I don't think I can... But I can prove (-y)(-x)=-(xy), and (yx)=-(xy) with it.
    Almost all the current theorems would be bassackwards, but wouldn't math still work?

    or is there some case in my logic that would end up with -1=1?
    hmm...
    -1*-1=-1=1*1
    multiply -1 on both sides
    (-1)(-1)(-1)=1*1(-1)
    (-1)(-1)=(-1)(-1)
    -1=-1

    how about
    multiply 1 on both sides
    1*(-1)(-1)=1*1*1
    1*-1=-1*1
    1=1

    I think the logic still works, but every function would be flipped along the y-axis, so for example the zeta function's solutions would be 2,4,6,... and it's non-trivial ones would all lie on real part -1/2, the line y=x would have a negative slope of 1, but symmetrical functions like even polynomials (y=x^n, where n=2,4,6,...) and the cosine function would all be the same. Am I interpreting this wrong, or is there a flaw, and therefore a reason why we've defined things to be the way they are?
     
    Last edited: Jul 23, 2005
  6. Jul 23, 2005 #5

    Hurkyl

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    "1*1 = -1"

    Note that this contradicts the fact that 1 is defined to be the multiplicative identity. :smile:

    Well, not exactly: it says that 1 = -1, which is a perfectly fine thing to say when working mod 2.
     
  7. Jul 23, 2005 #6
  8. Jul 23, 2005 #7
    Well that kinda gets into the very logic of what a negative is. Think of it as logic "not" operations. Think about using these in sentences and you will get the idea.

    -1*-1=1 ; No + No = Yes
    1*1=1 ; Yes + Yes = Yes
    -1*1=-1 ; No + Yes = No
    1*-1=-1 ; Yes + No = No

    For instance, if I said...
    I did not not go to the store. That really means, you DID go to the store. -1 * -1 = 1 ; No + No = Yes ... and so forth.

    You kinda see how there is a relationship between mathematics and grammer? lol Im not sure if this is an official way to present it, but I just though it up a while back.
     
  9. Jul 23, 2005 #8
    so are all words considered to be multiplied together in a sentance? I've always thought they were a sum.

    hmm...

    -1*-1=1 ; No + No = Yes => if someone says no twice, they are unsure of what they really want because they feel the need to back up thier first answer, so therefore they really mean yes... eek! sounds like the guy that never took no for an answer!

    1*1=1 ; Yes + Yes = Yes => if someone says yes twice, they really mean it!

    -1*1=-1 ; No + Yes = No => if someone says both no and yes to something, then lets look on the pessimistic side and assume they meant no. where's our sense of adventure?!

    1*-1=-1 ; Yes + No = No => same reason as above
     
    Last edited: Jul 23, 2005
  10. Jul 23, 2005 #9

    well, could there be a different multiplicitive identy?

    hmm... I'll have to study "multiplicitive identy" for a while to fully follow your argument. For now I'll say that you may have the reason why it won't work
     
  11. Jul 23, 2005 #10
    Yes, they would most definitly have to be multiplied in a sentence. Let me give an example.

    John Kerry didnt not lose the election. == John Kerry lost the election.

    In the first sentence, the words "Didnt", "not", and "lose" all count as logical "not" operations.

    In the second sentence, the word "lost" counts as the only "not" operator. So notice that when you multiply them, they equal the same thing

    -1 * -1 * -1 == -1

    If you got sum, you would get -3 == -1 :surprised what would that mean lol

    Edit :: I say that "lost" is a not operator, and therefore equals -1 because "lost" would be the same as "didnt win".
     
    Last edited: Jul 23, 2005
  12. Jul 23, 2005 #11
    poor Kerry...

    but it's worse for the nation...

    poor us...

    but who doesn't not know how not bad Kerry would not have not not been?
     
  13. Jul 23, 2005 #12
    but who doesn't not know how not bad Kerry would not have not not been?
    But who -1 * -1 know how -1 * -1 Kerry would have -1 * -1 * -1 been?
    Just multiply them, cancel out the 1's, and w00t

    But who knows how good Kerry would not have been? :wtf:

    Yea I was going for Kerry too. Too bad...
     
    Last edited: Jul 23, 2005
  14. Jul 23, 2005 #13

    matt grime

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    yes and no. there may be a different multiplicative identity but it would be for a different multiplicatve structure. all of the algebraic properties of the numbers you are thinking about follow from extending those on the natural numbers and are simple and natural consequences.
     
  15. Jul 23, 2005 #14
    so, would the consequences be that all functions are a transform of all known functions? ie. a mirror image about the y-axis?

    y=x would have a -1 slope?
    imaginary numbers would deal with the sqrt of 1 rather than -1?
    the left side of the y-axis would be the right and vice versa?

    would the linear transformation in [tex]\Re^2[/tex] then be:

    [tex] \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)[/tex]
    ?
     
    Last edited: Jul 23, 2005
  16. Jul 23, 2005 #15

    matt grime

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    what on earth?
     
  17. Jul 23, 2005 #16

    Hurkyl

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    I think he's saying that the multiplication you get by using -1 as the multiplicative identity is isomorphic to the ordinary reals through the map x → -x.
     
  18. Jul 23, 2005 #17

    matt grime

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    one more that i can add to the list of questions that i cannot even begin to answer.
     
  19. Jul 23, 2005 #18
    well, I'm just trying to understand why we've defined multiplication rules between negatives and positives, negatives and negatives, and positives and positives the way we did. I proposed different rules:
    -1*-1=-1
    1*1=-1
    -1*1=1
    1*-1=1
    and I'm trying to find the consequences of how this change would change all the math that is built on multiplication rules, and I'm thinking that everything would be backwards, but math would still work. Negatives would be treated like how we treat positives and positives would be treated like how we treat negatives.

    yes, no?

    but the main question at hand, that is if math would still work but just "backwards", is... Why did people define things the way they are rather than doing it the other way? OR, if this whole idea isn't logical, then that would be the reason why, and I can put it to rest.

    eh, nevermind...

    this is probably more of an anthropology question. or psychology, or evolution of ideas type question or something. I sound like a bumbling fool. :rolleyes:
     
    Last edited: Jul 23, 2005
  20. Jul 23, 2005 #19

    matt grime

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    We (well, not me) already explained this to you.

    0*x=0 for any number, right?

    now 0=a-a for any a, right?

    so put these together and it follows that

    0=0*x=(a-a)*x

    now we want multiplication to be distributive so that

    0=a*x+(-a)*x

    or

    (-a)*x=-(a*x)

    right?

    so you want to undo some or all of the following:

    the definition off additive inverses (ie x-x is no longer 0), the distributivity of multiplication over addition (or that 2=1*2=1*(1+1)= 1*1+1*1 is no longer true, or that 2 is not 1+1), or that 0*x is not 0, but this again follows from the fact that 1*0=(1*(0+0)=2*0, so you would be again denying that 1 is the identity or that multiplication is distributive, or that 0 is the additive identity or that additive inverses exist).

    people defined things the way they are in order to consistently extend the operations that we can think of easily for positive whole numbers. in order to do it differently we would have to redefine most everything and create many special subcases such as a(b+c)=ab+ac if a,b and c are all positive, but not necessarily otherwise. and of cuorse you would then need to define every single possible variation if we are to have nunderlying universal rules.
     
  21. Jul 23, 2005 #20
    I don't follow the reasoning from this:
    (-a)*x=-(a*x)
    to the idea that 1-1 is no longer equal to zero, and 1+1 is no longer equal to 2 or 0*x doesn't equal zero.
    because if:
    1*x=x, and -1*x=-x, then
    (-a)*x=-(a*x)
    0=a*x+(-a)*x
    0=(1)a*(1)x + (-1)a*(1)x
    0=(1*1)*a*x + (-1)*(1)*a*x
    0=(-1)*a*x + (1)*a*x
    which is still true, even with diferent multiplication rules of neg and pos:
    -1*-1=-1
    1*1=-1
    -1*1=1
    1*-1=1

    man, I'm tired of this, are you? I give up, I'll trust that there is a logical reason that you're trying to show me, but I just don't get it because I just don't catch on. btw, I've already pretty much figured that my interpretation that all functions would be flipped across the y-axis is incorrect, y=x would have a slope of 1. I'm bettin that I just don't get it right now so pardon me for being headlong.
     
    Last edited: Jul 23, 2005
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