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Commutative property of multiplication

  1. Sep 24, 2011 #1
    This is similar to the question

    However, it is slightly different. It's probably more of a philosophy of math topic, however, I posted this on a philosophy forum and can't find many people interested in math there. I hope it hasn't been overly discussed. Thank you for reading even if you don't respond or think I'm losing a few marbles.

    I was talking about imaginary numbers with someone, and thinking it over. I understand how to use and apply them, but I have a problem with the operations themselves, but couldn't really express myself. After a lot of thought, the exact point where there was a disconnect came into view.

    Should multiplication be commutative?

    Granted it is for positive numbers like when the Peano axioms were written, but it was taken as a given for negative as well. It seems to me that it was forced commutative, because that's the way it had been and it worked. Fortunately, how we do it works, but unfortunately, it does change the outcome slightly.

    EDIT: the negative are represented by "`" in this example

    Say we already had real numbers, but had never touched on negatives. If we had thought of them as opposite of positives, this is what we could have ended up with:

    Code (Text):

    `3*`2 =`6
     3* 2 = 6
     3*`2 = ?
    `3* 2 = ?
    The problem was with the cross over of negative and positive. There are many solutions that could come from this, but the one we ended up with changed what I see as the natural outcome of multiplication.

    Code (Text):

    `3*`2 = 6
     3* 2 = 6
     3*`2 =`6
    `3* 2 =`6
    It seems to work fine because it's commutative just like the axioms that were written when negative numbers weren't even being used. I currently disagree with this outcome.

    If you look at where the multiplication was going you could make rules for when it crosses zero:

    Code (Text):

     3* 2 = 6
     3* 1 = 3
     3* 0 = 0
    So the next natural step (IMO):

    Code (Text):

     3*`1 =`3
     3*`2 =`6
    And for negative:

    Code (Text):

    `3*`2 =`6
    `3*`1 =`3
    `3* 0 = 0
    `3* 1 = 3
    `3* 2 = 6
    It's more complex in many ways (commutative doesn't work across the board, distributive is different, and probably associative is different also), but to me it seems cleaner (for lack of a better word).

    It would change many things, and no one will probably ever accept it. However, does this make sense to anyone else?

    FYI: the proofs I see online won't do for this, because they assume distributive property as well which is closely related (I think derived from this property actually). For instance:

    But that breaks on this line (for this way of thinking): "(-1)(-1) + 1(-1) = 0(-1) . distributive property".

    Because that already doesn't equal 0*`1 with the argument given. According to what is said above (`1)(`1) + 1(`1) would equal `1 + `1 = `2.

    but that would break down at (-x)*(-y) + (-(x*y)) + x*y = x*y because (`x)*y with what is said above would probably be -(x*`y)

    EDIT: `x*y when removing the negative would probably be closer to `x*y=`x*-`y=-(`x*`y)

    This breaks at (-a)(-b) + a(-b) = (-a + a)(-b) = 0 * b = 0, because you can't pull out the "-b" like that. The distributive property is different. The same with distributing the negative in the "hence".

    ... This is why I said things are more complex, because if you think it makes sense, then a lot of the other properties can't be assumed off the bat either.

    Another way it's been brought up is `1+`1 can mean a debt added by one debt, but `1*2 is a debt multiplied by 2. The problem I have with this, is that it's multiplied by what? The surplus? If you use this argument and just say it's multiplied by the quantity, then why isn't that a debt quantity or basically saying `1*`2 means a debt times 2 of debt? This might sound odd if you look at it from the debt side, but if you were, it would be 1 surplus times 2 of surplus. From the surplus side it's 1 debt times 2 of debt. The opposite multiplying a debt by a surplus or multiplying a surplus by a debt would have to be defined and explained why it would be that way, wouldn't it?

    Am I sounding too much like a loon with this line of thinking?
    Last edited: Sep 25, 2011
  2. jcsd
  3. Sep 24, 2011 #2


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    Your thesis breaks here. We define the notation -3 as the additive inverse of 3, that is, -3 represents the element which satisfies (-3) + (3) = 0.

    (-3)(-2) = -6 = -(3)(2)
    by factoring 6. But we have (-1)(3) = -3, so we can divide by -3 and we are left with -2 = 2. So every number has the property x = -x, and hence you have thrown out the concept of negatives altogether.

    Let x represent a transaction, if x = 3, it means that I give you three dollars. If x = -4, it means that you give me four dollars. Then 2x is same transaction twice. So 2(-1) means you give mean one dollar and then you give me another dollar. Hence you have given me two dollars, which is -2. The transaction has been multiplied by repeating the same transaction.

    Another option is the following: consider velocity. A car travels at 5 meters per second. Now a car can go forwards or backwards. So -5 m/s means reverse. Let's use the latter case. Take a camcorder and record the car. If I press play for 2 seconds the car travels 10 meters backwards on my TV. But if I press rewind for 2 seconds the car appears to travel forwards 10 meters. This is why negatives multiplied by negatives should be positive.
    Last edited: Sep 24, 2011
  4. Sep 24, 2011 #3


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    I don't see quite what your point is. Yes, we could define (-2)(-3)= -6 but to what purpose? As you say, we would immediately lose commutativity. We would also lose the "distributive law"- (-3)(-2+ 2) would be equal to (-3)(0)= 0 while (-3)(-2)+ (-3)(2)= -6- 6= -12.

    Rather than extend the definition of "times" to negative integers in a way that destroys some of the basic rules of arithmetic we learned for positive integers, it makes much more sense to define it in such a way as to keep those laws- which is what we do when we define "negative times negative is positive".

    I must say that the original question, "Since (-3)+(-2)=-5, why doesn't (-3)*(-2)=-6?", seems very strange to me. It implies that the one follows from the other and that is not true. It's like asking, "Since rabbits are gray, why are cows yellow?".
    Last edited by a moderator: Sep 27, 2011
  5. Sep 24, 2011 #4
    No, however, it is different. The negative are kind of like a different number line, but not thrown out. -3 would represents the element which satisfies 0 - 3. It would be a translation from one line to the other.

    For (-3)(-2) = -6 = -(3)(2) to work, it would have to be defined as such. Not saying my definition is good, but it seems natural so far. (-3)(-2) = -6 =/= -(3)(2) = 6 and (-1)(3) = 3. I'm just saying that this concept is more complex, I'm not trying to throw out negatives.

    Still fits actually with it up front. 2*-1 = -1 in both systems if you define it the way I did.

    When you play it forward you are not changing the outcome of the numberline:

    -5 * -1 = -5

    When you play it backward, you are flipping the operation to the other side of the numberline:

    -5 * -(-1) = 5

    So it appears forward.
  6. Sep 24, 2011 #5
    (-3)(-2)+ (-3)(2)= -6 + 6= 0
    (-3)(-2)+ (-3)(-2)= -6 - 6= -12
    (-3)(-2)+ (3)(-2)= -6 - 6= -12

    It is more complex though, so not saying it should be taken. However, I'm not sure why commutativity is assumed true other than that it fit the definitions that were already in place

    I would agree that it makes things easier, and in that it makes sense. However, it changes the outcome slightly. I'm not even suggesting we throw out the current system, but want to know if this would work as a system in itself.

    Oh, as far as why. Because it seems more natural. Not from the perspective of someone who already knows it. Like me, if I had told myself the same thing a few weeks ago, I would have thought I was insane. I think that there might be something to it and may show another side to operations that we might be missing.
    Last edited: Sep 24, 2011
  7. Sep 24, 2011 #6


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    Oh, wait you have
    3(-1) = -3
    (-3)(1) = 3

    missed that.

    Wait what? Do you mean 2(-1) = -2?

    So what is -(-5)?
    And is that different from (-1)(-5)?
  8. Sep 24, 2011 #7
    -(-5) would be an operation that changes the number line from negative to positive.

    (-1)(-5) would be and operation of 1*5 on the same number line.

    Basically, if you viewed negative numbers with a "`" it may be easier to see how it seems:

    `1*`5 = `5

    -`5 = 5

    That's how it's different.

    In multiplication `1 * 5 = `1 * -`5 = - (`1*`5) = -(`5) = 5
  9. Sep 25, 2011 #8
    It won't let me edit the OP, so here is how distribution of multiplication over addition would work in this system. I hadn't really thought it fully through when I made the first post, so didn't know the outcome at that time.

    b(a + c) = ba + bc

    (a + c)b = ab + cb

    which means:

    `1 * 5 = -`1 * 5 = 1*5 = 5

    `1 * 5 = `1 * -`5 = (`1*`5)*1 = `5*1 = `5*1 = 5

    Edit: outside of trying to figure out the structure, let me outline the thinking just a little more

    This is how I see current vs the way this is.

    going in a negative or positive direction and not changing direction of movement
    current: -1*1
    this: `1*`1

    current: 1*1
    this: 1*1

    going in a negative or positive direction and changing the direction of movement
    current: -1*-1
    this: `1*-`1 = `1*1

    current: 1*-1
    this: 1*-1 = 1*`1

    If you wanted to change this system from amount of movement to a positional system, then the first number would be a coordinate in which to move from. This is even more complex, so that's why I didn't even bring it up.

    If a and b are the same sign:

    a*-b = a-(a*b)

    EDIT: Working on real problems makes for a complete mess of things. Multiplying from the right doesn't work for distribution, so the rules need to be fully solidified (at least for the multiplication and addition) before I can even try to push this any more. Thank you for comments.
    Last edited: Sep 25, 2011
  10. Sep 26, 2011 #9
    Ah, I found the flaw in thinking that I was looking for. It was literally the word "movement" that showed me it. Well, that and trying on real world problems. I really was trying to find a movement system and make it work outside of itself.

    Either way, I found that the only way to really make it work was to use it exactly as you use addition and subtraction: from the base. That still means it's not commutative, but it could work (maybe) even though it really is better just to use the current system.

    Code (Text):

    movement            current
    `3 * `1 = `6 = -6 = -3 + (-3 * 1)
    `3 *  1 =  0 =  0 = -3 - (-3 * 1)
     1 * `3 = `2 = -2 =  1 - ( 1 * 3)
     a *  0 =  a =  a =  a - ( a * 0) = a + ( a * 0)
    I'd still like to write it up one day, just so it can be pointed to when someone asks why a negative and a negative can't be negative.

    EDIT: I would have seen it earlier if I had thought about what I called "positioning" or the problem "a*-b = a-(a*b)".

    This means that for every x: x*0=x, x*S(n) = x+x*n or something like that. My grasp of logic language isn't the best in the world.

    An example of conversion from one system to the other I think would be something like (just to show that this isn't really saying anything more than what's already out):

    with this system: f(x) = C*x

    would be normally: f(x) = (C+(C*x)) and (-C+(C*x)) if y is negative
    and for powers on up, it would be even more complicated.

    EDIT2: I just noticed something, I think this is kind of a reverse complex number system. I think that the conversion to normal might be closer to f(x) = (C+(C*x)) and (C-(C*x)) rather than what I had... Maybe I should look into complex analysis if I want to actually make any headway in this.
    Last edited: Sep 26, 2011
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