|Sep24-11, 08:05 PM||#1|
commutative property of multiplication...
This is similar to the question
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:
`3*`2 =`6 3* 2 = 6 3*`2 = ? `3* 2 = ?
`3*`2 = 6 3* 2 = 6 3*`2 =`6 `3* 2 =`6
If you look at where the multiplication was going you could make rules for when it crosses zero:
3* 2 = 6 3* 1 = 3 3* 0 = 0
3*`1 =`3 3*`2 =`6
`3*`2 =`6 `3*`1 =`3 `3* 0 = 0 `3* 1 = 3 `3* 2 = 6
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:
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.
EDIT: `x*y when removing the negative would probably be closer to `x*y=`x*-`y=-(`x*`y)
... 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?
|Sep24-11, 08:32 PM||#2|
(-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.
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.
|Sep24-11, 08:33 PM||#3|
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?".
|Sep24-11, 09:20 PM||#4|
commutative property of multiplication...
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.
-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.
|Sep24-11, 09:28 PM||#5|
(-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
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.
|Sep24-11, 09:34 PM||#6|
Oh, wait you have
3(-1) = -3
(-3)(1) = 3
And is that different from (-1)(-5)?
|Sep24-11, 09:56 PM||#7|
(-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
|Sep25-11, 08:10 AM||#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
`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
going in a negative or positive direction and changing the direction of movement
this: `1*-`1 = `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.
|Sep26-11, 05:20 AM||#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.
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)
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.
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