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February 14th, 2011, 10:14 AM   #1
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Dividing by zero

hello,
i believe that it is in fact possible to divide by zero. you see, dividing sees how many times a number can be subtracted from a different number. so therefore anything divided by zero would equal to infinity. lets give an example of this
2 divided by 0, would equal infinity because zero can go into two an infinite amount of times.
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February 14th, 2011, 12:37 PM   #2
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Re: Dividing by zero

What, precisely, do you mean by 'infinity'?
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February 14th, 2011, 01:35 PM   #3
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Re: Dividing by zero

Quote:
Originally Posted by mattpi
What, precisely, do you mean by 'infinity'?
That what I want to know, too. Is this infinity equal to itself? Equal to its own sum? Its own product? Its own quotient? Its negation? Its own power? For each of the "no"s, what does that operation yield?
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February 14th, 2011, 02:19 PM   #4
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Re: Dividing by zero

Apologies for just intertwining in this thread, but below is an attempt I made a few years ago to sort the division by zero. It works very well, but there is a 'nasty' little snag I found in the end. Guess you guys will spot it immediately...


Here's the thought:

There is a difference between 0/0 and 1/0. The first is "undertermined" and the second is "undefined". For 1/0 = 0 no definable solution exist since 0 x 0 doesn't equal 1. But analogiously to , why not try to define the undefinable as follows:

j = 1 / 0 -> 0j = 1 -> 1/j = 0

Calculation rules to test if this leads to a consistent model:

Addition:
aj + bj = (a + b) j for all a<>0 and b<>0
aj + bj = 1 + bj for a=0, b <> 0
aj + bj = aj + 1 for b=0, a <> 0
aj + bj = 2 for a=0 and b=0

Substraction:
aj bj = (a - b) j for all a<>0 and b<>0 and a<>b
aj bj = 0j = 1 for all a=b
aj bj = 1 - bj for a=0, b <> 0
aj bj = aj - 1 for b=0, a <> 0
aj bj = 1 - 1 = 0 for a=0 and b=0

Division:
aj / bj = (aj * 1/j) * 1/b = (aj * 0) * 1/b = 1/b for all a<>0 and b<>0
aj / bj = (aj * 1/j) * 1/b = (0j * 0) * 1/b = 0 for a=0, b <> 0
aj / bj = (aj * 1/j) * 1/b = (aj * 0) * 1/0 = j for b=0, a <> 0
aj / bj = (aj * 1/j) * 1/b = (0j * 0) * 1/0 = 1 for a=0 and b=0

Multiplication:
aj x bj = (a*b)j^2 for all a<>0 and b<>0
aj x bj = bj for all a=0, b <>0
aj x bj = aj for all b=0, a <> 0
aj x bj = 1 for all a=0 and b=0

Powers: (j^a)
j^a * j^b = j^(a+b) for all (a+b)>0
j^a * j^b = 1 for all (a+b) = 0 (a=-b or (a=0 and b=0))
j^a * j^b = 1/j^(a+b) for all (a+b) < 0
j^a / j^b = j^(a-b) for all (a-b)>0
j^a / j^b = 1 for all (a-b) = 0 (a=b or (a=0 and b=0))
j^a / j^b = 1/j^(a-b) for all (a-b) < 0

Powers: (a^j)
a^bj * a^cj = a^((b+c)j) for all (b+c) <>0
a^bj * a^cj = a for all (b+c) = 0
a^bj / a^cj = a^((b-c)j) for all (b-c) <>0
a^bj / a^cj = 1 for all (b-c) = 0

Goniometrical:
tan (90 + n*180) = j (n = 0, 1, 2, ...)
sin (90 + n * 180) = j * cos (90 + n * 180)
cotan (90 + n*180) = 1/j = 0

Commutativity?
For all a b is is true that:
aj + bj = bj + aj
e.g. 2j + 0j = 0j + 2j (= 1 + 2j)

aj * bj = bj * aj
e.g. 1j * 2j = 2j * 1j (= 2j^2)

Distributivity over addition?
For all a b is is true that:
(aj + bj) * cj = aj * cj + bj * cj.
e.g. (0j + 0j) * 0j = 1*1 + 1*1 (= 2)

(1j + 1j) * 1j = 1j*1j + 1j*1j (= 2j2)
e.g. (0j + 1j) * 0j = 0j * 0j + 1j * 0j (= 1 + j)

Associativity?
For all a b is is true that:
aj + (bj + cj) = (aj + bj) + cj
e.g. 0j * (1j * 2j) = (0j * 1j) * 2j (= 2j2)

aj * (bj * cj) = (aj * bj) * cj
e.g. 0j * (1j * 2j) = (0j * 1j) * 2j (= 2j2)
e.g. 2j * (3j * 4j) = (2j * 3j) * 4j (= 24j3)

Who spots the exception where the whole thing fails?
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February 14th, 2011, 03:51 PM   #5
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Re: Dividing by zero

Quote:
Originally Posted by Agno
Who spots the exception where the whole thing fails?
The system is inconsistent:
2 = 2 * 1 = 2 * 0j = 0 * 2 * j = 0 * j = 1
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February 14th, 2011, 05:11 PM   #6
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Re: Dividing by zero

Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by Agno
Who spots the exception where the whole thing fails?
The system is inconsistent:
2 = 2 * 1 = 2 * 0j = 0 * 2 * j = 0 * j = 1
Spot on (I actually hadn't even found that one). The other inconsistency is:

1 = 0j = (0 + 0)j = 0j + 0j = 1 + 1 = 2 (and this also implies 1 = 2,3,4...+n).

Impressed by the speed of your brain, CR !
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February 14th, 2011, 05:24 PM   #7
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Re: Dividing by zero

Quote:
Originally Posted by Agno
The other inconsistency is:

1 = 0j = (0 + 0)j = 0j + 0j = 1 + 1 = 2 (and this also implies 1 = 2,3,4...+n).
I saw that, too, but I happened to 'formalize' the other one first.
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February 14th, 2011, 09:55 PM   #8
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Re: Dividing by zero

Why would we want to divide by zero?
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February 16th, 2011, 06:40 AM   #9
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Re: Dividing by zero

Quote:
Originally Posted by DLowry
Why would we want to divide by zero?
for the lols
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February 16th, 2011, 07:01 AM   #10
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Re: Dividing by zero

If division by zero is consistent, then this implies that all elements are equal to 0. Indeed, if a/0 = x, then a = 0*x=0. Thus we can at most divide zero by zero. But 0/0 = x implies that 0=0*x. So x can be anything. So all elements are equal to 0/0, thus they are all equal to each other.

In summary, there is only one number system where division by zero is permitted. It's the number system {0} consisting of only one object. In this number system, all binary operations (including division) have the result 0.

Final observation: Allowing division by zero is useless.
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