February 14th, 2011, 10:14 AM  #1 
Newbie Joined: Feb 2011 Posts: 3 Thanks: 0  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. 
February 14th, 2011, 12:37 PM  #2 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: Dividing by zero
What, precisely, do you mean by 'infinity'?

February 14th, 2011, 01:35 PM  #3  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Dividing by zero Quote:
 
February 14th, 2011, 02:19 PM  #4 
Senior Member Joined: Jan 2011 Posts: 120 Thanks: 2  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^(ab) for all (ab)>0 j^a / j^b = 1 for all (ab) = 0 (a=b or (a=0 and b=0)) j^a / j^b = 1/j^(ab) for all (ab) < 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^((bc)j) for all (bc) <>0 a^bj / a^cj = 1 for all (bc) = 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? 
February 14th, 2011, 03:51 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Dividing by zero Quote:
2 = 2 * 1 = 2 * 0j = 0 * 2 * j = 0 * j = 1  
February 14th, 2011, 05:11 PM  #6  
Senior Member Joined: Jan 2011 Posts: 120 Thanks: 2  Re: Dividing by zero Quote:
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 !  
February 14th, 2011, 05:24 PM  #7  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Dividing by zero Quote:
 
February 14th, 2011, 09:55 PM  #8 
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Dividing by zero
Why would we want to divide by zero?

February 16th, 2011, 06:40 AM  #9  
Newbie Joined: Feb 2011 Posts: 3 Thanks: 0  Re: Dividing by zero Quote:
 
February 16th, 2011, 07:01 AM  #10 
Senior Member Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0  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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