actual solution to 0/0
- To: mathgroup at smc.vnet.net
- Subject: [mg37465] actual solution to 0/0
- From: clillian at math.ucla.edu (chad)
- Date: Thu, 31 Oct 2002 04:41:51 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
There are some problems with the "Meaningful Solution" *In response to the first argument: Yes it is true that any number multiplied by zero is zero, infinity is not truly a number so that rule does not apply. The rules of +,-,*,/ do not apply to infinity like they do to numbers (infinity +1=infinity, infinity*2=infinity etc.). Therefore it cannot be said that infinity*0=0 (infinity * anything =infinity, but likewise anything * 0= 0 we encounter a contradiction for inf*0) *In response to the second argument: The claim is that 100/100=1, this is true. Now he claims that you cancel off the zeros (0/0) of the ones place which results in 10/10 followed by the zeros of the tens place resulting in 1/1=1. That is not what is happening. Using that same method to divide 25/125 we can cancel off the 5/5 leaving us with 2/12 =1/6 where 25/125=1/5 (1/6 not equal to 1/5). There is obviously something wrong with that method of cancellation (the problem is that he would cancel the 5s from (20+5)/(120+5) which is illegal). This is what really happens: 100/100=(10*10)/(10*10) =(10/10)*(10/10) =1*(10/10) =10/10 =1 Likewise: 25/125=(5*5)/(5*25) =(5/5)*(5/25) =1*(5/25) =5/25 =5/(5*5) =(5/5)*(1/5) =1*(1/5) =1/5. *In response to the final argument: Yes it is very true that any number divided by itself yeilds one. It is also true that zero divided by any number is zero. Likewise any number divided by zero is undefined (infinite). So we have 3 competing definitions for 0/0: 1. 0/0=1 since any number divided by itself =1 2. 0/0=0 since zero divided by any number =0 3. 0/0=inf since any number divided by zero is infinite or undefined So what is 0/0? It is not defined (because of the three competing definitions), by introducing limits the value of 0/0 can be any finite number, or infinity (but for this we need to be at the calculus level).