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Dividing 0 by 0by repugno
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#1
Mar2604, 11:53 AM

P: 79

Greetings all,
What happens if you divide 0 by itself? I had an argument with my math teacher about this; she reasons that there could be 2 answers to this question.  If you divide a number by itself it is always one  If you divide by 0 it is infinity The first explanation I believe is illogical since you cannot have nothing divided by nothing giving you 1. I can also not imagine how dividing 0 by 0 can give you an infinite number. What I can understand is, if I have nothing then I divide this into 0 parts I will end up with nothing, because there was nothing in the first place. Which is considered to be correct? Also, why if we divide a number by 0 we get infinity? I have an object like an apple, I then decide to cut it up into no parts. How could I then have an infinite number of apple pieces? Any help would be much appreciated. Thank you. 


#2
Mar2604, 01:15 PM

P: 383

The idea of "dividing into parts" gets a little slipshod.
For example, what does "divide into 0 parts" mean? What does "divide into 1 part" mean? What does "divide into 2 parts" mean? Working up from the last question, "divide into 2 parts" seems to mean "divide in half". So, 4/2 = 2 seems correct, since 2*2 = 4. If that is correct, then "divide into 1 part" shouldn't mean the same thing, should it? This must really mean "don't divide at all". So, 4/1 = 4, exactly because 4*1 = 4. No division took place at all. Now for "divide into 0 parts". That should be different from "divide into 1 part", shouldn't it? Why not try using a return multiply to confirm an answer, just as was done to verify the prior answers? OK, so what number * 0 should equal 4? No number will work. Whatever number you come up with (call it x), then x*0 can't be 4. So 4/0 must have no answer. The reason for that is that x*0 = 0 no matter what x is. Try it. What is x*0 + x* 0? A fundamental arithmetic postulate says that multiplication distributes over addition, and it undistributes too. So x*0 + x*0 = x*(0 + 0). What is 0 + 0? No problem: it is 0. So, x*0 + x*0 = x*0. x*0 = x*0 + 0. So x*0 + x*0 = x*0 + 0. It's not hard to arrive at the conclusion x*0 = 0, and that is true no matter what number x is, as long as we try to preserve fundamental postulates of arithmetic, like distribution of multiplication over addition. Finally, what about 0/0? Pick any number x whatsoever. Then x*0 = 0, so any x whatsoever would be a correct answer. Why do we get such an answer? We expect that by multiplying by larger and larger numbers, we will always get one too big to render the dividend. So 3*2 = 6 is bigger than 4, while 1*2 = 2 is smaller than 4. In fact,exactly one number between 1 and 3 yields the dividend 4 when multiplying by the divisor 2. Therefore 2 is the unique quotient of 4 and 2. But no number x is large enough to make x*0 larger than 0. So we don't have a unique answer to 0/0. There is another matter where 0/0 shows up. That is in a ratio of limits of functions. Suppose f and g are numericvalued functions over a number ring or field. Then if for some number a, limit f(x) = 0 and limit g(x) = 0 when x > a. Then f(a)/g(a) is written symbolically for the limit of f/g as x > a. In this case, we show 0/0. But that 0/0 is not supposed to represent a number value. It is the limit (supposing there is one!) of the ratio f(x)/g(x) as x > a. But separating the numerator and denominator limits is not valid in this case, so this symbol can't be considered "an answer". Other characteristics of the constituent functions, like continuity and differentiability in a neighborhood of a can offer clues about existence of a limit of the ratio and its value. The actual values of f and g don't matter. This is all actually a different matter (limits of functions), but it might explain some of the varied answers to your query about 0/0. 


#3
Mar2604, 04:34 PM

Sci Advisor
PF Gold
P: 2,226

Just to show why you shouldn't take a limits of functions necessarily as real values:
[tex]x \rightarrow \infty \ , x^\frac{1}{x} \rightarrow 1 \ , x \in [1,\infty)[/tex] If we now make a new number, infinity, taking equalities from limits as you have done and assuming we can perform basic algebra on it: [tex]\infty^\frac{1}{\infty} = 1 \therefore 1^\infty = \infty[/tex] but: [tex]x \rightarrow \infty \ , 1^x \rightarrow 1[/tex] 


#4
Apr1404, 07:04 AM

P: 1,210

Dividing 0 by 0
It is easy to show that:



#5
Apr1404, 08:43 AM

P: n/a

The reason you can't divide any number by zero is because there is no number, real or imaginary, which could be used to represent the answer without creating paradoxes. Let's say you define X = 0/0 and leave the value of X unknown. That means X is just a replacement for 0/0. Harmless enough, right? However, you can't do anything with X. You can't multiply it, divide it, add it, subtract it. The result of 0/0 is a useless number. That's why you can't do it. Last but not least, it's not true that a finite number divided by zero equals infinity. Strictly speaking, division by zero is undefined. It's not a knowable quantity. 


#6
Apr1404, 09:45 AM

P: 1,210

I want to say that x/0 is undifined or useless if we understand it through Boolean Logic, Fuzzy Logic or any other logical system which is basd on 0 XOR 1 connective.
But there can be different forms of logic, which are not 0 XOR 1 connective, for example: http://www.geocities.com/complementarytheory/BFC.pdf 


#7
Apr1504, 04:50 AM

P: 112

The reason we have no zero divisors is because if there was a zero divisor then:
[tex]\exists 0^^1 : 0^^1 \cdot 0 = 1[/tex] But of course this contradicts the fact that: [tex]a \cdot b=0 \Leftrightarrow a \vee b=0[/tex] 


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