Author Topic: Why doesn't zero divided by zero work (Read 18717 times)

Dingus · « **Reply #30 on:** February 25, 2012, 10:20:42 am »

That has been asked and answered. Did you read my post? BTW, the opposite of infinity is minus infinity.

Quote

Anyway I personally have no clue why 0/0 doesn't work, though, but what I think is that 0 by itself is nothing, so basically you divide nothing by nothing. However I think it's also because you multiply nothing by the infinity, but I'm not sure if the opposite of "infinity" is "nothing".

aeTIos · « **Reply #31 on:** February 25, 2012, 10:25:02 am »

So that implies 0/0=1 which is mathemagically correct. (x/x=1)
hmmm...

Dingus · « **Reply #32 on:** February 25, 2012, 11:18:24 am »

Quote from: Jonius7 on February 24, 2012, 03:51:44 am

Yeah, we're learning Calculus (mostly differential atm) in Year 12, and we had a long discussion about limits.
The limit technically does not exist at the given value e.g.: lim h->0 f(x)=3x^2+7x-5.
At h=0 the equation is undefined as we're dividing by 0 (The answer could be infinity, for all we know, or it could be 9000.00000032), but the point of Calculus (well at least this part of calculus) is that we are to find the theoretical derivative of the equation when h=0, even though it technically does not exist.

Hope that made some sense, and I may have made a mistake.

Huh? For continuous functions, i.e., functions whose graphs don't have a break in them and can be drawn without lifting your pencil from the paper, the limit of f(x) as x approaches a is f(a), the value of the function at a. Furthermore, all polynomials are continuous. So technically what you said is incorrect for polynomials but can be correct for other functions which have a discontinuity where the limit is being taken. For example, the rational function (x^2-x)/(x-1) has a discontinuity at x=1 because the function is not defined and doesn't exist there because of division by zero (the denominator is zero), however the limit as x approaches one does exist and is one. You can see from the graph, that as x approaches one from either direction it gets closer and closer to one but "at" one we know the function doesn't exist. On the other hand, the polynomial function x^2-x exists at and is continuous at x=1 so the limit of f(x)=x^2-x as x approaches one is f(1) (the value of the function AT the value that x is approaching) which is zero.

ACagliano · « **Reply #33 on:** February 25, 2012, 11:22:24 am »

Would there be a way to create a calc hook that enables it to automatically figure out a limit if the denominatior = 0

Dingus · « **Reply #34 on:** February 25, 2012, 11:28:13 am »

Quote from: aeTIos on February 25, 2012, 10:25:02 am

So that implies 0/0=1 which is mathemagically correct. (x/x=1)
hmmm...

No, 0/0 is undefined therefore x/x is undefined (does not exist) at x=0! Division by zero is not a valid mathematical operation and a zero denominator is division by zero. In a course on groups, rings and fields you can learn why. For now, just accept that division by zero is just not a valid mathematical operation.

aeTIos · « **Reply #35 on:** February 25, 2012, 11:28:52 am »

AngelFish · « **Reply #36 on:** February 25, 2012, 11:45:35 am »

Quote from: Hot_Dog on February 23, 2012, 12:37:49 pm

Guys, please no "divide by zero" funny pictures on this topic.

Somebody gave me a good reason that 0/0 isn't valid. But I can't remember what it was.

Let's represent 0/0 as something else, namely 0*(1/0). This makes it obvious what the main problem is. Division by 0, of *any* number, whether a real or a rational number, simply isn't defined in the standard system of reals. It's not immediately obvious why this is so. 0 doesn't appear to be a special number (besides being the null quantity) until we look at the definition of a field, which the set of real numbers forms when combined with the operations of addition, negation, multiplication, and multiplicative inversion (Finding A=B^-1). Basically, 0 is what is called the additive identity, which means that any number from the field can be added to it and you will get the same number back out. A+additive identity=A in any field F. The important part of being defined as the additive identity is that by definition, division by the additive identity is undefined in any non-trivial field. It's just a matter of definition. Things go funky when you allow it and inconsistencies creep in, such as being able to "prove" that 1=2 in the real numbers.

TL;DR 0/0 is undefined because convention says it's undefined to make things work out properly. You can say 0/0 is whatever you want, but as soon as you do so, everything else falls apart.

Also, now that I've written this, I see that adriweb beat me to it with a well timed Wikipedia quote.

Dingus · « **Reply #37 on:** February 25, 2012, 07:51:36 pm »

Quote from: Qwerty.55 on February 25, 2012, 11:45:35 am

Quote from: Hot_Dog on February 23, 2012, 12:37:49 pm
Guys, please no "divide by zero" funny pictures on this topic.

Somebody gave me a good reason that 0/0 isn't valid. But I can't remember what it was.

Let's represent 0/0 as something else, namely 0*(1/0). This makes it obvious what the main problem is. Division by 0, of *any* number, whether a real or a rational number, simply isn't defined in the standard system of reals. It's not immediately obvious why this is so. 0 doesn't appear to be a special number (besides being the null quantity) until we look at the definition of a field, which the set of real numbers forms when combined with the operations of addition, negation, multiplication, and multiplicative inversion (Finding A=B^-1). Basically, 0 is what is called the additive identity, which means that any number from the field can be added to it and you will get the same number back out. A+additive identity=A in any field F. The important part of being defined as the additive identity is that by definition, division by the additive identity is undefined in any non-trivial field. It's just a matter of definition. Things go funky when you allow it and inconsistencies creep in, such as being able to "prove" that 1=2 in the real numbers.

TL;DR 0/0 is undefined because convention says it's undefined to make things work out properly. You can say 0/0 is whatever you want, but as soon as you do so, everything else falls apart.

Also, now that I've written this, I see that adriweb beat me to it with a well timed
Quote

Excellent explanation. Thank you.

Jonius7 · « **Reply #38 on:** February 26, 2012, 03:38:02 am »

I'm confused, the whole thing has just been quoted...

Oh, I see at the very bottom inside the quote.

In response to Qwerty's explanation, 0/0 does not fit in with the concepts of mathematics and therefore "everything else" (the mathematical concepts) falls apart when you try to take 0/0 into account. That is why 0/0 is undefined.
Great explanation.

phenomist · « **Reply #39 on:** February 27, 2012, 09:34:02 pm »

Was going to throw in L'Hopital's rule, but there's already an enlightening section on that on page 2.

It all depends on the context of the 0's, really, and simply putting "0" isn't giving the context. Hence its undefinability.

Jonius7 · « **Reply #40 on:** February 28, 2012, 12:00:32 am »

Well 0 is a number but when you try and divide by itself the answer cannot be defined simply in mathematical terms.
Hence, it's kinda out of the realm of mathematics that we understand.

Author Topic: Why doesn't zero divided by zero work (Read 18717 times)

Dingus

Re: Why doesn't zero divided by zero work

aeTIos

Re: Why doesn't zero divided by zero work

Dingus

Re: Why doesn't zero divided by zero work

ACagliano

Re: Why doesn't zero divided by zero work

Dingus

Re: Why doesn't zero divided by zero work

aeTIos

Re: Why doesn't zero divided by zero work

AngelFish

Re: Why doesn't zero divided by zero work

Dingus

Re: Why doesn't zero divided by zero work

Jonius7

Re: Why doesn't zero divided by zero work

phenomist

Re: Why doesn't zero divided by zero work

Jonius7

Re: Why doesn't zero divided by zero work