Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Hot_Dog on February 23, 2012, 12:37:49 pm
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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.
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Because anything multiplied by 0 equals 0, you piece of shit.
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Warning, fake Juju.
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Because anything multiplied by 0 equals 0
Technically that means that 0/0 should be "all real numbers."
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0/0, 0 can still go into 0 and ambiguous number of times. It can go into 0 1,2,3... ∞ number of times.
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Because 0/# should always be zero, but #/0 should always approach infinity. It's very undefined.
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Note: The fake juju knows nothing about math. Division by zero is not a defined operation because doing so can cause incorrect results. Take for example
x(x-x)=x^2-x^2. Factor the right side and divide both sides by x-x and you have x(x-x)/(x-x)=(x+x)(x-x)/(x-x) Then canceling results in x=x+x which is x=2x. Finally, divide both sides by x and you have 1=2! The only mistake made was dividing by x-x which is zero, i.e., I divided by zero which caused the disastrous result of 1=2!
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Note: The fake juju knows nothing about math. Division by zero is not a defined operation because doing so can cause incorrect results. Take for example
x(x-x)=x^2-x^2. Factor the right side and divide both sides by x-x and you have x(x-x)/(x-x)=(x+x)(x-x)/(x-x) Then canceling results in x=x+x which is x=2x. Finally, divide both sides by x and you have 1=2! The only mistake made was dividing by x-x which is zero, i.e., I divided by zero which caused the disastrous result of 1=2!
Now THAT makes sense.
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Note: The fake juju is actually some idiot troll from Juju's school. He often trolls his blogs and Youtube channel with random weird stuff but it seems he found this site now. <_<
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Hahahaha, fake juju.
(http://blackicepro.com/pics/divideby0.jpg)
Maybe that was the pic you were looking for Hot Dog :P
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Ephan :P
Guys, please no "divide by zero" funny pictures on this topic.
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".
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This is how I always knew it. Let's say you have 6 candies to divide by 3 people, well, each person gets 2 candies.
But if you have 6 candies to divide by 0 people, well, that can't be done. It's not 0 candies for each person, because there are no people to give candies, so it's just impossible.
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It's infinite candies for nobody!!!! YAY CANDY CLONER
Edit (to give this post more value):
Dingus post is very interesting and is almost a proof that you can't devide by zero.
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Oh right, I guess ephan's example kinda clears it up.
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I just read that on Wikipedia, and it's pretty good :)
(warning : quite high level math...)
Abstract algebra
Any number system that forms a commutative ring — for instance, the integers, the real numbers, and the complex numbers — can be extended to a wheel in which division by zero is always possible; however, in such a case, "division" has a slightly different meaning.
The concepts applied to standard arithmetic are similar to those in more general algebraic structures, such as rings and fields. In a field, every nonzero element is invertible under multiplication; as above, division poses problems only when attempting to divide by zero. This is likewise true in a skew field (which for this reason is called a division ring). However, in other rings, division by nonzero elements may also pose problems. For example, the ring Z/6Z of integers mod 6. The meaning of the expression should be the solution x of the equation 2x = 2. But in the ring Z/6Z, 2 is not invertible under multiplication. This equation has two distinct solutions, x = 1 and x = 4, so the expression is undefined.
In field theory, the expression is only shorthand for the formal expression ab−1, where b−1 is the multiplicative inverse of b. Since the field axioms only guarantee the existence of such inverses for nonzero elements, this expression has no meaning when b is zero. Modern texts include the axiom 0 ≠ 1 to avoid having to consider the trivial ring or a "field with one element", where the multiplicative identity coincides with the additive identity.
( http://en.wikipedia.org/wiki/Division_by_zero#Linear_algebra (http://en.wikipedia.org/wiki/Division_by_zero#Linear_algebra) )
Also :
To show how division by zero does not work with the rules or mathematics we can use the associative law of multiplication and the fact that 0x2=0
We can use the rules of arithmetic to show 0x2=0 as follows:
given 2 is defined as 1+1=2 then
M1 tells us that 0x2=2x0
D tells us that (1+1) x0 = 1x0 + 1x0
But M3 tells us that 1x0=0
so this equals 0+0
finally A3 implies 0+0=0
Now having proved 0x2=0 we can use this with M2 to give
1 = inf x 0 = inf x (0x2) = (inf x 0) x 2 = 1 x 2 = 2
Since 1 does not equal 2 then an inconsistency with the most basic rules of numbers and as such division by zero does not work with these rules.
M1: The commutative law of multiplication states:
ab=ba
for any two numbers a and b
M2: The associative law of multiplication states:
a(bc) = (ab)c
for any 3 numbers a, b and c
M3: The multiplicative identity:
1a = a
for any number a
D: The distributive law states:
(a+b)c = ac + bc
for any three numbers a, b and c
A3: The additive identity:
0+a=a
for any number a
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The division of zero is the reason why L'Hôpital's rule (http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule) is necessary for finding some limits in calculus.
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This is also why L'Hôpital's rule (http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule) is necessary for finding some limits in calculus.
which is awesome :)
(but sadly rarely usable on complex cases :( )
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(but sadly rarely usable on complex cases :( )
On what sort of cases is it unusable?
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These :D
(see attached : it's a math homework about calculating limits of the functions written)
(maybe some of them are doable, but probably not all)
(since you can't always take the derivative (easily))
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These are tricky (especially since I'm not very familiar with the hyperbolic trig functions), so I cheated and used Wolfram Alpha. http://www.wolframalpha.com/input/?i=lim_x-%3E0%281%2Fsin%28x%29%5E2-1%2Fsinh%28x%29%5E2%29
If you look in the steps, it uses L'Hôpital's rule.
Many functions need to be rewritten to properly use L'Hôpital's rule though.
I don't really understand the ones in Exercice 5-7 since I don't read french very well yet. They sort of look like integrals though.
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Well it seems that all of the fancy and complex responses have been taken... The simplest way it was described to me was that you can't take nothing from nothing, because there's nothing to give and you're really not getting anything in return.
inb4 someone finds some logical fallacy with that.
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These are tricky (especially since I'm not very familiar with the hyperbolic trig functions), so I cheated and used Wolfram Alpha. http://www.wolframalpha.com/input/?i=lim_x-%3E0%281%2Fsin%28x%29%5E2-1%2Fsinh%28x%29%5E2%29
If you look in the steps, it uses L'Hôpital's rule.
Indeed, but that's because it's well-adapted for this function (simple trig). Or not ? (Look at all the steps... it works but it's kind of ridiculous :P)
But it's probably not the easiest/fastest way to solve this limit...
Anyway, for trigs functions I know it can be quite helpful, but for what I'm studying so far at school, we don't do these anymore, since it's always the same thing, but rather do limits of series and difficult things :(
(Edit : I love WolframAlpha 9001%, but it sometimes fails, and can't always give you steps ( :( ) for example : http://www.wolframalpha.com/input/?i=lim_x-%3E1+ln%28x%29%5E%28x-1%29 (http://www.wolframalpha.com/input/?i=lim_x-%3E1+ln%28x%29%5E%28x-1%29) )
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I understand what you're saying. I learn about series next year.
Wolfram Alpha is great but it feels frustrating that it now costs money to use all of its cool new features (and a couple of its old features).
Since it's still getting new things added to it, it's fine that it doesn't always show exactly what I want, yet. (I'll wait until it can read minds.)
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Yeah, I had to cope with divisions by zero in limits in my math courses, we just had to move the x's around until the division by zero disappears.
And yeah, any number multiplied by 0 is 0, so 0 divided by 0 should be any number, which makes no sense, therefore 0/0 is undefined.
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The interesting thing is when you start working with limits. Lets say you have two equations that both equal zero at X=1, and lets say we divide one equation by the other. If we find the ratio at X=1, obviously that is undefined. But if we get reeeeallly close to X=1 (take the limit as X->1) the ratio can actually approach any number! Depending on the equation of course, I once devised an equation that was undefined at X=0, but the limit approached 42 :D
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0/0?
You can't even write that out ;P
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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.
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I hate limits.
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Think about this....
Take a number, n. Subtract 0 from it and keep going until n=0. Let me know when you're done. ;)
If you think about it that way, you get infinity. Technically, infinity IS an undefined number (the textbook definition of infinity is the inability to assign a definite quantity).
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There is one flaw in that tho, what if n were 0 to start off with then taking 0 from n would still equal 0 and I'd be done!!! :D. It's a good simple example tho, not taking the flaw into account.
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That has been asked and answered. Did you read my post? BTW, the opposite of infinity is minus infinity.
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".
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So that implies 0/0=1 which is mathemagically correct. (x/x=1)
hmmm...
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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.
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Would there be a way to create a calc hook that enables it to automatically figure out a limit if the denominatior = 0
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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.
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Ok.
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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.
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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
Excellent explanation. Thank you.
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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.
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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.
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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.