# 0 divided by 0

#### Apoloto

##### Eat your veggies now SHEWT EP!
Registered Senior Member
Okay, this might be a stupid thread, but I am just wondering: what do you think the outcome would be if you divided 0 by 0?
So basically, this is a thread created for, discussing 0 divided by 0. And yes, I am bored right now.:yawn:

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Undefined.
There is no answer because any answer would seem to be correct.

Does zero exist to be devided by itself?

Okay, this might be a stupid thread, but I am just wondering: what do you think the outcome would be if you divided 0 by 0?
So basically, this is a thread created for, discussing 0 divided by 0. And yes, I am bored right now.:yawn:

Apoloto, this is actually not a stupid question. When working in physics, mathematics, and especially calculus, you could definitely run into (0/0) conditions.

And fortunately for us, a very smart guy in the 17th century by the name of, Guillaume de l'Hôpital, solved this problem.

It is called the L'Hopital's rule: See L'Hopital's rule

See: Discussion on L'Hopital's rule on (0/0)

Best

Thanks Magneto...interesting information.

I bet homeboy was a real ladies' man back in the day with this killer hairstyle!

Thanks Magneto...interesting information.

I bet homeboy was a real ladies' man back in the day with this killer hairstyle!

Actually, he was rumored to be a ladies man and a thief. History tells us that he used a lot of Bernoulli's ideas, and claimed them as his own.

See: Guillaume de l'Hôpital and Bernoulli Controversy

Using someone elses ideas and claiming them as your own is similar to someone that post frequently on the internet, whos moniker ends in "Sight"

Thanks, Magneto 1, but though L'Hopital's rule looks interesting, I'm sorry, but this kind of math isn't my forte. Also, who were you referencing at the end of your reply to MacGyver1968?

.... Also, who were you referencing at the end of your reply to MacGyver1968?

It wasn't you!

Then who was it? PLZZZ

Thanks, Magneto 1, but though L'Hopital's rule looks interesting, I'm sorry, but this kind of math isn't my forte. Also, who were you referencing at the end of your reply to MacGyver1968?

You could try using the search and get a list you can play with. It seems you are getting more play out 0/0 than I would have thought.

You could try using the search and get a list you can play with. It seems you are getting more play out 0/0 than I would have thought.
Yeah, I've been getting a lot of play from ALL of my threads! Except for my Sci-Fi book recommendations, which I want to get replied to the most :l

You could try using the search and get a list you can play with.

Yes...by doing this you would not be Far from the answer!

Yes...by doing this you would not be Far from the answer!

The Reals are defined in terms of addition and multiplication. Division and subtraction are not different operators but short hand for inverses under these binary operations. For instance 1/3 is not "1 divided by 3" it is "The number which when you multiple it by 3 you get 1". Thus the notion of 1/x is generalised to "The number which when you multiple it by x you get 1". There is no number in the Reals such that when you multiple it by 0 you get 1 thus 1/0 doesn't exist, thus asking "What is 0/0" is meaningless because 0/0 = 0*(1/0) and the second term isn't a number. You might as well be asking "What is blue*elephant?".

L'Hopital's work doesn't come into it since the whole point is that it is defined by limits because direct evaluation is meaningless.

L'Hopital's work doesn't come into it since the whole point is that it is defined by limits because direct evaluation is meaningless.

I am sure that you already know this Dr. Professor.

But the short and skinny of L'Hopital's rule is this.

If you have a function that results in f(x) = (0/0), then you take the differential of that function. If that function after differentiating still ends up, f'(x) = (0/0), then you take the differential of that function again. f''(x) = (0/0). And you keep doing this until you get a function that is not (0/0); f'''(x) = 1.

If you keep taking the differentials and you keep getting f'''''''''(x) = (0/0) then that equation is undefined and has no limits!!

Best.

I am sure that you already know this Dr. Professor.
I love how you try to make fun of me for having a doctorate when earlier in the week you were patronising to CptBork for being 'only' a grad student and he should study harder. So someone who is working for a PhD you try to insult and someone who has finished a PhD you try to insult?! That's consistent! Particularly when you've not studied for nor do you have a PhD.

Insulting people for their academic record when they have a better academic record than you is just thick, Mr Kemp.

But the short and skinny of L'Hopital's rule is this.

If you have a function that results in f(x) = (0/0), then you take the differential of that function. If that function after differentiating still ends up, f'(x) = (0/0), then you take the differential of that function again. f''(x) = (0/0). And you keep doing this until you get a function that is not (0/0); f'''(x) = 1.
Firstly the principle of L'Hopital's rule is high school stuff. Secondly it does nothing to retort what I said. Thirdly it assumes the 0/0 is generated by the ratio of two differentiable functions. Simply asking "What is 0/0?" makes no reference to functions and limits so you can't make any use of the rule. Secondly it is possible to consider an f(x) and g(x) such that f(0) = 0 and g(0) but such that $$\frac{f(x)}{g(x)}$$ at x=0 is anything. For instance $$f(x) = \sin(kx)$$ and $$g(x) = x$$ gives the ratio as k at x=0.

If you keep taking the differentials and you keep getting f'''''''''(x) = (0/0) then that equation is undefined and has no limits!!
Not necessarily. The use of L'Hopital's rule is essentially the application of a Taylor expansion for the numerator and denominator functions and considering the first non-zero term. Some smooth non-zero functions can have Taylor expansions where each and every coefficient is zero. For example $$f(x) = e^{-\frac{1}{x^{2}}}$$ at x=0 has Taylor coefficients of 0 to all orders, yet clearly if I consider g(x) = 2f(x) then the ratio $$\frac{f(x)}{g(x)}$$ at x=0 is simply 1/2, despite L'Hopital's rule being inapplicable yet the functions are both infinitely differentiable at x=0.

You should spend less time trying to insult people more competent at physics and maths than you and more time reading actual physics and maths. Then you'd not need to mindlessly parrot Wikipedia articles you don't understand. Perhaps you might even stop putting your foot in your mouth every time you post, Mr Kemp.

Undefined.
There is no answer because any answer would seem to be correct.

Not all infinities are created equal.

Capt. Obvious says: If we assume f and g are scalar functions and set f = g in L'Hopital's equation we get 0/0 = 1 and multiply f by a constant a we get any scalar.

to see this just use g = 1/x:

6(0.00000000001)/(0.00000000001) = 6 and it don't matta' how many 0's you want to put after that decimal point ...