Mathematical Proof there Is No God

[Hijiri Byakuren]

[Casparov]

Division by zero is illegitimate, hence n/0=∞ is not true for any n. It's simply mathematical nonsense. ∞ isn't even a legit number in any field (which the real numbers form).

Disclaimer: I do not actually employ this as a good argument for the existence god.

That being said, it can be argued that n/0=undefined is just a way of getting around the actual conclusion that n/0=∞.

the equation: 9/3=X can be written out as: How many times can 9 be divided by 3? Or: How many 3's can you fit inside of 9? The answer of course, no matter how you ask the question, is 3. You can fit 3 3's into a 9, and you can divide 9 by 3 exactly 3 times. Therefore, the answer to 9/3=X is X=3. So 9/3=3.

Similarly, the equation: 9/0=X can be written out as: How many times can 9 be divided by 0? Or: How many 0's can you fit inside of 9? The answer of course, no matter how you ask it, is ∞. You can fit an infinite amount of zero's into a 9, and you can divide 9 by 0 an infinite amount of times.

Therefore, the true answer to 9/0=X is X=∞. So n/0=∞.

Mathematicians avoid ∞ these days as much as ancient mathematicians avoided 0, that's why they say that n/0=undefined. It's analogous to saying "You can't ask that question!" But of course we can, they can't tell us what to do.  O0

[PickelledEggs]

Seconded.

Plus another one is a thirded. And further mathmatical proof that god doesn't exist. 2+1 =3 therefore no god.

[aitm]

Division by zero is illegitimate, hence n/0=∞ is not true for any n. It's simply mathematical nonsense. ∞ isn't even a legit number in any field (which the real numbers form).

The null set is a legitimate set, and it exists in any universe of discourse involving sets. Hence it is by definition not the universal set because the null set doesn't contain anything, including itself, and the universal set has to contain everything by definition. Furthermore, the null set is NOT by definition "that which does not exist" â€" it's just the set with no elements as members. The set of "that which does not exist" would be chock full of elements. The set containing the empty set is a distinct set, not the empty set â€" it contains a single member: the empty set.

So both of your lines of reasoning fail, and you've here demonstrated another two subjects you don't know: set theory and real analysis.

:undecided:

oh yeah....well..............well...........er...


air handlers under 5 ton don't need duct smoke detectors by code........so there mother fucker!

[Hakurei Reimu]

That being said, it can be argued that n/0=undefined is just a way of getting around the actual conclusion that n/0=∞.

No it's not. ∞, with the stated properties, destroys the field properties.

For any two numbers a and b, there is a unique number x that is the solution to x/a = b. This is derivable completely from the field properties and is a theorem. Yet you claim that n/0=∞ for all n in the field. There's only one way to resolve the two statements â€" that there is but one element in the field. It is the only way your statement can be resolved with the field properties.

We've just proven that the real numbers only contain one element, call it s. Yet for any field, we have two unique elements, denoted 1 and 0 (the additive and multiplicative identities) that must also be in the field to be a field. So we've reached a contradiction. If s = 1, s ≠ 0, and vice versa. Therefore, ∞ cannot belong in the field, and since division produces a member of the field, ∞ cannot be the answer to any expression involving the real numbers.

[Casparov]

No it's not. ∞, with the stated properties, destroys the field properties.

For any two numbers a and b, there is a unique number x that is the solution to x/a = b. This is derivable completely from the field properties and is a theorem. Yet you claim that n/0=∞ for all n in the field. There's only one way to resolve the two statements â€" that there is but one element in the field. It is the only way your statement can be resolved with the field properties.

We've just proven that the real numbers only contain one element, call it s. Yet for any field, we have two unique elements, denoted 1 and 0 (the additive and multiplicative identities) that must also be in the field to be a field. So we've reached a contradiction. If s = 1, s ≠ 0, and vice versa. Therefore, ∞ cannot belong in the field, and since division produces a member of the field, ∞ cannot be the answer to any expression involving the real numbers.

I am very interested in what you are saying here, can you explain in more detail? And can you point in the direction of more information about this? I am not familiar with Field Theory. I am interested in why you say, "there is but one element in the field. It is the only way your statement can be resolved with the field properties." Perhaps there really is only one element in the field.

[aileron]

No it's not. ∞, with the stated properties, destroys the field properties.

Some mathematical systems, such as the IEEE standard in computation do define division by zero as infinity.  The main reason for this is that computational devices can try to preserve some information in the case of an arithmetic overflow.  The IEEE also has the concept of a signed zero to preserve information in case of an underflow, so dividing by zero actually results in either positive or negative infinity depending on the sign of the zero.  Math purists may gag at all of this, but it does serve a purpose in programming.

[Hakurei Reimu]

I am very interested in what you are saying here, can you explain in more detail? And can you point in the direction of more information about this?

http://digitalcommons.trinity.edu/mono/7/

I am interested in why you say, "there is but one element in the field. It is the only way your statement can be resolved with the field properties." Perhaps there really is only one element in the field.

Suppose c and d are solutions to x/a = b. Then c = d and the solution must be unique. For ba = ba (by definition), (c/a)a = (d/a)a (substitution), c(1/a a) = d(1/a a) (association), c 1 = d 1 (definition of reciprocal), c = d (multiplicative identity). Given only that c and d are both solutions to x/a = b, we prove that c = d. Therefore, the solution to x/a = b is unique.

If n/0=∞ were legit, then it fits the form x/a = b, so the theorem we just proved applies. Therefore, n/0=∞ for all n in the field, and at the same time, there is one unique element in the field, n, that is the solution to this equation â€" the implication is that if you pluck out any two elements from the field, you are always guaranteed that they will be the same element. If there were more than one element in the field to choose from, then you could pluck two distinct elements from the field, which would either violate our unique solution theorem, or falsify our hypothesis.

But to be a field, it must have two distinct elements (0 and 1) by property D in the reference I just gave you. The field thus contains at least two elements. Since the theorem is true by the field properties, it must be our hypothesis that is false. There is no element ∞ in the field with the stated properties. I'm not going to get into why division by zero is illegitimate in general this early in the morning.

Some mathematical systems, such as the IEEE standard in computation do define division by zero as infinity.

The floating point arithmetic used by computers doesn't form a field.

[Casparov]

The floating point arithmetic used by computers doesn't form a field.

So my response to your refutation could be, "n/0=∞ in Floating Point Arithmetic which doesn't form a field."??

[Contemporary Protestant]

Whats Casaprov's first language?

[sasuke]

So my response to your refutation could be, "n/0=∞ in Floating Point Arithmetic which doesn't form a field."??

Infinities are part of math, but not in the crude way that you think about them.  For example certain sets have infinite sizes (cardinals) and those cardinals can be compared.  They even have their own arithmetic.
In calculus for example, the limit of 1/n as n approaches infinity is 0.  This just means that no matter how small a number (call it epsilon) you challenge me with, I can always find a number (n) large enough that will make 1/n smaller than epsilon.
I think that Hakurei stipulated that it has to be a part of a field because he (rightfully imo) assumed that you were doing "standard" arithmetic, but I could be wrong.  Joe, Hakurei, and others here have a better grasp of this stuff than I do, and they'll hopefully be here soon to clarify.

[Hakurei Reimu]

So my response to your refutation could be, "n/0=∞ in Floating Point Arithmetic which doesn't form a field."??

That was to aileron. Look at the quotes.

[stromboli]

So much for my looking for a yes or no answer........

[Casparov]

That was to aileron. Look at the quotes.

Hakurei Reimu, I am sure you know what a "qubit" is in Quantum Information Theory. If there existed such a number that is a superposition of all numbers, a probability distribution of being any one number but simultaneously existing as all of them at once, a kind of "qubit of numbers", would it not be true that:

If; n/0=∞, Then; n="qubit of numbers"  ?????

Stated another way: 0*∞="qubit of numbers"  ????

[Solitary]

 You can't divide any number by infinity, no matter what your teacher told you. And it sure couldn't be infinity then. Infinity squared plus infinity divided by two = 1+2+3+4---------Aleph, infinity of infinities. Or cubits if you wish. He! He!Solitary