0/0=?

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No, its not a number. Its just a reference point, just like all the other numbers. But actually, there is only one true number, the distance between two points, that is the number one. The other numbers are imposters, con artists. They are but groups of ones masquerading themselves a single entity. After Highlander: There can only be one.

Way to make sense.
 
I know they always say that 0/0 is an indeterminated form, you can't find any result, but why? How ?

x/0, while x is different than 0, =+oo or -oo, or 0 if x=0,something
0/1=0, while 1/0=+oo, why not 0/0=0? while 0x0=0
as 1/0, will get, 0 x (something) = to try to reach 1, but it will never do, it keeps like 0x0x0x0x0x0x0x0x....; so +oo (that's how I think about it), while 0/0, will get us 0x0x0x0x0x0x0x0... that's already 0, 0 x (something) =0

I just know that 0/0=error, you can't do it, why?

We know , what is 1 .

We know , what is 2 .

We know , what is 3 .

Do we know , what is 0 ?
 
infinity times zero is not defined in the real numbers since infinity is not a number.

In analysis, if $$\lim_{x\to 0} f(x) = \infty$$ $$\lim_{x\to 0} g(x) = 0$$ that still isn't enough information to solve what $$\lim_{x\to 0} f(x) g(x)$$ is.

Even in systems where there are infinite numbers, division by zero is not allowed to produce a number and zero times an infinite number is still zero in the Cardinal Numbers, The Ordinal Numbers, the Hypercomplex Numbers or the Surreal Numbers.
 
infinity times zero is not defined in the real numbers since infinity is not a number.

In analysis, if $$\lim_{x\to 0} f(x) = \infty$$ $$\lim_{x\to 0} g(x) = 0$$ that still isn't enough information to solve what $$\lim_{x\to 0} f(x) g(x)$$ is.

Even in systems where there are infinite numbers, division by zero is not allowed to produce a number and zero times an infinite number is still zero in the Cardinal Numbers, The Ordinal Numbers, the Hypercomplex Numbers or the Surreal Numbers.

Thanks for your clarification .

If x is tending towards 0 ; that means x can never become 0 . X is either 0+ or 0- .

Then if we say y = 0 ; what does y stands for ? ... I mean what could be the physical aspect of y ?
 
This has to be the SILLIEST thread on this forum. And that is saying something!

So, pace rpenner, let's try this:

Assume we are in a regime where it is defined that if $$\frac{x}{y} = z$$ then $$x = z \times y$$

Further let's assume that, under this regime that, for any $$x$$ that $$\frac{x}{x} =1$$. Then by the above we have that $$x = 1 \times x = x$$, a very reassuring result.

Now suppose that $$x=0$$. Then surely $$\frac{x}{x}=1 \Rightarrow \frac{0}{0} = 1$$? (BTW the funny looking arrow means "implies that")

From the above we have that $$\frac{0}{0} = 1 \Rightarrow 0 = 1 \times 0 = 0$$ again a very nice result.

Running the argument in reverse, again in a regime where this is allowed, gives us that, since $$2 \times 0 =0 \Rightarrow \frac{0}{0} = 2 \Rightarrow 2 =1$$

Which is not a very nice result at all.
 
infinity times zero is not defined in the real numbers since infinity is not a number.

In analysis, if $$\lim_{x\to 0} f(x) = \infty$$ $$\lim_{x\to 0} g(x) = 0$$ that still isn't enough information to solve what $$\lim_{x\to 0} f(x) g(x)$$ is.

True, you need to apply some extra rules in order to resolve the this type of limit.


Even in systems where there are infinite numbers, division by zero is not allowed to produce a number

Correct, for example in IEEE 754 the above creates a NaN.

and zero times an infinite number is still zero in the Cardinal Numbers, The Ordinal Numbers, the Hypercomplex Numbers or the Surreal Numbers.

This depends, in IEEE 754 the above produces a NaN.
 
Tach -- QuarkHead is working with new axioms and working to a contradiction -- a classical mathematical argument that the new definition of division is flawed.
 
This depends, in IEEE 754 the above produces a NaN.

I'm well aware of that. My company once hosted a lecture by the inventor of IEEE 754.

But IEEE does not support the full range of operations on Infinity, for example 2^Infinity = Infinity which does not make sense mathematically. 2^63 + 1 = 2^63 in IEEE math as well.

Code:
$ perl -e 'print 2**63 + 1 - 2**63, "\n";'
0

IEEE infinity stands for any number larger in magnitude than their mantissa-exponent format can represent. IEEE has signed zeros to represent numbers smaller in magnitude than their format can represent. NaN is a catch-all condition to represent results that don't fit in it's approximate number line.

Code:
$ perl -e 'print 2**6300, "\n";'
inf
$ perl -e 'print 0* 2**6300, "\n";'
nan
$ perl -e 'print 2**-6300, "\n";'
0
$ perl -e 'print 2**55 * (1/5 + 1/70 - 1/7 - 1/14), "\n";'
1
 
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The message of this question is that zero and infinitely close to zero are very different things.
 
Tach -- QuarkHead is working with new axioms and working to a contradiction -- a classical mathematical argument that the new definition of division is flawed.

Tach, like WL, doesn't deal well with hypotheticals, or people examining the implications of hypotheticals.

I'm not surprised that he would fail to recognize a Proof by Contradiction.
 
.

Hi AlphaNumeric, QuarkHead. :)

Dividing by zero isn't at all like multiplying by one. You are taking how a specific computer program has an if then check for dividing by zero and returns the input if so. Mathematics views them as entirely different. Multiplication by 1 is essential in most number systems but division by 0 is explicitly excluded usually.


This has to be the SILLIEST thread on this forum. And that is saying something!

So, pace rpenner, let's try this:

Assume we are in a regime where it is defined that if $$\frac{x}{y} = z$$ then $$x = z \times y$$

Further let's assume that, under this regime that, for any $$x$$ that $$\frac{x}{x} =1$$. Then by the above we have that $$x = 1 \times x = x$$, a very reassuring result.

Now suppose that $$x=0$$. Then surely $$\frac{x}{x}=1 \Rightarrow \frac{0}{0} = 1$$? (BTW the funny looking arrow means "implies that")

From the above we have that $$\frac{0}{0} = 1 \Rightarrow 0 = 1 \times 0 = 0$$ again a very nice result.

Running the argument in reverse, again in a regime where this is allowed, gives us that, since $$2 \times 0 =0 \Rightarrow \frac{0}{0} = 2 \Rightarrow 2 =1$$

Which is not a very nice result at all.


Thank you both, AN and QH, for your time and trouble (and kind patience) so far in elucidating the axiomatic/mathematical aspects involved, and the inconsistencies you point to for O/O etc in that purely mathematical context.

I would still ask for anyone interested to provide comments/perspectives on the purely 'physical' aspects involved, which I also pointed to by using 'action' and 'non-action' to describe the division/multiplication comparative cases using zero/One respectively.

I should perhaps have also stressed in my original post the physical DIGITAL nature of the comparative cases I used. Hence why I am also still very interested in your/others' contributions/perspectives regarding especially the YES (action) or NO (non-action) aspect itself.

In this context, QuarkHead, I am specifically limiting the 'multiplier (action) to 'I'. Use of any other number is OK for the purely axiomatic/mathematical example you provided (2=1, good one!) to illustrate the absurdity of O/O=I in purely axiomatic/mathematical constructs, but there is also the purely digital/physical aspect which I would like to explore further via reading your/others' takes on O/O and the observation I made regarding comparative equality of 'outcome' when multiplying by "I" (read: digita/physical "action") or dividing by "O" (read: digital/physical "non-action"). Hence the limiting to "I" as the multiplier and "O" as the divisor, to indicate the 'yes/no" action/non-action states/cases only (purely physical rather than further mathematical comparison is also what I am interested in getting perspectives on). :)


All further contributions appreciated. Thanks. Back in a couple of days if I can. Cheers!

.
 
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Tach, like WL, doesn't deal well with hypotheticals, or people examining the implications of hypotheticals.

I'm not surprised that he would fail to recognize a Proof by Contradiction.

You continue to troll my posts. Please cease and desist.
 
You continue to troll my posts. Please cease and desist.
:rolleyes:

It is simply another example of your inability to recognize a valid solution.

It's a tautology that you failed to recognize the post as a proof by contradiction.
 
Tach -- QuarkHead is working with new axioms and working to a contradiction -- a classical mathematical argument that the new definition of division is flawed.

I have to agree with Quarkhead, this thread does qualify for one of the silliest ones. But, then again, there are so many silly ones.... hard to choose the silliest of them all.
 
The thread subject is elementary, but not silly.
A division is an amount divided by an amount.
You have to have something to devide, and you have to have something to divide it by.
QED as my long suffering maths teacher might have said.
 
Sorry, I cannot be bothered to respond to these last two posts, except to say they seem to embody my original notion of "silliness"

[mod]this no longer refers to the previous 2 posts - the ones QH refers to have been deleted.[/mod]

But RealityCheck, I really don't understand what you are getting, at - I dare say it is me, not you at fault.

Let me try this, though as I am not, and never claimed to be, a physicist, wait for the big guns to express a view.

I suggest (no more!) that if something is mathematically impossible, then it is also physically impossible.

However, the inverse is false: things that are mathematically possible may well be physically impossible. Does that help? Is it even true?

BTW, the reason Tach failed to understand the nature of my "proof" was because, in his response immediately following my post, he assumed the result that was the object of that proof.

This is not only the cardinal sin in mathematical logic, it suggests that the sinner will never hope to understand the purpose of a proof by contradiction.
 
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