Is the problem with dividing by zero recognizing different infinities?

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Is the problem with dividing by zero recognizing different infinities?

to me 1/0 is clearily a infinitely large number(x).

and 2/0 should be 2x.

so x *0=1 assuming for examples sake(0/0=1)
so 2x*0=2

I guess for this to work some abstract value of infinity must be defined in the beginning.

Please note I am just trying to get a better grasp on the problem of division by zero.

P.S. is there a system of measuring values of infinity already?
 
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klij said:
Is the problem with dividing by zero recognizing different infinities?

to me 1/0 is clearily a infinitely large number(x).
No, it's undefined, which is a completely different thing. The best way to look at this is by using the concept of limits. Instead of actually dividing by zero, which is not defined, look at the quotient 1/x, where x gets successively closer to zero.

Here are values of 1/x, for a few values.
1/.1 = 10
1/.01 = 100
1/.001 = 1000
1/.0001 = 10,000
And so on.

The problem is that if x is a negative number that is getting close to zero, 1/x gets more and more negative.
1/-.1 = -10
1/-.01 = -100
1/-.001 = -1000
1/-.0001 = -10,000
And so on.

The point is - as x approaches zero, 1/x gets larger and larger without bound, or it gets more and more negative, depending on whether x is positive or negative. So it is not true to say that 1/0 is an infinitely large number.

What is safe to say is that 1/0 is undefined, as are 2/0, -5/0, and so on.

From a different perspective, if want to define 1/0 as being equal to some number, say y, then you have 1/0 = y, which implies that 1 = 0 * y. There is no number that you can multiply by 0 to get 1.




klij said:
and 2/0 should be 2x.

so x *0=1 assuming for examples sake(0/0=1)
so 2x*0=2
0/0 is likewise meaningless, but for a different reason. If you think that 0/0 = 1, and are not uncomfortable multiplying both sides of this equation by zero, you get 0 = 0 * 1, which is a true statement. However, can't you also use the same reasoning to say that 0/0 = 2, reasoning also that 0 = 0 * 2, which is also a true statement?

The trouble is that you could define 0/0 to be any number. In arithmetic, we like division problems to have one answer, so we have a problem here with too many answers.
klij said:
I guess for this to work some abstract value of infinity must be defined in the beginning.

Please note I am just trying to get a better grasp on the problem of division by zero.

P.S. is there a system of measuring values of infinity already?
Sort of. There is a part of mathematics that deals with numbers that are infinitely large infinity - the transfinite numbers. The smallest of these is called aleph null, \aleph_0. Aleph null is the cardinality of the natural numbers, {1, 2, 3, ...}, where cardinality refers to the size of this set. Surprisingly, it is also the cardinality of the even positive integers, {2, 4, 6, 8, ...} and the rationals. There are sets that have larger cardinalities, such as the set of real numbers, which has cardinality aleph one, or \aleph_0.
 
thank you for the insight I never even considered the fact zero is neither positive or negative.
 
Mark44 said:
0/0 is likewise meaningless, but for a different reason. If you think that 0/0 = 1, and are not uncomfortable multiplying both sides of this equation by zero, you get 0 = 0 * 1, which is a true statement. However, can't you also use the same reasoning to say that 0/0 = 2, reasoning also that 0 = 0 * 2, which is also a true statement?

The trouble is that you could define 0/0 to be any number. In arithmetic, we like division problems to have one answer, so we have a problem here with too many answers.

I've heard 0/0 called an indeterminate form, meaning that it is determined by constraints.

For instance, in the case of x/y=2, in the limit as x-->0, then 0/0=2. For the case x/y=3, then 0/0=3. Is this correct?
 
Mark44 said:
There are sets that have larger cardinalities, such as the set of real numbers, which has cardinality aleph one, or \aleph_0.
Apart from the typo (2^{\aleph_0}?) the real numbers have cardinality \aleph_\alpha for some \alpha\geq 1 unless you assume CH without proof.

It should perhaps be mentioned that if \mathfrak{a} is any cardinal (infinite or otherwise) 0\times\mathfrak{a}=0, so for cardinals division by zero remains undefined.
 
klij said:
Is the problem with dividing by zero recognizing different infinities?

Not at all.

The problem has to do with multiplication by zero. Multiplying by zero erases information. (Mathematically, we say multiplication is not http://en.wikipedia.org/wiki/Injective_function" ). Imagine a simple game:

Suppose your friend picks a number. He says he won't tell you what the number is, but if you give him a second number, he'll tell you what the product of the two numbers is.

If you want to know what number he picked, all you have to do is give him a number then divide by it. The easiest case is 1, (because the product of 1 * x = x). However, you could give him 2, then halve whatever the product is to guess his original number. Or you could give him 3, then divide whatever product he gives you by three to guess his original number.

But if you give him zero, this trick doesn't work. The product of 0 * x = 0. So if you tell him zero, he will tell you zero right back. You have NO INFORMATION left about what number he started with.

Infinities don't help at all here, either.

"Infinity!"

"No...", your friend tells you, "I said I would pick a real number. (Which of course, infinity is not)."

So the reason division by zero is "undefined" or "disallowed" is much simpler than you think. It's just evidence that multiplication by zero throws away useful information about what numbers you started with.
 
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Division by zero can be defined without that much hassle, what is undefined is just dividing zero by zero.

So for example R/0 is infinity -> 1/0=2/0, but if you try to multiply both sides with zero to cancel the zeroes you would have to perform 0/0 which we didn't define.
 
Martin Rattigan said:
Apart from the typo (2^{\aleph_0}?) the real numbers have cardinality \aleph_\alpha for some \alpha\geq 1 unless you assume CH without proof.

It should perhaps be mentioned that if \mathfrak{a} is any cardinal (infinite or otherwise) 0\times\mathfrak{a}=0, so for cardinals division by zero remains undefined.
Yeah, I meant \aleph_1. I wrote it out in words but missed changing the subscript that I copied and pasted.
 

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