0 * Infinity = 0 - where is the problem in the below mentioned proof?

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pradyumna1974

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Hi,
Here is an interesting derivation.

(0 * ꝏ) = ?

(As per https://en.wikibooks.org/wiki/Calculus/Infinite_Limits/Infinity_is_not_a_number )

ꝏ= Lim ((x=>0+) (1/x)) , where 0+ is a non-zero real number very close to zero, but never becomes zero.

0 * ꝏ= 0 * lim (x=>0+) (1/x)

=> 0 * (lim(x=>0+) (1) / lim (x=>0+) (x))

=> 0 * 1 / (lim (x=> 0+) (x))

=> (0/ (lim (x=> 0+) (x))) * 1 (Because (Lim (x=>a) x = a))

=> (0/ 0+) = 0 (zero is divided by 0+, which is non-zero real number, that would never become zero. so the result must be zero.)

Looking forward for comments...
 
One problem, among others, is that your "0+", "a non-zero real number very close to zero, but never becomes zero" doesn't exist (in the real number system). The notation doesn't mean that there is such a number, but just that we are approaching 0 from the right.
 
I would say that, since "\(\displaystyle \infty\)" isn't a real number, the very statement "\(\displaystyle 0\cdot\infty\)" doesn't make sense until you have defined what it means to multiply a real number by \(\displaystyle \infty\)!

In terms of "limits", as Dr. Peterson suggests, given any number, "c", we can always find a sequence, \(\displaystyle \{a_n\}\), converging to 0 and a sequence, \(\displaystyle \{b_n\}\), "converging to \(\displaystyle \infty\)" (increasing without bound) such that \(\displaystyle \{a_nb_n\}\) converges to c!
 
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There are so many things wrong with the OP that one scarcely knows where to begin.

Multiplication is a binary operation or function; it takes two numbers as input.

Thus, the very first line of 0 * = ? makes absolutely no sense.

I cannot say I like the cited wikipedia article. I have no objection to finitistic mathematics, but to assume that Cantor did not put transfinite numbers on a logical basis, in fact not to mention Cantor at all, strikes me as misleading. HOI is much clearer: infinity is not a "real" number. Therefore you cannot work with infinity as though it is a real number. Sheep and platypus are both mammals, but sheep do not lay eggs.

We teach students about extending number systems to solve new kinds of problems. Each time we do so, the rules of arithmetic become a bit more complex. For example, when we deal with integers, division is frequently not defined. When we deal with real numbers, division by every real number except zero is defined. It is possible to extend the real number system to include infinity (either one or two infinities). Then a real number divided by zero is defined unless the real number being divided is ZERO.

The OP seems to think that [MATH]0^+[/MATH] designates a real number not equal to zero. Under that assumption, it is obvious that

[MATH]0 * \dfrac{1}{0^+} = 0.[/MATH]
This is not standard notation so we might as well call it e. Therefore using his weird notation

[MATH]\lim_{x \rightarrow e} \dfrac{1}{x} = \dfrac{1}{e} \implies\\ 0 *[/MATH]
 
Respected Dr. Peterson,

First of all , thank you for taking time and helping by sharing your thoughts. With all due respect, I have a different understanding of real number set. I agree that "0+" is a notation to represent infinitely large number of real numbers that could be present between given real number "c" and absolute "0", where x tends from c to 0+ (x => 0+)

But it we take any real number from the set , that is denoted by "0+", it will still be a real number and would exist, possibly with infinite precision. May be, my derivation is not clear enough. But my thought is that 0 divided by any real number that exists in that set and denominator in that fraction , denoted by "0+", no matter what the value of the denominator real number is. This could mean that uncertainty introduced by ꝏ is not relevant in this context, as numerator is 0. So why can't we hypothesise that 0 * ꝏ = 0 ? Why do we put so much emphasis on the uncertainty of ꝏ and ignore the effect of multiplication with 0 (i.e. 0 in numerator) ? It looks like there is a higher priority / focus on uncertainty of ꝏ in these derivations than multiplication with 0.

@HallsofIvy, thank you for sharing your thoughts as well!
 
@JeffM, thank you for sharing your thoughts as well! I appreciate your example on mammals.

The first line was a problem statement. I think I should have elaborated that statement clearly. I am sorry for not being clear.
 
I am still not sure what the problem statement is. The notation

[MATH]\lim_{x \rightarrow 0^+}[/MATH]
usually means we are considering only real values of x that are greater than 0 but smaller than an arbitrary constant.

Is that what you mean?

It seems to me that you are saying that 0 * a real number equals 0. Yes it does.

But if you are trying to argue that 0 * (1/0) = 0, that is not a meaningful statement in any number system I'm aware of.
 
Problem statement : what is the value of (0 * ꝏ) ?

I am not trying to say that 0+ is equal to 0. In fact, what I am saying is that 0+ is a symbol, that represents a real number <> 0, but very very close to 0 and 0+ > 0. Hence 0+ is a symbol to represent any real number, that meets the following criteria.

1. 0+ <> 0
2. 0+ > 0
3. 0+ is very very close to 0 with really positive minimal gap between 0+ and 0.
4. x => 0+, means x will never touch 0 at all. x will always tend to move towards 0, but never becomes 0, when x is reduced by infinitesmall decrement.

so IMO, 0/0+ = 0.
 
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pradyumna1974 said:
Problem statement : what is the value of (0 * ꝏ) ?

I am not trying to say that 0+ is equal to 0. In fact, what I am saying is that 0+ is a symbol, that represents a real number <> 0, but very very close to 0 and 0+ > 0. Hence 0+ is a symbol to represent any real number, that meets the following criteria.

1. 0+ <> 0
2. 0+ > 0
3. 0+ is very very close to 0 with really positive minimal gap between 0+ and 0.
4. x => 0+, means x will never touch 0 at all. x will always tend to move towards 0, but never becomes 0, when x is reduced by infinitesmall decrement.

so IMO, 0/0+ = 0.
Click to expand...

You're talking about "infinitesimals", which are not part of the real number system. That's the error in your work.

These can be defined carefully, but you have to define it (not just assume its existence) before using it in a proof. See, for example, here: https://en.wikipedia.org/wiki/Infinitesimal
 
pradyumna1974 said:
Problem statement : what is the value of (0 * ꝏ) ?
I am not trying to say that 0+ is equal to 0. In fact, what I am saying is that 0+ is a symbol, that represents a real number <> 0, but very very close to 0 and 0+ > 0. Hence 0+ is a symbol to represent any real number, that meets the following criteria.
1. 0+ <> 0
2. 0+ > 0
3. 0+ is very very close to 0 with really positive minimal gap between 0+ and 0.
4. x => 0+, means x will never touch 0 at all. x will always tend to move towards 0, but never becomes 0, when x is reduced by infinitesmall decrement.
Click to expand...
There material that you are looking for is found in In Elementary Calculus: An Infinitesimal Approach by Jerome Keisler in chapter 1. The chapters and/or the whole book are a free down-load at http://www.math.wisc.edu/~keisler/.
 
I think the easiest way I picture in my mind and how to teach someone else is the following:

In my understanding, ∞ can any number above zero. When someone uses infinity they are defining numbers from as (infinitesimally small) --> ∞
Thus, infinity is just a test going from the smallest number to the highest number. Testing many numbers to see if they fit. Let me ask you this way: Can the following be true? 0.1 = 3? No. Thus, infinity can't be used as a number

Thus, you cannot use it as a real number. We use infinity to test functions to see where they will end in a long period of time. That allows us to make better decisions in the workforce. Math is a tool box thus Consider infinity just a tool like taking a derivative.

Please correct me if I am incorrect. Thanks!
 
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JeffM said:
I am still not sure what the problem statement is. The notation

[MATH]\lim_{x \rightarrow 0^+}[/MATH]
usually means we are considering only real values of x that are greater than 0 but smaller than an arbitrary constant.

Is that what you mean?

It seems to me that you are saying that 0 * a real number equals 0. Yes it does.

But if you are trying to argue that 0 * (1/0) = 0, that is not a meaningful statement in any number system I'm aware of.
Click to expand...

I really don't think you should relate ∞ to 1/0. That is incorrect. 1/0 is undefined. Understand that again range being (infinitesimally small ) --> ∞
I understand where you are coming from for explaining however. I really appreciate your input.
 
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Dr. Peterson and @pka,

Thank you for sharing thoughts on infinitesimal numbers and hyper real number system.

I revised the derivation using hyper real infinitesimal number system. From hyper real number set, Using ε <> 0, (but an infinitesimal number)
the formula for infinity can be rewritten as follows.

ꝏ = Lim ( (x=>ε) (1/x) ) = ω (in hyper real number system or surreal number system)

(where ε is a hyper real number / infinitesimal number, close to 0, but never equal to 0 and as ε gets smaller in hyper real numbers set, the ꝏ gets larger as hyper real infinity).

0 * ꝏ => 0 * Lim ( (x=>ε) (1/x) )
=> 0 * 1/ ε
=> 0 / ε

According to my understanding, 0 / ε <> 0 / 0, since ε is non-zero positive hyper real infinitesimal number, that will be very close to zero.
Therefore 0 / ε => 0, because 0 divided by any non-zero number results in 0.

0/0 is definitely undefined. But 0/ ε should be defined, because ε is never equal to 0. 0 is also an infinitesimal real number according to this book (http://www.math.wisc.edu/~keisler/). So the division operation between 0 and ε is division between two different infinitesimal numbers, which is a valid arithmetic operation and not undefined, according my understanding.

Is there still a problem with this revised version of derivation too? Is 0 / ε undefined in hyper real infinitesimal number system ?

Please share your thoughts.
 
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Here is an elaborate derivation of the above discussed approach and based on my learning from this book http://www.math.wisc.edu/~keisler/.

In hyper real number system or surreal number system, let ε be an infinitesimal number, very close to 0, but not equal to 0. Let ω be infinity in hyper-real number system. Then we can define ω as

ω = Lim (x => ε) (1/x) where ε ≠ 0 and ε is non-zero hyper-real infinitesimal number.

As ω is infinity in hyper-real number set *R , we hypothesize ω ≈ ∞. How is this correct ?

*R hyper-real number set is super-set for R (real number set).
ω is unbounded number in *R
∞ is unbounded number in R

As both are unbounded in their respective number sets and R (real number set) is subset of *R (hyper-real number set), therefore ω ≈ ∞.

0 * ∞ ≈ 0 * ω = 0 * Lim (x => ε) (1/x) where ε ≠ 0
=> 0 * 1/ ε
=> 0 / ε

0 / ε ≠ 0/0 because ε≠0.

0/0 is undefined. 0/ε cannot be undefined because ε is non-zero infinitesimal number.

0 is also a member of hyper-real number set and and also infinitesimal number. 0 / ε is an arithmetic operation in hyper-real number set, which follows similar rules of real number set.

0 / ε = 0 / non-zero hyper-real number = 0 / any real number = 0.

So we can conclude that 0 * ω = 0, hence 0 * ∞ ≈ 0.

Since ω is unbounded and infinity in hyper-real number set *R (i.e. super-set of R – real number set), ω cannot be less than ∞. There fore ∞ <= ω. But ∞ is unbounded by definition and cannot be less than any other number that could exist. Using these two inferences, we can conclude that ∞ = ω.

With substitution, we can conclude that

0 * ω = 0 * ∞ = 0

Are there any errors in the above derivation ? Please share your thoughts.

P.S. I am a mathematics learner . I don't claim myself to be expert. I would like to try to solve various problems in my own way. Please help me learn more of the magic of mathematics, especially Calculus.

Thank you all!
 
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A second revision of above stated derivation.

In hyper real number system or surreal number system, let ε be an infinitesimal number, very close to 0, but not equal to 0. Let ω be infinity in hyper-real number system. Then we can define ω as

ω = Lim (x => ε) (1/x) where ε ≠ 0 and ε is non-zero hyper-real smallest infinitesimal number.

As ω is infinity in hyper-real number set *R , ω ≈ ∞. How is this correct ?

*R hyper-real number set is super-set for R (real number set).

ω is unbounded number (infinity) in *R

∞ is unbounded number (infinity) in R

As both are unbounded in their respective number sets and R (real number set) is subset of *R (hyper-real number set), therefore ω ≈ ∞.

0 * ∞ ≈ 0 * ω = 0 * Lim (x => ε) (1/x) where ε ≠ 0
=> 0 * 1/ ε
=> 0/ ε => 0

0 / ε ≠ 0/0 because ε≠0.

0/0 is undefined. 0/ε cannot be undefined because ε is non-zero smallest hyper-real infinitesimal number.
0 is also a member of hyper-real number set and and also infinitesimal number. 0 / ε is an arithmetic
operation in hyper-real number set, which follows similar rules of real number set.

Hence 0/ε = 0 / non-zero hyper-real number = 0 / any number = 0.

So we can conclude that 0 * ω = 0, hence 0 * ∞ ≈ 0.

Since ω is unbounded and infinity in hyper-real number set *R (i.e. super-set of R – real number set), ω cannot be less than ∞. There fore ∞ <= ω. But ∞ is unbounded and cannot be less than any other real number that could exist. Using these two inferences, we can conclude that ∞ = ω or ∞ = ω - ε, because ε would be infinitesimal positive variation from ∞ (infinity in R) to ω (infinity in *R), where ε > 0 and ε is the smallest infinitesimal hyper-real number.

With substitution, we consider a case of ∞ = ω, then 0 * ∞ = 0 * ω = 0

if we consider the case of ∞ = ω - ε.
0 * ∞ = 0 * (ω - ε) = 0 * ω – 0 * ε

We know that 0 * ε = 0, as ε≠0.
We already proved that 0 * ω = 0 / ε = 0

So 0 * ∞ = 0 * ω – 0 * ε = 0 / ε - 0 * ε = 0 – 0 = 0

Hence we can infer the following statements.

1. ( ꝏ Є R ) <= (ω Є *R)
2. And ∞ <= ω, then 0 * ∞ = 0 as proved above.

So we can conclude that 0 * ∞ = 0

Are there any problems with the above derivation ?
 
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The problem what you are saying is that ε = x; 0<x<1. It can be any number in that range.
ω = ∞ 1<y<∞ can be any number in that range.
Infinity and infinitesimally small have the same properties
 
With regards to 0*∞:

Lets say 2*x = 3*y where x=0; y=0

x=(3/2)y x=0 y=0

0=0 meaning ∞

Meaning infinity is just infinity.
 
I have been thinking about this more and maybe I am not correct.
I don't really don't know.

2x=3x ; x=0
2=3? if you cross it out. or 0=0?

or

0*(1/ε) where ε ≈ 0 which is not 0. Can you say that 0 and ε ≈ 0 can be labeled as the same amount and call it x
where x*(1/x)=1

Can 0*∞ =1?
 
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