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Posted

Infinity can't be defined, therefore 1 divided by infinity can't be defined. It's just assumed as a limiting factor to be zero, when used in calculus for example.

Posted

It isn't that infinity cannot be defined, but that it has no place in a discussion about arithmetic in the real numbers. Use of the word should be avoided altogether except when it is an explicitly and carefully defined object in the mathematical sense. The question, as stated, is just unmathemaitcal. There are "points at infinity" but this is a careful mathematical construction. We say 1/x tends to infinity as x tends to zero as a shorthand that means EXACTLY that the modulus of 1/x grows without bound. Some people, including wolfram I believe, abuse notation to say that 1/0 =infinity, but that isn't good mathematical practice. Every statement involving limits and infinity can, and should, be stated in terms of finite things until such time as people accept that there is no sich thing as "infinity" as a point in the real number system.

Posted

0 represents an infentisimally small number as well as nothing.

 

Hence the reciprocal of an infinitly large number will be 0.

Posted

But you are going outside the realms of the reals and into the hyperreals which means that you have all kinds of set theoretic problems (coming up to the set of all sets) to deal with. One need not even pass to the notion of infintesimals to have a system where "1/infinity = 0". The extended complex plane will suffice, with a little topology, and "infinity" is properly the point at infinity. I'm not sure that even in the hyper-reals your statement is actually true, ed84c, since it presumes that 1/H, where H is infinte, is real, and infinitesimal, which I don'y believe is actually true - infintesimal yes, but real, no.

  • 2 weeks later...
Posted

You can't find the value of [math]\frac{1}{\infty}[/math], because infinity is not defined in the set of complex numbers, therefore you can't perform any arithmatic on it. The only thing you can do that is similar would be to find [math]\lim_{x\to\infty}\frac{1}{x}[/math] which equals 0.

  • 1 month later...
Posted

well i think there is two ways of doing things you can use calculus where it aproaches zero, or you can use calculatorus where the highest thing is one over tenduotrigintillion (1 with 100 zero's) and you get 0 again

Posted

according to my calculator,

 

1/infinity = error

 

which may possibly be the most appropriate answre, as infinity isnt a real number. however, when doing lower-calibre maths, iv never come across an occasion where it was incorrect to pretend that 1/infinity = 0

Posted

You know, it's really a question that can't be truly answered.

Infinity isn't really a real number, it's more like a concept that there is no end number. 1 over infinity on the other hand, isn't a number at all, since Infinity isn't a number at all.

 

So really, the answer is none, because there is no such number as infinty.

  • 1 month later...
Posted

1/infinity, why do you care? 1 over a big +number is a small +number, 1 over a bigger +number is a smaller +number, but remember that zero is not positive so no, it is not zero, it is infitessimal, which belongs to the same class as infinity, leave it at that.

 

Infinity does not really mean absolute biggest but big enough to be considered so. Infitessimal does not really mean absolutely zero but close enough to be consider so.

Posted

Except of course in the extended complex plane, or extended real line (one point compactifications), when 1/infinity is defined to be zero, and 1/0 is defined to be infinity, and all the other rules that ought to be true hold as well. See, eg, the Moebius transformations of Cu{infinity}

Posted

Infiniety is a mathmatical construct and has no real application when working with equations, it's like a +++DIVIDE BY CUCUMBER ERROR+++

Posted

Does it not? So presumably the integral round a closed path of a meromorphic function being equal the the number of poles in the interior counted with residue (up to a constant) isn't a "real use of it in an equation"?

Posted
Does it not? So presumably the integral round a closed path of a meromorphic function being equal the the number of poles in the interior counted with residue (up to a constant) isn't a "real use of it in an equation"?

 

 

Matt, it's been a long time since I've used poles and residues.

 

I remember the method of images, and some other things here and there, but its been a long time.

 

One question....

 

What is a meromorphic function, and how does it tie into the concept of infinity?

Posted
What is a meromorphic function, and how does it tie into the concept of infinity?

 

It's a function in the complex plane that is holomorphic (analytic) except at singularities.

Posted

If you do wish to have a long discussion about them Johnny would you mind starting another thread on it? I've nothing agains talking about them, but here isn't the appropriate place.

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