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Andrew26

When calculating 1/infinity, you will yield the indeterminate form infinity/infinity

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If 1/infinity is the number that when multiplied by infinity equals 1, then the number is 1/infinity. 

To arrive at you would multiply (infinity)*(1/infinity)=infinity/infinity which does technically equal 1, but it does also technically equal 2 or and any other positive real. There 1/infinity should be treated as undefined.

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5 minutes ago, Andrew26 said:

If 1/infinity is the number that when multiplied by infinity equals 1, then the number is 1/infinity. 

Infinity is not a number, therefore 1/infinity is not a number.

5 minutes ago, Andrew26 said:

There 1/infinity should be treated as undefined.

It is.

 

What is your point?

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1 minute ago, mathematic said:

Usually 1/infinity is treated as zero, not undefined.

Is it? I didn't know that. (I have to assume you are right, based on your username :) )

1 hour ago, Andrew26 said:

If 1/infinity is the number that when multiplied by infinity equals 1, then the number is 1/infinity. 

To arrive at you would multiply (infinity)*(1/infinity)=infinity/infinity which does technically equal 1, but it does also technically equal 2 or and any other positive real. There 1/infinity should be treated as undefined.

You need to use limits to evaluate something like 1/infinity. You can then show that as the divisor gets arbitrarily close to infinity, the result gets arbitrarily close to zero. That is why, as mathematic says, it is possible to choose to define it as 0.

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14 minutes ago, Strange said:

Is it? I didn't know that. (I have to assume you are right, based on your username :) )

You need to use limits to evaluate something like 1/infinity. You can then show that as the divisor gets arbitrarily close to infinity, the result gets arbitrarily close to zero. That is why, as mathematic says, it is possible to choose to define it as 0.

You also need to distinguish between the use of the entity


[math]\frac{1}{\infty }[/math]


in loose form by applied mathematicians (like me)

and proper limits like this


[math]\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}};x > 1,n \ge 0[/math]


[math]\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{x^n}}};x > 1,n \ge 0[/math]


where if you pick the debarred values of x or n you will converge to 1 or 1/0

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On 11/24/2019 at 10:02 PM, Andrew26 said:

If 1/infinity is the number that when multiplied by infinity equals 1, then the number is 1/infinity. 

Yes, you are right. 

And if \(1/\infty\) is a number, other than \(-1\) or \(+1\),  that when multiplied by infinity produces \(1\), then your aunt is yellow, has four wheels and is a schoolbus.

Anything follows from a false antecedent, whether it is true or false does not matter.

If you follow standard definitions, then \(a/\infty = 0\) is true for every real number \(a\). And \(0\) multiplied by \(\infty\) is most often defined to be \(0\). When you combine those usual definitions you get \( (a/\infty)\times \infty = 0\) for all real values \(a\). Always \(0\), never \(1\).

If you have non-standard definitions, you have to present them, that will be fine.

 

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On 1/4/2020 at 10:35 AM, taeto said:

Yes, you are right. 

And if 1/ is a number, other than 1 or +1 ,  that when multiplied by infinity produces 1 , then your aunt is yellow, has four wheels and is a schoolbus.

Anything follows from a false antecedent, whether it is true or false does not matter.

If you follow standard definitions, then a/=0 is true for every real number a . And 0 multiplied by is most often defined to be 0 . When you combine those usual definitions you get (a/)×=0 for all real values a . Always 0 , never 1 .

If you have non-standard definitions, you have to present them, that will be fine.

How dare you speak about my aunt like that!

 

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10 minutes ago, Country Boy said:

How dare you speak about my aunt like that!

If used to dozens of unruly schoolchildren daily, then she is probably not taking too much offence. 

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1 hour ago, Country Boy said:

Are you saying that you are an unruly schoolchild?

If that is what I am saying, then yes.

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