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I suppose these are the equations with infinity that are universally considered correct:

  1. ∞ = ∞
  2. ∞ + n = ∞
  3. ∞ * n = ∞
  4. n/∞ = 0

Where n can be any possible value.

These equations can be rearranged to give the following results:

  1. ∞ - ∞ = 0
  2. ∞ - ∞ = n
  3. ∞ / ∞ = n
  4. ∞ * 0 = n

Where n can be any possible value.

But can n also be infinite?

If so the following final derivations can be made(in no particular order):

  1. ∞ + ∞ = ∞
  2. ∞ - ∞ = n        (where -∞ <= n <= ∞)
  3. ∞ * ∞ = ∞
  4. ∞ / ∞ = n        (where -∞ <= n <= ∞)

Are these statements valid? Also the interesting thing here is that in the final equations, 1. and 3. both have a single value of infinity whereas 2. and 4. Can have any possible value, including -∞, 0 and ∞. Does that mean 1. and 3. are not undefined?

Nk07
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    "But the question is, can $n$ also be infinite?" Yes, $2n-n=n$ and $n\to\infty$ gives $\infty-\infty=\infty$ in your way of thinking. – Dietrich Burde Jan 01 '21 at 14:32
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    related – Peter Jan 01 '21 at 14:39
  • I had another question though. it seems like ∞−∞ results in ∞, but it can also be 0 or any other value. I suppose this is why ∞ - ∞ is considered undefined as it may result in many possible values. but ∞ + ∞ always gives one value, that is ∞. So does that mean ∞ + ∞ is not undefined and can be considered to be ∞ in all cases? – Nk07 Jan 01 '21 at 14:40
  • Treating $\infty$ as a number (with which we can do the usual manipulations) leads inevitably to contradictions. Such equations should be considered as thumb-rules how to deal with limits. In this sense (and only this sense) , we can say , for example $\infty+\infty=\infty$ , but $\infty-\infty$ cannot be defined even in this sense because we cannot conclude the limit. – Peter Jan 01 '21 at 14:44
  • See https://math.stackexchange.com/questions/3102813/is-infty-undefined/3102868#3102868 – Michael Hoppe Jan 01 '21 at 19:09

1 Answers1

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Let us follow the convention that an expression with $\infty$ is "defined" (in the extended reals) if: when you replace each $\infty$ with any function/sequence whose limit is $\infty$, and each real number with any function/sequence with that limit, the limit of the entire expression is always the same real number or divergence to $\infty$ or $-\infty$.

Then you are right that $\infty+\infty$ and $\infty*\infty$ are defined and equal $\infty$. And you are right that $\infty-\infty$ and $\infty/\infty$ (and $0*\infty$) are undefined. (If you need proofs for these and can't find them elsewhere, let me know.)

Because of this, it is wrong to write something like "$\infty-\infty=n$ where $n$ is any number", because $\infty-\infty$ is undefined. It's not equal to anything. And we don't usually use the equals sign in a way that would allow multiple values at once.

But can n also be infinite?

I will interpret this as "are there functions where $\infty-\infty$, $\infty/\infty$ and $\infty*0$ correspond to limits that diverge to infinity?" Yes. For examples: ${\displaystyle \lim_{n\to\infty}}(2n)-(n)=\infty$, ${\displaystyle \lim_{n\to\infty}}(n^2)/(n)=\infty$, and ${\displaystyle \lim_{n\to\infty}}(n^2)*(1/n)=\infty$.

Mark S.
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  • I see, thanks for the answer. however I had one question, in your final sentence don't the examples reduce to ∞-∞ and ∞/∞ which are undefined because they result in many possible values and so don't always diverge to ∞ ? – Nk07 Jan 01 '21 at 22:01
  • @Neelim It depends what you mean by "they result in many possible values". "$\infty-\infty$" is undefined because, for instance, "${\displaystyle \lim_{n\to\infty}}(n+17)-(n)=17$ and ${\displaystyle \lim_{n\to\infty}}(n)-(n^2)=-\infty$". But the particular limit ${\displaystyle \lim_{n\to\infty}}(2n)-(n)$ has just one extended real value: $\infty$. – Mark S. Jan 01 '21 at 22:12
  • it seems to me like $lim n→∞ (2n)−(n) = 2∞ - ∞ = ∞ - ∞ $ which is undefined and not just one value i.e. ∞ – Nk07 Jan 01 '21 at 22:21
  • @Neelim That is incorrect. You can't always replace $n$s by $\infty$s. ${\displaystyle \lim_{n\to\infty}}(2n)-(n)={\displaystyle \lim_{n\to\infty}}n=\infty$. But you can only replace the $n$s with $\infty$ *if* you're not in an undefined case like $\infty-\infty$. – Mark S. Jan 01 '21 at 22:23
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    oh I see, so the expressions with n are reduced first and then replaced with ∞ – Nk07 Jan 01 '21 at 22:24
  • @Neelim Yes. In simple calculations, that is straightforward and works fine. In more complicated cases like ${\displaystyle \lim_{n\to\infty}}n+2-\sqrt{n^2+3n+1}=1/2$, you need to know Calculus or just the right algebra trick to try. – Mark S. Jan 01 '21 at 22:29