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In many places I have heard that $\frac 10 = \infty$. While I do believe this to be a flawed concept, and there are many posts on this, I wanted to investigate some properties of $\frac 10$. First I would like to state some properties of infinity:
$$1)\infty + 1 = \infty$$
$$2)k* \infty = \infty$$
These can be hard to prove with most other expressions for $\infty$ but with this very flawed description, both these properties can be shown!
$1) LHS= \frac 10 + 1= \frac{1+1*0}{0} = \frac 10 = RHS$
$2) LHS= k*\frac 10 = \frac k0 = \frac{1}{0/k} = \frac 10 = RHS$
However sometimes it is simply wrong. For example:
$\frac 10 = \infty$
$1 = \infty * 0$
$1 = 0$

So I would like to know if it actually is a plausible solution or not. I know there are many posts about this but I hope no one minds if these properties are investigated.
Thanks in advance!

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    $\begingroup$ No, none of this can be handled this way. You can't treat $\infty$ as something you can do arithmetic on. Limits are needed to handle these concepts correctly. $\endgroup$ – Randall Dec 26 '17 at 3:33
  • $\begingroup$ Thanks for the heads up! $\endgroup$ – Mohammad Zuhair Khan Dec 26 '17 at 3:34
  • $\begingroup$ Also, certainly you agree that if $x$ is very very close to $y$, that $1/x$ is also very very close to $1/y$. But then you have a problem with $1/(-0.0000001)$ which should be reasonably close to your $1/0=\infty$, but the former quotient is actually negative $\endgroup$ – Randall Dec 26 '17 at 3:36
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    $\begingroup$ You can discuss idealized points at infinity but they do not behave well arithmetically. For example you can meaningfully discuss the point at infinity on the Riemann Sphere so that $1/0 = \infty$ but multiplying and adding infinities still runs you into difficulties. $\endgroup$ – CyclotomicField Dec 26 '17 at 3:37
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    $\begingroup$ There is the notion of the extended real number line, which is a setting in which you can do arithmetic with $\infty$. However, in this number system, almost none of the standard arithmetic properties hold unchanged. Even in this setting division by $0$ remains undefined $\endgroup$ – eepperly16 Dec 26 '17 at 4:30
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This is not true. Period. Why you ask? Because $0$ doesn't have a multiplicative inverse. You yourself have noted that it produces contradictions. $\infty$ is more of an upper bound for reals than a number itself. While there are fields of algebra where it's treated as a number, most of the time, it's not. BUT, in some fields outside mathematics e.g. Physics, for all intents and purposes ${1 \over 0}=\infty$ as they only need approximations for real life applications. But, in mathematics, $$\lim_{\alpha \to 0+}{1 \over \alpha}=\infty$$ This is as much as you can say.

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    $\begingroup$ That needs to be a right-hand limit to be true. $\endgroup$ – Randall Dec 26 '17 at 4:23
  • $\begingroup$ @Randall Yeah. My bad. $\endgroup$ – SinTan1729 Dec 26 '17 at 9:45
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The "properties" you mentioned

  • $\infty+1=\infty$
  • $k\cdot\infty=\infty$ (for $k>0$)

are called arithmetic operations for the extended real numbers. They are true by definition. With such definition, lots of theorems in real analysis can be stated in a neat way. If one is talking about the set $\overline{\mathbb R}$ of extended real numbers, then there two different "infinities": $\pm\infty$. Note that $\overline{\mathbb R}$ is defined as $$ \overline{\mathbb R}=\{\mathbb R\}\cup\{-\infty,+\infty\} $$ where ${\mathbb R}$ denotes the set of real numbers. In this case, the plus sign in $+\infty$ is usually not omitted. On the other hand, one could also talk about the extended nonnegative real numbers $[0,\infty]$, which appears a lot especially in integration theory.

However, one should be careful that the arithmetic operations of ${\mathbb R}$ can be only partially extended to $\overline{\mathbb R}$ or $[0,\infty]$. For instance $\frac{1}{0}$ is not defined in $\overline{\mathbb R}$: it is neither $-\infty$ nor $+\infty$. On the other hand, $\frac{1}{0}=\infty$ (as well as $\frac{1}{\infty}=0$) is used in the statement of the Cauchy-Hadamard Theorem since with such definition, the radius of convergence of any power series $\sum a_nz^n$ can be written as $$ R=\frac{1}{\limsup_{n}|a_n|^{1/n}}. $$

See also a discussion on the extended reals in this set of lecture notes by Terry Tao. Here is an excerpt:

Most of the laws of algebra for addition, multiplication, and order continue to hold in this extended number system; for instance addition and multiplication are commutative and associative, with the latter distributing over the former, and an order relation ${x \leq y}$ is preserved under addition or multiplication of both sides of that relation by the same quantity. However, we caution that the laws of cancellation do not apply once some of the variables are allowed to be infinite; for instance, we cannot deduce ${x=y}$ from ${+\infty+x=+\infty+y}$ or from ${+\infty \cdot x = +\infty \cdot y}$. This is related to the fact that the forms ${+\infty - +\infty}$ and ${+\infty/+\infty}$ are indeterminate (one cannot assign a value to them without breaking a lot of the rules of algebra). A general rule of thumb is that if one wishes to use cancellation (or proxies for cancellation, such as subtraction or division), this is only safe if one can guarantee that all quantities involved are finite (and in the case of multiplicative cancellation, the quantity being cancelled also needs to be non-zero, of course). However, as long as one avoids using cancellation and works exclusively with non-negative quantities, there is little danger in working in the extended real number system.

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  • $\begingroup$ Thanks a ton for the help Jack! $\endgroup$ – Mohammad Zuhair Khan Dec 26 '17 at 15:36
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I have always known infinity as a concept not as a no. Dealing with infinity the way we deal with no. Is wrong ,we can't really do that. 1/0 is not equal to infinity just because it's a pattern that it goes larger consider lim $1/x$ as $x$ goes to zero but if you see from what direction we are going that matters a lot left hand limit comes to be different from the right hand limit hence limit actually doesn't exist.

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    $\begingroup$ I don't think MZK's ideas should be dismissed altogether. If you extend the reals to the real projective line, then it does make sense to say that 1/0 = infinity. See e.g. the Wikipedia article: en.wikipedia.org/wiki/Projective_line $\endgroup$ – Michael Behrend Dec 26 '17 at 8:28
  • $\begingroup$ Thanks @MichaelBehrend. The wikipedia article helped a lot! I will be giving my O levels in January and I have only the basic knowledge of stuff like these so please forgive my apparent stupidity. $\endgroup$ – Mohammad Zuhair Khan Dec 26 '17 at 15:40
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The appropriate setting for $\frac{1}{0}=\infty$ is the complex numbers. Here the field $\mathbb C$ is extended to the Riemann sphere $\hat{\mathbb C}=\mathbb C \cup \{\infty\}$. Such an extension is extremely useful in complex analysis when one works with holomorphic functions, and is a first step toward the modern approach to Riemann surfaces.

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    $\begingroup$ I always thought it fell into the category of extended real numbers. $\endgroup$ – Mohammad Zuhair Khan Dec 27 '17 at 9:42
  • $\begingroup$ Anyway thanks for pointing this out. $\endgroup$ – Mohammad Zuhair Khan Dec 27 '17 at 9:42
  • $\begingroup$ The extended real numbers usually involve the addition of a pair of infinities, $-\infty$ and $\infty$, whereas your question is most naturally interpreted as involving the addition of a single point. $\endgroup$ – Mikhail Katz Dec 27 '17 at 9:44
  • $\begingroup$ Okay thanks! Does $\mathbb C \cup \{\infty\}$ denote the set of extended complex numbers? $\endgroup$ – Mohammad Zuhair Khan Dec 27 '17 at 9:45
  • $\begingroup$ Yes. The notation $\hat{\mathbb C}$ is quite common in complex analysis. Note that obviously $\hat{\mathbb{C}}$ is not a field, so referring to this as "numbers" needs to be taken with a grain of salt. $\endgroup$ – Mikhail Katz Dec 27 '17 at 9:49

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