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Let $a \in \mathbb{C}$. Ahlfors says we let $a + \infty = \infty$ and $a \cdot \infty = \infty$. But we cannot define $\infty + \infty$ without violating the laws of arithimetic (i.e. field axioms).

I don't see why this is. Don't we have $\infty + \infty = \infty$ by applying the distributive law to $2\cdot \infty$? What am I misunderstanding?

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    In terms of set theory, it is true that for any infinite power K:k+k=k*k=k. note that for a=0 : a*k=0 and not infinity2012-03-19
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    @Belgi: This $\infty$ is not a set-theoretic cardinality at all.2012-03-19

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It has to do with Dimension I think. For example, we can let cos(75 degrees) be cos 45 degrees plus cos 30 degrees, because functions act different way than whole numbers, we can not factor them out. Same thing with Infinity, does it act the same way with functions and whole numbers ? The answer is no. And are the infinity are all the same. What if one infinity is in area, one in linear and another one in cube. Therefore, infinity plus infinity square equals to infinity cube ? It doesn't work that way. If we can solve this problem out, I think we can create a wormhole to go from one side to another side in a shortcut. And infinity has to do with time too, does infinity has the same time mass, we do not know ?

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The misunderstanding is assuming that the distributive law applies without exception to the extended arithmetic you're defining here. It doesn't, and your reasoning shows why.

Without defining $\infty+\infty$ you can make the field axioms hold to the extent that the expressions in the axioms are defined at all. The point of not defining $\infty+\infty$ is exactly to avoid expressions for which different applications of the field axioms would give multiple inconsistent values.

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Ah, but are you sure you have the distributive law? :)

The proof of non-existence is that the arithmetic operations are defined by continuous extension. So, we need to check whether or not the following limit exists: $$ \lim_{(x,y) \mapsto (\infty, \infty)} x + y $$ If it existed, we could compute it by taking the limit a particular path. The first chooses $x=y$, and the second $x=-y$: $$ \lim_{(x,y) \mapsto (\infty, \infty)} x + y = \lim_{x \mapsto \infty} x + x = \infty $$ $$ \lim_{(x,y) \mapsto (\infty, \infty)} x + y = \lim_{x \mapsto \infty} x - x = 0 $$

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    How does $y \to \infty$ if $-y = x \to \infty$?2012-03-19
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    (Certainly if $(x,y) \to (\infty,\infty)$ then $x+y \to \infty$ as well.)2012-03-19
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    He's working in the projective complex numbers, I believe. $-\infty$ and $\infty$ are the same thing, just like the projective real numbers... and different than the extended real numbers, where there is a different infinite number at each end.2012-03-19
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Assume you had such a $\infty$ consistant with field axioms. Then $ \infty = \infty + \infty \Longrightarrow \infty - \infty = 0$. This follows from the fact that in a field, a unique additive inverse exists for every element, so that also exists for $ \infty$. You wanted to define $\infty = \infty + \infty$, and if we add the additive inverse on both sides, we get $ 0 = \infty + (-\infty) = (\infty + \infty) + (-\infty)$. By associativity, we can swap the brackets on the right to couple $\infty$ with its additive inverse and get $\infty + 0 = \infty$

But then $ 1 = 1 + \infty + (-\infty) = \infty + (-\infty) = 0$, a contradicition.

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    How do you derive $\infty - \infty = 0$?2012-03-19
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    I do not see why this is a problem with $\infty + \infty$. The trouble comes when you try to discuss $\infty - \infty$.2012-03-19
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    @Mark, but that's the same since $-\infty=\infty$ by the first law applied to $a=-1$.2012-03-19
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    @TMM subtract $\infty$ from both sides of the equation.2012-03-19
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    I don't see how, and Desiato's proof skips a step. Following the logic there, and using field axioms, we can only arrive at the conclusion that $0\cdot\infty = \infty$, which is surprising, but not actually contradictory.2012-03-19
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    And I should add you must avoid extending the field axioms to the point of saying that $\infty$ has either an additive or a multiplicative inverse. I think you can have distributivity if you avoid trying to give $\infty$ an inverse.2012-03-19
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    @azarel: Even if that is allowed, subtracting $\infty$ from both sides should give $\infty = 0$. But if you can subtract $\infty$ then from $a + \infty = \infty$ it follows that $a = 0$ for all $a$...2012-03-19
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    @TMM $a-a=0$ holds in any group. So, in particular, it should hold for $a=\infty$ if we had such a $\infty$ constant with the field axioms.2012-03-19
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    @azarel: $a - a := a + (-a)$ is *defined* only if $a$ has an additive inverse $-a$, with $a + (-a) = 0$. No definition of the additive inverse of $\infty$ is given. Also see Henning's answer.2012-03-19
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    I would read the quote from the question "But we cannot define $ \infty + \infty $ without violating the laws of arithimetic (i.e. field axioms)." to imply that the definition claimed not to exist would then satisfy the field axioms and thus the existance of additive inverse. Otherwise, it is rather trivial that you can add any object you like to $ \mathbb{R} $.2012-03-19