Infinity plus Infinity
$begingroup$
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 $2cdot infty$? What am I misunderstanding?
complex-analysis
asked Mar 19 '12 at 19:54
richyrichy
5813
$endgroup$
add a comment |
$begingroup$
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 $2cdot infty$? What am I misunderstanding?
complex-analysis
asked Mar 19 '12 at 19:54
richyrichy
5813
$endgroup$
$begingroup$
In terms of set theory, it is true that for any infinite power K:k+k=kk=k. note that for a=0 : ak=0 and not infinity
$endgroup$
– Belgi
Mar 19 '12 at 19:58
4
$begingroup$
@Belgi: This $infty$ is not a set-theoretic cardinality at all.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:00
add a comment |
$begingroup$
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 $2cdot infty$? What am I misunderstanding?
complex-analysis
asked Mar 19 '12 at 19:54
richyrichy
5813
$endgroup$
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 $2cdot infty$? What am I misunderstanding?
complex-analysis
complex-analysis
asked Mar 19 '12 at 19:54
richyrichy
5813
asked Mar 19 '12 at 19:54
richyrichy
5813
asked Mar 19 '12 at 19:54
richyrichy
5813
asked Mar 19 '12 at 19:54
richyrichy
5813
asked Mar 19 '12 at 19:54
richyrichy
5813
5813
$begingroup$
In terms of set theory, it is true that for any infinite power K:k+k=kk=k. note that for a=0 : ak=0 and not infinity
$endgroup$
– Belgi
Mar 19 '12 at 19:58
4
$begingroup$
@Belgi: This $infty$ is not a set-theoretic cardinality at all.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:00
add a comment |
$begingroup$
In terms of set theory, it is true that for any infinite power K:k+k=kk=k. note that for a=0 : ak=0 and not infinity
$endgroup$
– Belgi
Mar 19 '12 at 19:58
4
$begingroup$
@Belgi: This $infty$ is not a set-theoretic cardinality at all.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:00
$begingroup$
In terms of set theory, it is true that for any infinite power K:k+k=kk=k. note that for a=0 : ak=0 and not infinity
$endgroup$
– Belgi
Mar 19 '12 at 19:58
$begingroup$
In terms of set theory, it is true that for any infinite power K:k+k=kk=k. note that for a=0 : ak=0 and not infinity
$endgroup$
– Belgi
Mar 19 '12 at 19:58
4
4
$begingroup$
@Belgi: This $infty$ is not a set-theoretic cardinality at all.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:00
$begingroup$
@Belgi: This $infty$ is not a set-theoretic cardinality at all.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:00
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 $x=-y$:
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x + x = infty $$
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x - x = 0 $$
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
$endgroup$
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
$endgroup$
$begingroup$
How do you derive $infty - infty = 0$?
$endgroup$
– TMM
Mar 19 '12 at 20:02
$begingroup$
I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
$endgroup$
– MJD
Mar 19 '12 at 20:03
4
$begingroup$
@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:04
$begingroup$
@TMM subtract $infty$ from both sides of the equation.
$endgroup$
– azarel
Mar 19 '12 at 20:05
$begingroup$
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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
$endgroup$
– MJD
Mar 19 '12 at 20:06
|
show 5 more comments
$begingroup$
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 ?
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x=-y$:
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x + x = infty $$
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x - x = 0 $$
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
$endgroup$
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
add a comment |
$begingroup$
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 $x=-y$:
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x + x = infty $$
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x - x = 0 $$
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
$endgroup$
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
add a comment |
$begingroup$
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 $x=-y$:
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x + x = infty $$
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x - x = 0 $$
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
$endgroup$
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 $x=-y$:
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x + x = infty $$
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x - x = 0 $$
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
answered Mar 19 '12 at 20:02
HurkylHurkyl
111k9120262
111k9120262
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
add a comment |
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
How does $y to infty$ if $-y = x to infty$?
$endgroup$
– TMM
Mar 19 '12 at 20:31
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
(Certainly if $(x,y) to (infty,infty)$ then $x+y to infty$ as well.)
$endgroup$
– TMM
Mar 19 '12 at 20:34
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
$begingroup$
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.
$endgroup$
– Hurkyl
Mar 19 '12 at 22:17
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
$endgroup$
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.
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
answered Mar 19 '12 at 20:04
Henning MakholmHenning Makholm
240k17305542
240k17305542
add a comment |
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
$endgroup$
$begingroup$
How do you derive $infty - infty = 0$?
$endgroup$
– TMM
Mar 19 '12 at 20:02
$begingroup$
I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
$endgroup$
– MJD
Mar 19 '12 at 20:03
4
$begingroup$
@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:04
$begingroup$
@TMM subtract $infty$ from both sides of the equation.
$endgroup$
– azarel
Mar 19 '12 at 20:05
$begingroup$
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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
$endgroup$
– MJD
Mar 19 '12 at 20:06
|
show 5 more comments
$begingroup$
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.
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
$endgroup$
$begingroup$
How do you derive $infty - infty = 0$?
$endgroup$
– TMM
Mar 19 '12 at 20:02
$begingroup$
I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
$endgroup$
– MJD
Mar 19 '12 at 20:03
4
$begingroup$
@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:04
$begingroup$
@TMM subtract $infty$ from both sides of the equation.
$endgroup$
– azarel
Mar 19 '12 at 20:05
$begingroup$
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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
$endgroup$
– MJD
Mar 19 '12 at 20:06
|
show 5 more comments
$begingroup$
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.
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
$endgroup$
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.
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
edited Mar 19 '12 at 20:26
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
answered Mar 19 '12 at 19:59
DesiatoDesiato
1,067920
1,067920
$begingroup$
How do you derive $infty - infty = 0$?
$endgroup$
– TMM
Mar 19 '12 at 20:02
$begingroup$
I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
$endgroup$
– MJD
Mar 19 '12 at 20:03
4
$begingroup$
@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:04
$begingroup$
@TMM subtract $infty$ from both sides of the equation.
$endgroup$
– azarel
Mar 19 '12 at 20:05
$begingroup$
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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
$endgroup$
– MJD
Mar 19 '12 at 20:06
|
show 5 more comments
$begingroup$
How do you derive $infty - infty = 0$?
$endgroup$
– TMM
Mar 19 '12 at 20:02
$begingroup$
I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
$endgroup$
– MJD
Mar 19 '12 at 20:03
4
$begingroup$
@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
$endgroup$
– Henning Makholm
Mar 19 '12 at 20:04
$begingroup$
@TMM subtract $infty$ from both sides of the equation.
$endgroup$
– azarel
Mar 19 '12 at 20:05
$begingroup$
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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
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– MJD
Mar 19 '12 at 20:06
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How do you derive $infty - infty = 0$?
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– TMM
Mar 19 '12 at 20:02
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How do you derive $infty - infty = 0$?
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– TMM
Mar 19 '12 at 20:02
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I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
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– MJD
Mar 19 '12 at 20:03
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I do not see why this is a problem with $infty + infty$. The trouble comes when you try to discuss $infty - infty$.
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– MJD
Mar 19 '12 at 20:03
4
4
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@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
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– Henning Makholm
Mar 19 '12 at 20:04
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@Mark, but that's the same since $-infty=infty$ by the first law applied to $a=-1$.
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– Henning Makholm
Mar 19 '12 at 20:04
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@TMM subtract $infty$ from both sides of the equation.
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– azarel
Mar 19 '12 at 20:05
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@TMM subtract $infty$ from both sides of the equation.
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– azarel
Mar 19 '12 at 20:05
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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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
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– MJD
Mar 19 '12 at 20:06
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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 $0cdotinfty = infty$, which is surprising, but not actually contradictory.
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– MJD
Mar 19 '12 at 20:06
|
show 5 more comments
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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 ?
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
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add a comment |
$begingroup$
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 ?
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
$endgroup$
add a comment |
$begingroup$
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 ?
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
$endgroup$
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 ?
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
answered Nov 21 '14 at 4:54
Tong CuiTong Cui
1
1
add a comment |
add a comment |
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In terms of set theory, it is true that for any infinite power K:k+k=kk=k. note that for a=0 : ak=0 and not infinity
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– Belgi
Mar 19 '12 at 19:58
4
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@Belgi: This $infty$ is not a set-theoretic cardinality at all.
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– Henning Makholm
Mar 19 '12 at 20:00