Infinity plus Infinity












8












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















8












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














8












8








8


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










4 Answers
4






active

oldest

votes


















5












$begingroup$

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 $x=-y$:
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x + x = infty $$
$$ lim_{(x,y) mapsto (infty, infty)} x + y
= lim_{x mapsto infty} x - x = 0 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















7












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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



















    0












    $begingroup$

    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 ?






    share|cite|improve this answer









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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 $x=-y$:
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x + x = infty $$
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x - x = 0 $$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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
















      5












      $begingroup$

      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 $x=-y$:
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x + x = infty $$
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x - x = 0 $$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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














      5












      5








      5





      $begingroup$

      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 $x=-y$:
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x + x = infty $$
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x - x = 0 $$






      share|cite|improve this answer









      $endgroup$



      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 $x=-y$:
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x + x = infty $$
      $$ lim_{(x,y) mapsto (infty, infty)} x + y
      = lim_{x mapsto infty} x - x = 0 $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Mar 19 '12 at 20:02









      HurkylHurkyl

      111k9118262




      111k9118262












      • $begingroup$
        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










      • $begingroup$
        (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











      7












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 19 '12 at 20:04









          Henning MakholmHenning Makholm

          239k17303540




          239k17303540























              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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
















              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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














              1












              1








              1





              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$



              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.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Mar 19 '12 at 20:26

























              answered Mar 19 '12 at 19:59









              DesiatoDesiato

              1,057920




              1,057920












              • $begingroup$
                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


















              • $begingroup$
                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
















              $begingroup$
              How do you derive $infty - infty = 0$?
              $endgroup$
              – TMM
              Mar 19 '12 at 20:02




              $begingroup$
              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






              $begingroup$
              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




              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




              $begingroup$
              @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




              $begingroup$
              @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




              $begingroup$
              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











              0












              $begingroup$

              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 ?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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 ?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 ?






                  share|cite|improve this answer









                  $endgroup$



                  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 ?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 '14 at 4:54









                  Tong CuiTong Cui

                  1




                  1






























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