How much bigger is 3↑↑↑↑3 compared to 3↑↑↑3?












2












$begingroup$


3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?






big-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger.
    $endgroup$
    – Yves Daoust
    Oct 22 '18 at 21:29
















2












$begingroup$


3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?






big-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger.
    $endgroup$
    – Yves Daoust
    Oct 22 '18 at 21:29














2












2








2





$begingroup$


3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?






big-numbers







share|cite|improve this question











$endgroup$




3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?






big-numbers




big-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 22 '18 at 21:48









anomaly

17.4k42664




17.4k42664










asked Oct 22 '18 at 21:22









Shaan PaymasterShaan Paymaster

111




111












  • $begingroup$
    No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger.
    $endgroup$
    – Yves Daoust
    Oct 22 '18 at 21:29


















  • $begingroup$
    No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger.
    $endgroup$
    – Yves Daoust
    Oct 22 '18 at 21:29
















$begingroup$
No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger.
$endgroup$
– Yves Daoust
Oct 22 '18 at 21:29




$begingroup$
No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger.
$endgroup$
– Yves Daoust
Oct 22 '18 at 21:29










2 Answers
2






active

oldest

votes


















1












$begingroup$

By definition, $3uparrowuparrowuparrowuparrow 3 = 3uparrowuparrowuparrow(3uparrowuparrowuparrow(3uparrowuparrowuparrow 3))$.
begin{matrix}
3uparrowuparrowuparrow 3= & underbrace{3^{3^{3^{3^{cdot^{cdot^{cdot^{cdot^{3}}}}}}}}} \
& mbox{7,625,597,484,987 copies of 3}
end{matrix}



which should make it a little clearer how much bigger $3uparrowuparrowuparrowuparrow 3$ is than $3uparrowuparrowuparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What's the name of the operator?
    $endgroup$
    – 0x90
    Oct 22 '18 at 22:16












  • $begingroup$
    I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
    $endgroup$
    – kcborys
    Oct 23 '18 at 16:25






  • 1




    $begingroup$
    Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
    $endgroup$
    – Simply Beautiful Art
    Jan 9 at 21:25



















0












$begingroup$

Knuth's up-arrow notation is defined so that we have



$$auparrow^kb=underbrace{auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a}_{btext{ many }atext{'s}}$$



where $uparrow^k=underbrace{uparrowuparrowdotsuparrowuparrow}_k~$ and we evaluate from right to left (e.g. $auparrow buparrow c=auparrow(buparrow c)ne(auparrow b)uparrow c$ in general).



This means we have:



$$3uparrowuparrowuparrowuparrow3=3uparrowuparrowuparrow3uparrowuparrowuparrow3$$



and we also have



$$3uparrowuparrowuparrow3=3uparrowuparrow3uparrowuparrow3uparrowuparrow3$$



and



$$3uparrowuparrow3=3uparrow3uparrow3=3^{3^3}=7,625,597,484,987$$



Using this result, we may further calculate $3uparrowuparrowuparrow3$ as



$$underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$



which is pretty large. Keep in mind that this is $3uparrowuparrowuparrow3$, or $3uparrowuparrow3uparrowuparrow3$. If we took this result and did $uparrowuparrow$ that many times, we'd basically arrive at $3uparrowuparrowuparrowuparrow3$, which is equivalent to:



$$3uparrowuparrowuparrow3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrowdotsuparrowuparrow3}_{3uparrowuparrowuparrow3}$$



which is much much much larger compared to the relatively puny



$$3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrow3}_3$$






share|cite|improve this answer









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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    active

    oldest

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    1












    $begingroup$

    By definition, $3uparrowuparrowuparrowuparrow 3 = 3uparrowuparrowuparrow(3uparrowuparrowuparrow(3uparrowuparrowuparrow 3))$.
    begin{matrix}
    3uparrowuparrowuparrow 3= & underbrace{3^{3^{3^{3^{cdot^{cdot^{cdot^{cdot^{3}}}}}}}}} \
    & mbox{7,625,597,484,987 copies of 3}
    end{matrix}



    which should make it a little clearer how much bigger $3uparrowuparrowuparrowuparrow 3$ is than $3uparrowuparrowuparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What's the name of the operator?
      $endgroup$
      – 0x90
      Oct 22 '18 at 22:16












    • $begingroup$
      I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
      $endgroup$
      – kcborys
      Oct 23 '18 at 16:25






    • 1




      $begingroup$
      Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
      $endgroup$
      – Simply Beautiful Art
      Jan 9 at 21:25
















    1












    $begingroup$

    By definition, $3uparrowuparrowuparrowuparrow 3 = 3uparrowuparrowuparrow(3uparrowuparrowuparrow(3uparrowuparrowuparrow 3))$.
    begin{matrix}
    3uparrowuparrowuparrow 3= & underbrace{3^{3^{3^{3^{cdot^{cdot^{cdot^{cdot^{3}}}}}}}}} \
    & mbox{7,625,597,484,987 copies of 3}
    end{matrix}



    which should make it a little clearer how much bigger $3uparrowuparrowuparrowuparrow 3$ is than $3uparrowuparrowuparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What's the name of the operator?
      $endgroup$
      – 0x90
      Oct 22 '18 at 22:16












    • $begingroup$
      I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
      $endgroup$
      – kcborys
      Oct 23 '18 at 16:25






    • 1




      $begingroup$
      Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
      $endgroup$
      – Simply Beautiful Art
      Jan 9 at 21:25














    1












    1








    1





    $begingroup$

    By definition, $3uparrowuparrowuparrowuparrow 3 = 3uparrowuparrowuparrow(3uparrowuparrowuparrow(3uparrowuparrowuparrow 3))$.
    begin{matrix}
    3uparrowuparrowuparrow 3= & underbrace{3^{3^{3^{3^{cdot^{cdot^{cdot^{cdot^{3}}}}}}}}} \
    & mbox{7,625,597,484,987 copies of 3}
    end{matrix}



    which should make it a little clearer how much bigger $3uparrowuparrowuparrowuparrow 3$ is than $3uparrowuparrowuparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.






    share|cite|improve this answer









    $endgroup$



    By definition, $3uparrowuparrowuparrowuparrow 3 = 3uparrowuparrowuparrow(3uparrowuparrowuparrow(3uparrowuparrowuparrow 3))$.
    begin{matrix}
    3uparrowuparrowuparrow 3= & underbrace{3^{3^{3^{3^{cdot^{cdot^{cdot^{cdot^{3}}}}}}}}} \
    & mbox{7,625,597,484,987 copies of 3}
    end{matrix}



    which should make it a little clearer how much bigger $3uparrowuparrowuparrowuparrow 3$ is than $3uparrowuparrowuparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 22 '18 at 22:14









    kcboryskcborys

    45728




    45728












    • $begingroup$
      What's the name of the operator?
      $endgroup$
      – 0x90
      Oct 22 '18 at 22:16












    • $begingroup$
      I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
      $endgroup$
      – kcborys
      Oct 23 '18 at 16:25






    • 1




      $begingroup$
      Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
      $endgroup$
      – Simply Beautiful Art
      Jan 9 at 21:25


















    • $begingroup$
      What's the name of the operator?
      $endgroup$
      – 0x90
      Oct 22 '18 at 22:16












    • $begingroup$
      I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
      $endgroup$
      – kcborys
      Oct 23 '18 at 16:25






    • 1




      $begingroup$
      Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
      $endgroup$
      – Simply Beautiful Art
      Jan 9 at 21:25
















    $begingroup$
    What's the name of the operator?
    $endgroup$
    – 0x90
    Oct 22 '18 at 22:16






    $begingroup$
    What's the name of the operator?
    $endgroup$
    – 0x90
    Oct 22 '18 at 22:16














    $begingroup$
    I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
    $endgroup$
    – kcborys
    Oct 23 '18 at 16:25




    $begingroup$
    I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation.
    $endgroup$
    – kcborys
    Oct 23 '18 at 16:25




    1




    1




    $begingroup$
    Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
    $endgroup$
    – Simply Beautiful Art
    Jan 9 at 21:25




    $begingroup$
    Correction: $3uparrow^k3=3uparrow^{k-1}3uparrow^{k-1}3$, and generally $auparrow^kb=auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a$ with $b$ many $a$'s.
    $endgroup$
    – Simply Beautiful Art
    Jan 9 at 21:25











    0












    $begingroup$

    Knuth's up-arrow notation is defined so that we have



    $$auparrow^kb=underbrace{auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a}_{btext{ many }atext{'s}}$$



    where $uparrow^k=underbrace{uparrowuparrowdotsuparrowuparrow}_k~$ and we evaluate from right to left (e.g. $auparrow buparrow c=auparrow(buparrow c)ne(auparrow b)uparrow c$ in general).



    This means we have:



    $$3uparrowuparrowuparrowuparrow3=3uparrowuparrowuparrow3uparrowuparrowuparrow3$$



    and we also have



    $$3uparrowuparrowuparrow3=3uparrowuparrow3uparrowuparrow3uparrowuparrow3$$



    and



    $$3uparrowuparrow3=3uparrow3uparrow3=3^{3^3}=7,625,597,484,987$$



    Using this result, we may further calculate $3uparrowuparrowuparrow3$ as



    $$underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$



    which is pretty large. Keep in mind that this is $3uparrowuparrowuparrow3$, or $3uparrowuparrow3uparrowuparrow3$. If we took this result and did $uparrowuparrow$ that many times, we'd basically arrive at $3uparrowuparrowuparrowuparrow3$, which is equivalent to:



    $$3uparrowuparrowuparrow3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrowdotsuparrowuparrow3}_{3uparrowuparrowuparrow3}$$



    which is much much much larger compared to the relatively puny



    $$3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrow3}_3$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Knuth's up-arrow notation is defined so that we have



      $$auparrow^kb=underbrace{auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a}_{btext{ many }atext{'s}}$$



      where $uparrow^k=underbrace{uparrowuparrowdotsuparrowuparrow}_k~$ and we evaluate from right to left (e.g. $auparrow buparrow c=auparrow(buparrow c)ne(auparrow b)uparrow c$ in general).



      This means we have:



      $$3uparrowuparrowuparrowuparrow3=3uparrowuparrowuparrow3uparrowuparrowuparrow3$$



      and we also have



      $$3uparrowuparrowuparrow3=3uparrowuparrow3uparrowuparrow3uparrowuparrow3$$



      and



      $$3uparrowuparrow3=3uparrow3uparrow3=3^{3^3}=7,625,597,484,987$$



      Using this result, we may further calculate $3uparrowuparrowuparrow3$ as



      $$underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$



      which is pretty large. Keep in mind that this is $3uparrowuparrowuparrow3$, or $3uparrowuparrow3uparrowuparrow3$. If we took this result and did $uparrowuparrow$ that many times, we'd basically arrive at $3uparrowuparrowuparrowuparrow3$, which is equivalent to:



      $$3uparrowuparrowuparrow3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrowdotsuparrowuparrow3}_{3uparrowuparrowuparrow3}$$



      which is much much much larger compared to the relatively puny



      $$3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrow3}_3$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Knuth's up-arrow notation is defined so that we have



        $$auparrow^kb=underbrace{auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a}_{btext{ many }atext{'s}}$$



        where $uparrow^k=underbrace{uparrowuparrowdotsuparrowuparrow}_k~$ and we evaluate from right to left (e.g. $auparrow buparrow c=auparrow(buparrow c)ne(auparrow b)uparrow c$ in general).



        This means we have:



        $$3uparrowuparrowuparrowuparrow3=3uparrowuparrowuparrow3uparrowuparrowuparrow3$$



        and we also have



        $$3uparrowuparrowuparrow3=3uparrowuparrow3uparrowuparrow3uparrowuparrow3$$



        and



        $$3uparrowuparrow3=3uparrow3uparrow3=3^{3^3}=7,625,597,484,987$$



        Using this result, we may further calculate $3uparrowuparrowuparrow3$ as



        $$underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$



        which is pretty large. Keep in mind that this is $3uparrowuparrowuparrow3$, or $3uparrowuparrow3uparrowuparrow3$. If we took this result and did $uparrowuparrow$ that many times, we'd basically arrive at $3uparrowuparrowuparrowuparrow3$, which is equivalent to:



        $$3uparrowuparrowuparrow3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrowdotsuparrowuparrow3}_{3uparrowuparrowuparrow3}$$



        which is much much much larger compared to the relatively puny



        $$3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrow3}_3$$






        share|cite|improve this answer









        $endgroup$



        Knuth's up-arrow notation is defined so that we have



        $$auparrow^kb=underbrace{auparrow^{k-1}auparrow^{k-1}dotsuparrow^{k-1}a}_{btext{ many }atext{'s}}$$



        where $uparrow^k=underbrace{uparrowuparrowdotsuparrowuparrow}_k~$ and we evaluate from right to left (e.g. $auparrow buparrow c=auparrow(buparrow c)ne(auparrow b)uparrow c$ in general).



        This means we have:



        $$3uparrowuparrowuparrowuparrow3=3uparrowuparrowuparrow3uparrowuparrowuparrow3$$



        and we also have



        $$3uparrowuparrowuparrow3=3uparrowuparrow3uparrowuparrow3uparrowuparrow3$$



        and



        $$3uparrowuparrow3=3uparrow3uparrow3=3^{3^3}=7,625,597,484,987$$



        Using this result, we may further calculate $3uparrowuparrowuparrow3$ as



        $$underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$



        which is pretty large. Keep in mind that this is $3uparrowuparrowuparrow3$, or $3uparrowuparrow3uparrowuparrow3$. If we took this result and did $uparrowuparrow$ that many times, we'd basically arrive at $3uparrowuparrowuparrowuparrow3$, which is equivalent to:



        $$3uparrowuparrowuparrow3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrowdotsuparrowuparrow3}_{3uparrowuparrowuparrow3}$$



        which is much much much larger compared to the relatively puny



        $$3uparrowuparrowuparrow3=underbrace{3uparrowuparrow3uparrowuparrow3}_3$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 21:35









        Simply Beautiful ArtSimply Beautiful Art

        50.5k578181




        50.5k578181






























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