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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2966679%2fhow-much-bigger-is-3%25e2%2586%2591%25e2%2586%2591%25e2%2586%2591%25e2%2586%25913-compared-to-3%25e2%2586%2591%25e2%2586%2591%25e2%2586%25913%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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
















    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2966679%2fhow-much-bigger-is-3%25e2%2586%2591%25e2%2586%2591%25e2%2586%2591%25e2%2586%25913-compared-to-3%25e2%2586%2591%25e2%2586%2591%25e2%2586%25913%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me
            Read more