Decoding the sign expansion of surreal numbers The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraAre surreal numbers actually well-defined in ZFC?Algorithm for Converting Rational Into Surreal NumberPseudo-Surreal numbers are analogous to?Surreal numbers in set theories other than ZFCRelatively tight upper and lower bounds for surreal numbersSurreal numbers whose final segment is an integer.Hexadecimal numbers that does not look like decimalConverting Roman Numbers to Decimal (Not Duplicate)Surreal arithmetic with $frac12omega$examples of sign expansion from normal forms of surreal numbers

Huge performance difference of the command find with and without using %M option to show permissions

Single author papers against my advisor's will?

Can a flute soloist sit?

Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)

Was credit for the black hole image misappropriated?

What to do when moving next to a bird sanctuary with a loosely-domesticated cat?

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

Would an alien lifeform be able to achieve space travel if lacking in vision?

Can I visit the Trinity College (Cambridge) library and see some of their rare books

How to determine omitted units in a publication

Why can't devices on different VLANs, but on the same subnet, communicate?

Can each chord in a progression create its own key?

Student Loan from years ago pops up and is taking my salary

What's the point in a preamp?

Using dividends to reduce short term capital gains?

What can I do to 'burn' a journal?

One-dimensional Japanese puzzle

Is every episode of "Where are my Pants?" identical?

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

"is" operation returns false even though two objects have same id

Match Roman Numerals

Mortgage adviser recommends a longer term than necessary combined with overpayments

60's-70's movie: home appliances revolting against the owners

What is the padding with red substance inside of steak packaging?



Decoding the sign expansion of surreal numbers



The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraAre surreal numbers actually well-defined in ZFC?Algorithm for Converting Rational Into Surreal NumberPseudo-Surreal numbers are analogous to?Surreal numbers in set theories other than ZFCRelatively tight upper and lower bounds for surreal numbersSurreal numbers whose final segment is an integer.Hexadecimal numbers that does not look like decimalConverting Roman Numbers to Decimal (Not Duplicate)Surreal arithmetic with $frac12omega$examples of sign expansion from normal forms of surreal numbers










1












$begingroup$


One way to represent surreal numbers is the sign expansion. Now Wikipedia describes how to compare them, how to convert them to the standard representation of left/right sets, how to negate them, and how to add and multiply them (however those operations are described basically by converting to left/right sets, adding them in that form, and converting back).



However I'm interested in "reading" them. My question is now twofold: First, I want to know if what I think I've already found out is correct, and second, I'd like to know how to "read" more of them (ideally, a scheme that at least in principle allows to "read" all of them). With "reading", I mean finding out a number/formula representation that represents the same number, without going through the recursion formula defining the surreal numbers.



Here's what I think I've already found out:



First, consider the numbers with finite domain. Those are effectively represented by a finite string of $+$ and $-$. Since a number gets negated by changing all signs at once, it suffices to consider those numbers which do not start with a $-$ (i.e. the nonnegative numbers).



The special case that the string contains no $-$ is simple: The number of $+$ in the string is the number represented.



Otherwise, since the string begins with a $+$, it contains somewhere the substring $+-$. The first such substring can be interpreted as "decimal point", separating the integer part before it and the fractional part after it. The integer part before it is interpreted as before: The number of $+$ is directly the integer part of the number. The fractional part is then decoded as follows: Whenever the same sign follows, a $0$ is appended to the number, and whenever a different sign follows, an $1$ is appended to the number. Finally, when the end of the string is reached, a final $1$ is appended. The digit string such obtained is interpreted in base 2.



For example, the digit string $+++--+--$ contains a $-$, therefore we identify the first $+-$ as "decimal point". That is, we get $++mathbf+--+--$. Before the $+-$ we have two $+$, so the integer part is $2$. The fractional part starts with $-$ which is the same as the previous $-$ from the "decimal point", so we start with $0$. Then we get a sign change to $+$, giving an $1$, followed by a second sign change to $-$ again, giving another $1$, and no sign change giving another $0$. Finally, another $1$ is appended, giving a total fractional part of $.01101_2 = .40625_10$. Therefore the number is $2.40625$.



Next, consider the numbers with domain $omega$. I think the same interpretation can be applied here, except that you of course never reach a final digit, and thus never reach an end.



However there are two special cases, namely the ones which would result in an infinite number of $0$s (i.e. the string ending either in $++++dots$ or $----dots$) and an infinite number of $1$s (i.e. a string ending in $+-+-dots$); those, interpreted as real numbers, would give the finite binary fractions again. I think those correspond to $x+epsilon$ and $x-epsilon$ with $x$ a dyadic fraction (because those are exactly the generation $omega$ numbers which are missing, and the ordering would also fit).



Now I want to know:



  • Is my interpretation of the sign expansion correct?

  • And how can the sign expansions with domain $>omega$ be interpreted?









share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    One way to represent surreal numbers is the sign expansion. Now Wikipedia describes how to compare them, how to convert them to the standard representation of left/right sets, how to negate them, and how to add and multiply them (however those operations are described basically by converting to left/right sets, adding them in that form, and converting back).



    However I'm interested in "reading" them. My question is now twofold: First, I want to know if what I think I've already found out is correct, and second, I'd like to know how to "read" more of them (ideally, a scheme that at least in principle allows to "read" all of them). With "reading", I mean finding out a number/formula representation that represents the same number, without going through the recursion formula defining the surreal numbers.



    Here's what I think I've already found out:



    First, consider the numbers with finite domain. Those are effectively represented by a finite string of $+$ and $-$. Since a number gets negated by changing all signs at once, it suffices to consider those numbers which do not start with a $-$ (i.e. the nonnegative numbers).



    The special case that the string contains no $-$ is simple: The number of $+$ in the string is the number represented.



    Otherwise, since the string begins with a $+$, it contains somewhere the substring $+-$. The first such substring can be interpreted as "decimal point", separating the integer part before it and the fractional part after it. The integer part before it is interpreted as before: The number of $+$ is directly the integer part of the number. The fractional part is then decoded as follows: Whenever the same sign follows, a $0$ is appended to the number, and whenever a different sign follows, an $1$ is appended to the number. Finally, when the end of the string is reached, a final $1$ is appended. The digit string such obtained is interpreted in base 2.



    For example, the digit string $+++--+--$ contains a $-$, therefore we identify the first $+-$ as "decimal point". That is, we get $++mathbf+--+--$. Before the $+-$ we have two $+$, so the integer part is $2$. The fractional part starts with $-$ which is the same as the previous $-$ from the "decimal point", so we start with $0$. Then we get a sign change to $+$, giving an $1$, followed by a second sign change to $-$ again, giving another $1$, and no sign change giving another $0$. Finally, another $1$ is appended, giving a total fractional part of $.01101_2 = .40625_10$. Therefore the number is $2.40625$.



    Next, consider the numbers with domain $omega$. I think the same interpretation can be applied here, except that you of course never reach a final digit, and thus never reach an end.



    However there are two special cases, namely the ones which would result in an infinite number of $0$s (i.e. the string ending either in $++++dots$ or $----dots$) and an infinite number of $1$s (i.e. a string ending in $+-+-dots$); those, interpreted as real numbers, would give the finite binary fractions again. I think those correspond to $x+epsilon$ and $x-epsilon$ with $x$ a dyadic fraction (because those are exactly the generation $omega$ numbers which are missing, and the ordering would also fit).



    Now I want to know:



    • Is my interpretation of the sign expansion correct?

    • And how can the sign expansions with domain $>omega$ be interpreted?









    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      One way to represent surreal numbers is the sign expansion. Now Wikipedia describes how to compare them, how to convert them to the standard representation of left/right sets, how to negate them, and how to add and multiply them (however those operations are described basically by converting to left/right sets, adding them in that form, and converting back).



      However I'm interested in "reading" them. My question is now twofold: First, I want to know if what I think I've already found out is correct, and second, I'd like to know how to "read" more of them (ideally, a scheme that at least in principle allows to "read" all of them). With "reading", I mean finding out a number/formula representation that represents the same number, without going through the recursion formula defining the surreal numbers.



      Here's what I think I've already found out:



      First, consider the numbers with finite domain. Those are effectively represented by a finite string of $+$ and $-$. Since a number gets negated by changing all signs at once, it suffices to consider those numbers which do not start with a $-$ (i.e. the nonnegative numbers).



      The special case that the string contains no $-$ is simple: The number of $+$ in the string is the number represented.



      Otherwise, since the string begins with a $+$, it contains somewhere the substring $+-$. The first such substring can be interpreted as "decimal point", separating the integer part before it and the fractional part after it. The integer part before it is interpreted as before: The number of $+$ is directly the integer part of the number. The fractional part is then decoded as follows: Whenever the same sign follows, a $0$ is appended to the number, and whenever a different sign follows, an $1$ is appended to the number. Finally, when the end of the string is reached, a final $1$ is appended. The digit string such obtained is interpreted in base 2.



      For example, the digit string $+++--+--$ contains a $-$, therefore we identify the first $+-$ as "decimal point". That is, we get $++mathbf+--+--$. Before the $+-$ we have two $+$, so the integer part is $2$. The fractional part starts with $-$ which is the same as the previous $-$ from the "decimal point", so we start with $0$. Then we get a sign change to $+$, giving an $1$, followed by a second sign change to $-$ again, giving another $1$, and no sign change giving another $0$. Finally, another $1$ is appended, giving a total fractional part of $.01101_2 = .40625_10$. Therefore the number is $2.40625$.



      Next, consider the numbers with domain $omega$. I think the same interpretation can be applied here, except that you of course never reach a final digit, and thus never reach an end.



      However there are two special cases, namely the ones which would result in an infinite number of $0$s (i.e. the string ending either in $++++dots$ or $----dots$) and an infinite number of $1$s (i.e. a string ending in $+-+-dots$); those, interpreted as real numbers, would give the finite binary fractions again. I think those correspond to $x+epsilon$ and $x-epsilon$ with $x$ a dyadic fraction (because those are exactly the generation $omega$ numbers which are missing, and the ordering would also fit).



      Now I want to know:



      • Is my interpretation of the sign expansion correct?

      • And how can the sign expansions with domain $>omega$ be interpreted?









      share|cite|improve this question











      $endgroup$




      One way to represent surreal numbers is the sign expansion. Now Wikipedia describes how to compare them, how to convert them to the standard representation of left/right sets, how to negate them, and how to add and multiply them (however those operations are described basically by converting to left/right sets, adding them in that form, and converting back).



      However I'm interested in "reading" them. My question is now twofold: First, I want to know if what I think I've already found out is correct, and second, I'd like to know how to "read" more of them (ideally, a scheme that at least in principle allows to "read" all of them). With "reading", I mean finding out a number/formula representation that represents the same number, without going through the recursion formula defining the surreal numbers.



      Here's what I think I've already found out:



      First, consider the numbers with finite domain. Those are effectively represented by a finite string of $+$ and $-$. Since a number gets negated by changing all signs at once, it suffices to consider those numbers which do not start with a $-$ (i.e. the nonnegative numbers).



      The special case that the string contains no $-$ is simple: The number of $+$ in the string is the number represented.



      Otherwise, since the string begins with a $+$, it contains somewhere the substring $+-$. The first such substring can be interpreted as "decimal point", separating the integer part before it and the fractional part after it. The integer part before it is interpreted as before: The number of $+$ is directly the integer part of the number. The fractional part is then decoded as follows: Whenever the same sign follows, a $0$ is appended to the number, and whenever a different sign follows, an $1$ is appended to the number. Finally, when the end of the string is reached, a final $1$ is appended. The digit string such obtained is interpreted in base 2.



      For example, the digit string $+++--+--$ contains a $-$, therefore we identify the first $+-$ as "decimal point". That is, we get $++mathbf+--+--$. Before the $+-$ we have two $+$, so the integer part is $2$. The fractional part starts with $-$ which is the same as the previous $-$ from the "decimal point", so we start with $0$. Then we get a sign change to $+$, giving an $1$, followed by a second sign change to $-$ again, giving another $1$, and no sign change giving another $0$. Finally, another $1$ is appended, giving a total fractional part of $.01101_2 = .40625_10$. Therefore the number is $2.40625$.



      Next, consider the numbers with domain $omega$. I think the same interpretation can be applied here, except that you of course never reach a final digit, and thus never reach an end.



      However there are two special cases, namely the ones which would result in an infinite number of $0$s (i.e. the string ending either in $++++dots$ or $----dots$) and an infinite number of $1$s (i.e. a string ending in $+-+-dots$); those, interpreted as real numbers, would give the finite binary fractions again. I think those correspond to $x+epsilon$ and $x-epsilon$ with $x$ a dyadic fraction (because those are exactly the generation $omega$ numbers which are missing, and the ordering would also fit).



      Now I want to know:



      • Is my interpretation of the sign expansion correct?

      • And how can the sign expansions with domain $>omega$ be interpreted?






      number-systems surreal-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 31 at 6:35









      Eric Wofsey

      193k14220352




      193k14220352










      asked Apr 16 '15 at 19:41









      celtschkceltschk

      30.4k755101




      30.4k755101




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Your interpretation of finite sign expansions is slightly off. There are essentially two equivalent approaches to interpreting these finite expansions: One makes conversion between binary representations easy, and one hinting at the general structure even for domains greater than $omega$.



          The easy-conversion rule begins the way you started, but after the first instance of "$+-$" you can simply read off the remaining symbols as bits with $+$ being 1 and $-$ being zero, and then append a final $1$. Therefore, $+++−−+−−$ would be $2+left(.01001_2right)$. The other way to think about it is that the first three $+$ signs take you up to $3$, and then the first (and only the first, for a finite expansion) sign change starts a sequence of diminishing returns: $+++−−+−−=1+1+1-frac12-frac14+frac18-frac116-frac132=2+frac14+frac132$.



          In any case, your guess about expansions with domain $omega$ is essentially correct. Every domain-$omega$ expansion that doesn't end in a tail of $-$ or $+$ corresponds to the real number with the corresponding binary expansion. If it's all $+$, then it's $omega$. Among the positives, if it contains a $-$ but ends in a tail of $+$, then that would be a tail of $1$s in the binary expansion for some real number $x$, but the sign expansion actually corresponds to $x-varepsilon=x-frac1omega$. Similarly, a tail of $-$ would be $x+varepsilon$.



          A short answer to your second question would be just to give you some examples, using $cdots$ to indicate a pattern of "length" $omega$. $+++cdots+=omega+1$, $+++cdots-=omega-1$, $+++cdots+++cdots=omega2$, $+++cdots---cdots=fracomega2$ $+--+-+-+-+-+cdots+=frac13+varepsilon$, $+----cdots+=2varepsilon$, $+----cdots-=varepsilon/2$, $+----cdots---cdots=varepsilon^2$.



          A thorough answer is given by Philip Ehrlich's "Conway names, the simplicity hierarchy and the surreal number tree", which is currently available at http://www.ohio.edu/people/ehrlich/ConwayNames.pdf






          share|cite|improve this answer









          $endgroup$













            Your Answer








            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%2f1237974%2fdecoding-the-sign-expansion-of-surreal-numbers%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Your interpretation of finite sign expansions is slightly off. There are essentially two equivalent approaches to interpreting these finite expansions: One makes conversion between binary representations easy, and one hinting at the general structure even for domains greater than $omega$.



            The easy-conversion rule begins the way you started, but after the first instance of "$+-$" you can simply read off the remaining symbols as bits with $+$ being 1 and $-$ being zero, and then append a final $1$. Therefore, $+++−−+−−$ would be $2+left(.01001_2right)$. The other way to think about it is that the first three $+$ signs take you up to $3$, and then the first (and only the first, for a finite expansion) sign change starts a sequence of diminishing returns: $+++−−+−−=1+1+1-frac12-frac14+frac18-frac116-frac132=2+frac14+frac132$.



            In any case, your guess about expansions with domain $omega$ is essentially correct. Every domain-$omega$ expansion that doesn't end in a tail of $-$ or $+$ corresponds to the real number with the corresponding binary expansion. If it's all $+$, then it's $omega$. Among the positives, if it contains a $-$ but ends in a tail of $+$, then that would be a tail of $1$s in the binary expansion for some real number $x$, but the sign expansion actually corresponds to $x-varepsilon=x-frac1omega$. Similarly, a tail of $-$ would be $x+varepsilon$.



            A short answer to your second question would be just to give you some examples, using $cdots$ to indicate a pattern of "length" $omega$. $+++cdots+=omega+1$, $+++cdots-=omega-1$, $+++cdots+++cdots=omega2$, $+++cdots---cdots=fracomega2$ $+--+-+-+-+-+cdots+=frac13+varepsilon$, $+----cdots+=2varepsilon$, $+----cdots-=varepsilon/2$, $+----cdots---cdots=varepsilon^2$.



            A thorough answer is given by Philip Ehrlich's "Conway names, the simplicity hierarchy and the surreal number tree", which is currently available at http://www.ohio.edu/people/ehrlich/ConwayNames.pdf






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              Your interpretation of finite sign expansions is slightly off. There are essentially two equivalent approaches to interpreting these finite expansions: One makes conversion between binary representations easy, and one hinting at the general structure even for domains greater than $omega$.



              The easy-conversion rule begins the way you started, but after the first instance of "$+-$" you can simply read off the remaining symbols as bits with $+$ being 1 and $-$ being zero, and then append a final $1$. Therefore, $+++−−+−−$ would be $2+left(.01001_2right)$. The other way to think about it is that the first three $+$ signs take you up to $3$, and then the first (and only the first, for a finite expansion) sign change starts a sequence of diminishing returns: $+++−−+−−=1+1+1-frac12-frac14+frac18-frac116-frac132=2+frac14+frac132$.



              In any case, your guess about expansions with domain $omega$ is essentially correct. Every domain-$omega$ expansion that doesn't end in a tail of $-$ or $+$ corresponds to the real number with the corresponding binary expansion. If it's all $+$, then it's $omega$. Among the positives, if it contains a $-$ but ends in a tail of $+$, then that would be a tail of $1$s in the binary expansion for some real number $x$, but the sign expansion actually corresponds to $x-varepsilon=x-frac1omega$. Similarly, a tail of $-$ would be $x+varepsilon$.



              A short answer to your second question would be just to give you some examples, using $cdots$ to indicate a pattern of "length" $omega$. $+++cdots+=omega+1$, $+++cdots-=omega-1$, $+++cdots+++cdots=omega2$, $+++cdots---cdots=fracomega2$ $+--+-+-+-+-+cdots+=frac13+varepsilon$, $+----cdots+=2varepsilon$, $+----cdots-=varepsilon/2$, $+----cdots---cdots=varepsilon^2$.



              A thorough answer is given by Philip Ehrlich's "Conway names, the simplicity hierarchy and the surreal number tree", which is currently available at http://www.ohio.edu/people/ehrlich/ConwayNames.pdf






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                Your interpretation of finite sign expansions is slightly off. There are essentially two equivalent approaches to interpreting these finite expansions: One makes conversion between binary representations easy, and one hinting at the general structure even for domains greater than $omega$.



                The easy-conversion rule begins the way you started, but after the first instance of "$+-$" you can simply read off the remaining symbols as bits with $+$ being 1 and $-$ being zero, and then append a final $1$. Therefore, $+++−−+−−$ would be $2+left(.01001_2right)$. The other way to think about it is that the first three $+$ signs take you up to $3$, and then the first (and only the first, for a finite expansion) sign change starts a sequence of diminishing returns: $+++−−+−−=1+1+1-frac12-frac14+frac18-frac116-frac132=2+frac14+frac132$.



                In any case, your guess about expansions with domain $omega$ is essentially correct. Every domain-$omega$ expansion that doesn't end in a tail of $-$ or $+$ corresponds to the real number with the corresponding binary expansion. If it's all $+$, then it's $omega$. Among the positives, if it contains a $-$ but ends in a tail of $+$, then that would be a tail of $1$s in the binary expansion for some real number $x$, but the sign expansion actually corresponds to $x-varepsilon=x-frac1omega$. Similarly, a tail of $-$ would be $x+varepsilon$.



                A short answer to your second question would be just to give you some examples, using $cdots$ to indicate a pattern of "length" $omega$. $+++cdots+=omega+1$, $+++cdots-=omega-1$, $+++cdots+++cdots=omega2$, $+++cdots---cdots=fracomega2$ $+--+-+-+-+-+cdots+=frac13+varepsilon$, $+----cdots+=2varepsilon$, $+----cdots-=varepsilon/2$, $+----cdots---cdots=varepsilon^2$.



                A thorough answer is given by Philip Ehrlich's "Conway names, the simplicity hierarchy and the surreal number tree", which is currently available at http://www.ohio.edu/people/ehrlich/ConwayNames.pdf






                share|cite|improve this answer









                $endgroup$



                Your interpretation of finite sign expansions is slightly off. There are essentially two equivalent approaches to interpreting these finite expansions: One makes conversion between binary representations easy, and one hinting at the general structure even for domains greater than $omega$.



                The easy-conversion rule begins the way you started, but after the first instance of "$+-$" you can simply read off the remaining symbols as bits with $+$ being 1 and $-$ being zero, and then append a final $1$. Therefore, $+++−−+−−$ would be $2+left(.01001_2right)$. The other way to think about it is that the first three $+$ signs take you up to $3$, and then the first (and only the first, for a finite expansion) sign change starts a sequence of diminishing returns: $+++−−+−−=1+1+1-frac12-frac14+frac18-frac116-frac132=2+frac14+frac132$.



                In any case, your guess about expansions with domain $omega$ is essentially correct. Every domain-$omega$ expansion that doesn't end in a tail of $-$ or $+$ corresponds to the real number with the corresponding binary expansion. If it's all $+$, then it's $omega$. Among the positives, if it contains a $-$ but ends in a tail of $+$, then that would be a tail of $1$s in the binary expansion for some real number $x$, but the sign expansion actually corresponds to $x-varepsilon=x-frac1omega$. Similarly, a tail of $-$ would be $x+varepsilon$.



                A short answer to your second question would be just to give you some examples, using $cdots$ to indicate a pattern of "length" $omega$. $+++cdots+=omega+1$, $+++cdots-=omega-1$, $+++cdots+++cdots=omega2$, $+++cdots---cdots=fracomega2$ $+--+-+-+-+-+cdots+=frac13+varepsilon$, $+----cdots+=2varepsilon$, $+----cdots-=varepsilon/2$, $+----cdots---cdots=varepsilon^2$.



                A thorough answer is given by Philip Ehrlich's "Conway names, the simplicity hierarchy and the surreal number tree", which is currently available at http://www.ohio.edu/people/ehrlich/ConwayNames.pdf







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jul 1 '15 at 15:50









                Mark S.Mark S.

                12.3k22772




                12.3k22772



























                    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%2f1237974%2fdecoding-the-sign-expansion-of-surreal-numbers%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

                    Triangular numbers and gcdProving sum of a set is $0 pmod n$ if $n$ is odd, or $fracn2 pmod n$ if $n$ is even?Is greatest common divisor of two numbers really their smallest linear combination?GCD, LCM RelationshipProve a set of nonnegative integers with greatest common divisor 1 and closed under addition has all but finite many nonnegative integers.all pairs of a and b in an equation containing gcdTriangular Numbers Modulo $k$ - Hit All Values?Understanding the Existence and Uniqueness of the GCDGCD and LCM with logical symbolsThe greatest common divisor of two positive integers less than 100 is equal to 3. Their least common multiple is twelve times one of the integers.Suppose that for all integers $x$, $x|a$ and $x|b$ if and only if $x|c$. Then $c = gcd(a,b)$Which is the gcd of 2 numbers which are multiplied and the result is 600000?

                    Ingelân Ynhâld Etymology | Geografy | Skiednis | Polityk en bestjoer | Ekonomy | Demografy | Kultuer | Klimaat | Sjoch ek | Keppelings om utens | Boarnen, noaten en referinsjes Navigaasjemenuwww.gov.ukOffisjele webside fan it regear fan it Feriene KeninkrykOffisjele webside fan it Britske FerkearsburoNederlânsktalige ynformaasje fan it Britske FerkearsburoOffisjele webside fan English Heritage, de organisaasje dy't him ynset foar it behâld fan it Ingelske kultuergoedYnwennertallen fan alle Britske stêden út 'e folkstelling fan 2011Notes en References, op dizze sideEngland

                    Հադիս Բովանդակություն Անվանում և նշանակություն | Դասակարգում | Աղբյուրներ | Նավարկման ցանկ