A weird number system (I made it myself)












6












$begingroup$


So, I'm an eccentric mathematician. You can see this with my number system. I'm pulling a sneaky on you, and you'll have to tell me what everything means. I'm not telling you a thing: figure it for yourself.



Here are some things to know about the system:



$1+1=1-1=1*1=1$



$1/1$ is $undefined$



$2+1=2$



$2+2=3$



$3/2$ would be $2$ in decimal



$3+3=4$



$4+3$ would be $6$ in decimal



$4*3=5$



$5-2$ would be $7$ in decimal



$6=5/3$



Every number after the radix point is normal decimal, AKA the part after the decimal point is in Base-10.



If it's impossible, I've been wrong all this time. If there are multiple solutions, I'm sticking to being weird as always. If it's normal, well, kudos to figuring it out. Now bugger off.



Bonus question: Represent 10 in this system. This isn't required but I'm weird with my judgement too, so you might want to gain more of my respect.






mathematics logical-deduction







share|improve this question









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$endgroup$

















    6












    $begingroup$


    So, I'm an eccentric mathematician. You can see this with my number system. I'm pulling a sneaky on you, and you'll have to tell me what everything means. I'm not telling you a thing: figure it for yourself.



    Here are some things to know about the system:



    $1+1=1-1=1*1=1$



    $1/1$ is $undefined$



    $2+1=2$



    $2+2=3$



    $3/2$ would be $2$ in decimal



    $3+3=4$



    $4+3$ would be $6$ in decimal



    $4*3=5$



    $5-2$ would be $7$ in decimal



    $6=5/3$



    Every number after the radix point is normal decimal, AKA the part after the decimal point is in Base-10.



    If it's impossible, I've been wrong all this time. If there are multiple solutions, I'm sticking to being weird as always. If it's normal, well, kudos to figuring it out. Now bugger off.



    Bonus question: Represent 10 in this system. This isn't required but I'm weird with my judgement too, so you might want to gain more of my respect.






    mathematics logical-deduction







    share|improve this question









    New contributor




    Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      6












      6








      6





      $begingroup$


      So, I'm an eccentric mathematician. You can see this with my number system. I'm pulling a sneaky on you, and you'll have to tell me what everything means. I'm not telling you a thing: figure it for yourself.



      Here are some things to know about the system:



      $1+1=1-1=1*1=1$



      $1/1$ is $undefined$



      $2+1=2$



      $2+2=3$



      $3/2$ would be $2$ in decimal



      $3+3=4$



      $4+3$ would be $6$ in decimal



      $4*3=5$



      $5-2$ would be $7$ in decimal



      $6=5/3$



      Every number after the radix point is normal decimal, AKA the part after the decimal point is in Base-10.



      If it's impossible, I've been wrong all this time. If there are multiple solutions, I'm sticking to being weird as always. If it's normal, well, kudos to figuring it out. Now bugger off.



      Bonus question: Represent 10 in this system. This isn't required but I'm weird with my judgement too, so you might want to gain more of my respect.






      mathematics logical-deduction







      share|improve this question









      New contributor




      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      So, I'm an eccentric mathematician. You can see this with my number system. I'm pulling a sneaky on you, and you'll have to tell me what everything means. I'm not telling you a thing: figure it for yourself.



      Here are some things to know about the system:



      $1+1=1-1=1*1=1$



      $1/1$ is $undefined$



      $2+1=2$



      $2+2=3$



      $3/2$ would be $2$ in decimal



      $3+3=4$



      $4+3$ would be $6$ in decimal



      $4*3=5$



      $5-2$ would be $7$ in decimal



      $6=5/3$



      Every number after the radix point is normal decimal, AKA the part after the decimal point is in Base-10.



      If it's impossible, I've been wrong all this time. If there are multiple solutions, I'm sticking to being weird as always. If it's normal, well, kudos to figuring it out. Now bugger off.



      Bonus question: Represent 10 in this system. This isn't required but I'm weird with my judgement too, so you might want to gain more of my respect.






      mathematics logical-deduction




      mathematics logical-deduction






      share|improve this question









      New contributor




      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 2 hours ago









      Embodiment of Ignorance

      1216




      1216






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      Check out our Code of Conduct.









      asked 6 hours ago









      AndrewAndrew

      311




      311




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      New contributor





      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
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          $begingroup$

          I imagine I'm missing something and am at least somewhat cheating, but the following seems to be sufficient...



          I'll use the following notation:




          $Nd$ is the number $N$ in decimal
          $Nw$ is the number $N$ in the "weird number system"

          E.g. from the clues, $3w/2w = 2d$




          More importantly, we have:




          $Nb$ is the number $N$ in binary.




          Then, we can define:




          $0w$ is undefined in decimal
          $1w..Nw$ are represented by an "oscillating" binary number:
          $1w = 0000b = 0d$
          $2w = 0001b = 1d$
          $3w = 0010b = 2d$
          $4w = 0100b = 4d$
          $5w = 1000b = 8d$
          $6w = 0100b = 4d$
          $7w = 0010b = 2d$

          ...etc...




          $1+1=1-1=1*1=1$




          $... = 1w = 0d = 0d+0d = 0d-0d = 0d*0d$




          $1/1$ is undefined




          $1w/1w = 0d/0d$ is undefined




          $2+1=2$




          $2w+1w = 1d+0d = 1d = 2w$




          $2+2=3$




          $2w+2w = 1d+1d = 2d = 3w$




          $3/2$ would be $2$ in decimal




          $3w/2w = 2d/1d = 2d$




          $3+3=4$




          $3w+3w = 2d+2d = 4d = 4w$




          $4+3$ would be $6$ in decimal




          $4w+3w = 4d+2d = 6d$




          $4*3=5$




          $4w*3w = 4d*2d = 8d = 5w




          $5-2$ would be $7$ in decimal




          $5w-2w = 8d-1d = 7d$




          $6=5/3$




          Since $5w$ is effectively the weird system's largest number, it needs to be defined in terms of the other numbers, not the other way around:
          $6w = 5w/3w = 8d/2d = 4d = 4w$ (which is equivalent, i.e. $4w=6w=0100b=4d$)




          So, for the bonus question (represent $10w$):




          $10w = 8w = 2w$

          or, in binary/decimal
          $10w = 0001b = 1d$







          share|improve this answer









          $endgroup$













          • $begingroup$
            I think he meant 10d in weird.
            $endgroup$
            – Jasen
            58 mins ago













          Your Answer





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          2












          $begingroup$

          I imagine I'm missing something and am at least somewhat cheating, but the following seems to be sufficient...



          I'll use the following notation:




          $Nd$ is the number $N$ in decimal
          $Nw$ is the number $N$ in the "weird number system"

          E.g. from the clues, $3w/2w = 2d$




          More importantly, we have:




          $Nb$ is the number $N$ in binary.




          Then, we can define:




          $0w$ is undefined in decimal
          $1w..Nw$ are represented by an "oscillating" binary number:
          $1w = 0000b = 0d$
          $2w = 0001b = 1d$
          $3w = 0010b = 2d$
          $4w = 0100b = 4d$
          $5w = 1000b = 8d$
          $6w = 0100b = 4d$
          $7w = 0010b = 2d$

          ...etc...




          $1+1=1-1=1*1=1$




          $... = 1w = 0d = 0d+0d = 0d-0d = 0d*0d$




          $1/1$ is undefined




          $1w/1w = 0d/0d$ is undefined




          $2+1=2$




          $2w+1w = 1d+0d = 1d = 2w$




          $2+2=3$




          $2w+2w = 1d+1d = 2d = 3w$




          $3/2$ would be $2$ in decimal




          $3w/2w = 2d/1d = 2d$




          $3+3=4$




          $3w+3w = 2d+2d = 4d = 4w$




          $4+3$ would be $6$ in decimal




          $4w+3w = 4d+2d = 6d$




          $4*3=5$




          $4w*3w = 4d*2d = 8d = 5w




          $5-2$ would be $7$ in decimal




          $5w-2w = 8d-1d = 7d$




          $6=5/3$




          Since $5w$ is effectively the weird system's largest number, it needs to be defined in terms of the other numbers, not the other way around:
          $6w = 5w/3w = 8d/2d = 4d = 4w$ (which is equivalent, i.e. $4w=6w=0100b=4d$)




          So, for the bonus question (represent $10w$):




          $10w = 8w = 2w$

          or, in binary/decimal
          $10w = 0001b = 1d$







          share|improve this answer









          $endgroup$













          • $begingroup$
            I think he meant 10d in weird.
            $endgroup$
            – Jasen
            58 mins ago


















          2












          $begingroup$

          I imagine I'm missing something and am at least somewhat cheating, but the following seems to be sufficient...



          I'll use the following notation:




          $Nd$ is the number $N$ in decimal
          $Nw$ is the number $N$ in the "weird number system"

          E.g. from the clues, $3w/2w = 2d$




          More importantly, we have:




          $Nb$ is the number $N$ in binary.




          Then, we can define:




          $0w$ is undefined in decimal
          $1w..Nw$ are represented by an "oscillating" binary number:
          $1w = 0000b = 0d$
          $2w = 0001b = 1d$
          $3w = 0010b = 2d$
          $4w = 0100b = 4d$
          $5w = 1000b = 8d$
          $6w = 0100b = 4d$
          $7w = 0010b = 2d$

          ...etc...




          $1+1=1-1=1*1=1$




          $... = 1w = 0d = 0d+0d = 0d-0d = 0d*0d$




          $1/1$ is undefined




          $1w/1w = 0d/0d$ is undefined




          $2+1=2$




          $2w+1w = 1d+0d = 1d = 2w$




          $2+2=3$




          $2w+2w = 1d+1d = 2d = 3w$




          $3/2$ would be $2$ in decimal




          $3w/2w = 2d/1d = 2d$




          $3+3=4$




          $3w+3w = 2d+2d = 4d = 4w$




          $4+3$ would be $6$ in decimal




          $4w+3w = 4d+2d = 6d$




          $4*3=5$




          $4w*3w = 4d*2d = 8d = 5w




          $5-2$ would be $7$ in decimal




          $5w-2w = 8d-1d = 7d$




          $6=5/3$




          Since $5w$ is effectively the weird system's largest number, it needs to be defined in terms of the other numbers, not the other way around:
          $6w = 5w/3w = 8d/2d = 4d = 4w$ (which is equivalent, i.e. $4w=6w=0100b=4d$)




          So, for the bonus question (represent $10w$):




          $10w = 8w = 2w$

          or, in binary/decimal
          $10w = 0001b = 1d$







          share|improve this answer









          $endgroup$













          • $begingroup$
            I think he meant 10d in weird.
            $endgroup$
            – Jasen
            58 mins ago
















          2












          2








          2





          $begingroup$

          I imagine I'm missing something and am at least somewhat cheating, but the following seems to be sufficient...



          I'll use the following notation:




          $Nd$ is the number $N$ in decimal
          $Nw$ is the number $N$ in the "weird number system"

          E.g. from the clues, $3w/2w = 2d$




          More importantly, we have:




          $Nb$ is the number $N$ in binary.




          Then, we can define:




          $0w$ is undefined in decimal
          $1w..Nw$ are represented by an "oscillating" binary number:
          $1w = 0000b = 0d$
          $2w = 0001b = 1d$
          $3w = 0010b = 2d$
          $4w = 0100b = 4d$
          $5w = 1000b = 8d$
          $6w = 0100b = 4d$
          $7w = 0010b = 2d$

          ...etc...




          $1+1=1-1=1*1=1$




          $... = 1w = 0d = 0d+0d = 0d-0d = 0d*0d$




          $1/1$ is undefined




          $1w/1w = 0d/0d$ is undefined




          $2+1=2$




          $2w+1w = 1d+0d = 1d = 2w$




          $2+2=3$




          $2w+2w = 1d+1d = 2d = 3w$




          $3/2$ would be $2$ in decimal




          $3w/2w = 2d/1d = 2d$




          $3+3=4$




          $3w+3w = 2d+2d = 4d = 4w$




          $4+3$ would be $6$ in decimal




          $4w+3w = 4d+2d = 6d$




          $4*3=5$




          $4w*3w = 4d*2d = 8d = 5w




          $5-2$ would be $7$ in decimal




          $5w-2w = 8d-1d = 7d$




          $6=5/3$




          Since $5w$ is effectively the weird system's largest number, it needs to be defined in terms of the other numbers, not the other way around:
          $6w = 5w/3w = 8d/2d = 4d = 4w$ (which is equivalent, i.e. $4w=6w=0100b=4d$)




          So, for the bonus question (represent $10w$):




          $10w = 8w = 2w$

          or, in binary/decimal
          $10w = 0001b = 1d$







          share|improve this answer









          $endgroup$



          I imagine I'm missing something and am at least somewhat cheating, but the following seems to be sufficient...



          I'll use the following notation:




          $Nd$ is the number $N$ in decimal
          $Nw$ is the number $N$ in the "weird number system"

          E.g. from the clues, $3w/2w = 2d$




          More importantly, we have:




          $Nb$ is the number $N$ in binary.




          Then, we can define:




          $0w$ is undefined in decimal
          $1w..Nw$ are represented by an "oscillating" binary number:
          $1w = 0000b = 0d$
          $2w = 0001b = 1d$
          $3w = 0010b = 2d$
          $4w = 0100b = 4d$
          $5w = 1000b = 8d$
          $6w = 0100b = 4d$
          $7w = 0010b = 2d$

          ...etc...




          $1+1=1-1=1*1=1$




          $... = 1w = 0d = 0d+0d = 0d-0d = 0d*0d$




          $1/1$ is undefined




          $1w/1w = 0d/0d$ is undefined




          $2+1=2$




          $2w+1w = 1d+0d = 1d = 2w$




          $2+2=3$




          $2w+2w = 1d+1d = 2d = 3w$




          $3/2$ would be $2$ in decimal




          $3w/2w = 2d/1d = 2d$




          $3+3=4$




          $3w+3w = 2d+2d = 4d = 4w$




          $4+3$ would be $6$ in decimal




          $4w+3w = 4d+2d = 6d$




          $4*3=5$




          $4w*3w = 4d*2d = 8d = 5w




          $5-2$ would be $7$ in decimal




          $5w-2w = 8d-1d = 7d$




          $6=5/3$




          Since $5w$ is effectively the weird system's largest number, it needs to be defined in terms of the other numbers, not the other way around:
          $6w = 5w/3w = 8d/2d = 4d = 4w$ (which is equivalent, i.e. $4w=6w=0100b=4d$)




          So, for the bonus question (represent $10w$):




          $10w = 8w = 2w$

          or, in binary/decimal
          $10w = 0001b = 1d$








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          AlconjaAlconja

          22k1293143




          22k1293143












          • $begingroup$
            I think he meant 10d in weird.
            $endgroup$
            – Jasen
            58 mins ago




















          • $begingroup$
            I think he meant 10d in weird.
            $endgroup$
            – Jasen
            58 mins ago


















          $begingroup$
          I think he meant 10d in weird.
          $endgroup$
          – Jasen
          58 mins ago






          $begingroup$
          I think he meant 10d in weird.
          $endgroup$
          – Jasen
          58 mins ago












          Andrew is a new contributor. Be nice, and check out our Code of Conduct.










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