How many digits in base 2 do I need to represent any odd integer from 1 to $sqrt{N}$?












2












$begingroup$


$newcommand{floor}[1]{leftlfloor #1 rightrfloor}$An example here seems best. How many digits in base 2 do I need to represent any odd integer from $1$ to $sqrt{77}$, inclusive? It seems to be essentially half the digits required for $77$ --- in base 2. We can represent $77$ with $7$ digits in base 2. Half of that is $3.5$, so we need $4$ digits in base 2. So let's change "essentially half" to "exactly $floor{n/2} + 1$, where $n$ is the number of digits in base 2 held by $N$.



Since I'm interested in the odd integers (from $1$ to $sqrt{77}$), this means that the last digit must be $1$. But if $4$ is the number of bits, then I get the wrong list in base 2: $$0001_2 = 1_{10}, 0011_2 = 3_{10}, 0101_2 = 5_{10}, 0111_2 = 7_{10}, 1001_2 = 9_{10}, ...$$



That's wrong because $9 > sqrt{77}$. So the number must be less than $4$ digits in base 2. When I try $3$, I get the correct list



$$001_2 = 1_{10}, 011_2 = 3_{10}, 101_2 = 5_{10}, 111_2 = 7_{10},$$



but I haven't found an argument to convince myself I'm right. (Will this always work? Why?)






elementary-number-theory







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$endgroup$

















    2












    $begingroup$


    $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$An example here seems best. How many digits in base 2 do I need to represent any odd integer from $1$ to $sqrt{77}$, inclusive? It seems to be essentially half the digits required for $77$ --- in base 2. We can represent $77$ with $7$ digits in base 2. Half of that is $3.5$, so we need $4$ digits in base 2. So let's change "essentially half" to "exactly $floor{n/2} + 1$, where $n$ is the number of digits in base 2 held by $N$.



    Since I'm interested in the odd integers (from $1$ to $sqrt{77}$), this means that the last digit must be $1$. But if $4$ is the number of bits, then I get the wrong list in base 2: $$0001_2 = 1_{10}, 0011_2 = 3_{10}, 0101_2 = 5_{10}, 0111_2 = 7_{10}, 1001_2 = 9_{10}, ...$$



    That's wrong because $9 > sqrt{77}$. So the number must be less than $4$ digits in base 2. When I try $3$, I get the correct list



    $$001_2 = 1_{10}, 011_2 = 3_{10}, 101_2 = 5_{10}, 111_2 = 7_{10},$$



    but I haven't found an argument to convince myself I'm right. (Will this always work? Why?)






    elementary-number-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$An example here seems best. How many digits in base 2 do I need to represent any odd integer from $1$ to $sqrt{77}$, inclusive? It seems to be essentially half the digits required for $77$ --- in base 2. We can represent $77$ with $7$ digits in base 2. Half of that is $3.5$, so we need $4$ digits in base 2. So let's change "essentially half" to "exactly $floor{n/2} + 1$, where $n$ is the number of digits in base 2 held by $N$.



      Since I'm interested in the odd integers (from $1$ to $sqrt{77}$), this means that the last digit must be $1$. But if $4$ is the number of bits, then I get the wrong list in base 2: $$0001_2 = 1_{10}, 0011_2 = 3_{10}, 0101_2 = 5_{10}, 0111_2 = 7_{10}, 1001_2 = 9_{10}, ...$$



      That's wrong because $9 > sqrt{77}$. So the number must be less than $4$ digits in base 2. When I try $3$, I get the correct list



      $$001_2 = 1_{10}, 011_2 = 3_{10}, 101_2 = 5_{10}, 111_2 = 7_{10},$$



      but I haven't found an argument to convince myself I'm right. (Will this always work? Why?)






      elementary-number-theory







      share|cite|improve this question









      $endgroup$




      $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$An example here seems best. How many digits in base 2 do I need to represent any odd integer from $1$ to $sqrt{77}$, inclusive? It seems to be essentially half the digits required for $77$ --- in base 2. We can represent $77$ with $7$ digits in base 2. Half of that is $3.5$, so we need $4$ digits in base 2. So let's change "essentially half" to "exactly $floor{n/2} + 1$, where $n$ is the number of digits in base 2 held by $N$.



      Since I'm interested in the odd integers (from $1$ to $sqrt{77}$), this means that the last digit must be $1$. But if $4$ is the number of bits, then I get the wrong list in base 2: $$0001_2 = 1_{10}, 0011_2 = 3_{10}, 0101_2 = 5_{10}, 0111_2 = 7_{10}, 1001_2 = 9_{10}, ...$$



      That's wrong because $9 > sqrt{77}$. So the number must be less than $4$ digits in base 2. When I try $3$, I get the correct list



      $$001_2 = 1_{10}, 011_2 = 3_{10}, 101_2 = 5_{10}, 111_2 = 7_{10},$$



      but I haven't found an argument to convince myself I'm right. (Will this always work? Why?)






      elementary-number-theory




      elementary-number-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 3 '18 at 12:54









      Luitpold AmbreLuitpold Ambre

      135




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          $begingroup$

          The number of digits of a number $N$ in base $b$ is $log_b N+1$. So the number of digits of $sqrt{N}$ in base $2$ is



          $$log_2 sqrt{N}+1 = frac{1}{2}log_2 N +1.$$



          The $frac{1}{2}$ shows why it takes half as many digits for $sqrt{N}$ as $N$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
            $endgroup$
            – Luitpold Ambre
            Dec 3 '18 at 13:34












          • $begingroup$
            You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
            $endgroup$
            – B. Goddard
            Dec 3 '18 at 16:52













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          $begingroup$

          The number of digits of a number $N$ in base $b$ is $log_b N+1$. So the number of digits of $sqrt{N}$ in base $2$ is



          $$log_2 sqrt{N}+1 = frac{1}{2}log_2 N +1.$$



          The $frac{1}{2}$ shows why it takes half as many digits for $sqrt{N}$ as $N$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
            $endgroup$
            – Luitpold Ambre
            Dec 3 '18 at 13:34












          • $begingroup$
            You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
            $endgroup$
            – B. Goddard
            Dec 3 '18 at 16:52


















          1












          $begingroup$

          The number of digits of a number $N$ in base $b$ is $log_b N+1$. So the number of digits of $sqrt{N}$ in base $2$ is



          $$log_2 sqrt{N}+1 = frac{1}{2}log_2 N +1.$$



          The $frac{1}{2}$ shows why it takes half as many digits for $sqrt{N}$ as $N$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
            $endgroup$
            – Luitpold Ambre
            Dec 3 '18 at 13:34












          • $begingroup$
            You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
            $endgroup$
            – B. Goddard
            Dec 3 '18 at 16:52
















          1












          1








          1





          $begingroup$

          The number of digits of a number $N$ in base $b$ is $log_b N+1$. So the number of digits of $sqrt{N}$ in base $2$ is



          $$log_2 sqrt{N}+1 = frac{1}{2}log_2 N +1.$$



          The $frac{1}{2}$ shows why it takes half as many digits for $sqrt{N}$ as $N$.






          share|cite|improve this answer









          $endgroup$



          The number of digits of a number $N$ in base $b$ is $log_b N+1$. So the number of digits of $sqrt{N}$ in base $2$ is



          $$log_2 sqrt{N}+1 = frac{1}{2}log_2 N +1.$$



          The $frac{1}{2}$ shows why it takes half as many digits for $sqrt{N}$ as $N$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 13:16









          B. GoddardB. Goddard

          18.6k21440




          18.6k21440












          • $begingroup$
            $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
            $endgroup$
            – Luitpold Ambre
            Dec 3 '18 at 13:34












          • $begingroup$
            You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
            $endgroup$
            – B. Goddard
            Dec 3 '18 at 16:52




















          • $begingroup$
            $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
            $endgroup$
            – Luitpold Ambre
            Dec 3 '18 at 13:34












          • $begingroup$
            You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
            $endgroup$
            – B. Goddard
            Dec 3 '18 at 16:52


















          $begingroup$
          $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
          $endgroup$
          – Luitpold Ambre
          Dec 3 '18 at 13:34






          $begingroup$
          $newcommand{floor}[1]{leftlfloor #1 rightrfloor}$I think you mean $floor{log_b N} + 1$. This is only part of the answer. I'm not interested in all integers from $0$ to $sqrt{N}$, but only the odd integers. In the example I gave, we could take one bit less than $frac{1}{2}floor{log_b sqrt{N}} + 1$. If we use this exact quantity, we go up to $15$ instead of $7$. I haven't understood why this happens.
          $endgroup$
          – Luitpold Ambre
          Dec 3 '18 at 13:34














          $begingroup$
          You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
          $endgroup$
          – B. Goddard
          Dec 3 '18 at 16:52






          $begingroup$
          You're just rounding funny. If the log comes out to be an integer, then the number is a power of 2 and hence even, so you don't have to worry about integer logs. So you might as well round up and use $lceil log_2 N rceil,$ And to get rid of your little endpoint problem, note that you might as well round the square-root down. So how about $lceil log_2 lfloor sqrt{N} rfloor rceil.$
          $endgroup$
          – B. Goddard
          Dec 3 '18 at 16:52




















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