Decimal expansions of rational numbers.











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One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).



Using the same method I have run into a couple of problems.




  1. $x = frac{1}{6}$


In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$



so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.





  1. $x = frac{1}{37}$.


In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.



Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?










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    up vote
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    down vote

    favorite












    One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).



    Using the same method I have run into a couple of problems.




    1. $x = frac{1}{6}$


    In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$



    so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.





    1. $x = frac{1}{37}$.


    In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.



    Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).



      Using the same method I have run into a couple of problems.




      1. $x = frac{1}{6}$


      In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$



      so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.





      1. $x = frac{1}{37}$.


      In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.



      Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?










      share|cite|improve this question













      One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).



      Using the same method I have run into a couple of problems.




      1. $x = frac{1}{6}$


      In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$



      so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.





      1. $x = frac{1}{37}$.


      In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.



      Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?







      modular-arithmetic rational-numbers decimal-expansion






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      asked Nov 3 at 17:10









      xAly

      192




      192






















          1 Answer
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          accepted










          For $dfrac {1}{6} dots$



          $begin {align}
          1operatorname {mod} 6 equiv 1 \
          10operatorname {mod} 6 equiv 4 \
          40operatorname {mod} 6 equiv 4 \
          end {align}$



          $4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.



          For $dfrac {1}{37} dots$



          $begin {align}
          1operatorname {mod} 37 equiv 1 \
          10operatorname {mod} 37 equiv 10 \
          100operatorname {mod} 37 equiv 26 \
          260operatorname {mod} 37 equiv 1 \
          end {align}$



          $1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.



          It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.






          share|cite|improve this answer























          • But then what about the example. 1 is subtracted, not 3.
            – xAly
            Nov 3 at 19:01










          • Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
            – bjcolby15
            Nov 3 at 20:20













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          1 Answer
          1






          active

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          active

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          up vote
          1
          down vote



          accepted










          For $dfrac {1}{6} dots$



          $begin {align}
          1operatorname {mod} 6 equiv 1 \
          10operatorname {mod} 6 equiv 4 \
          40operatorname {mod} 6 equiv 4 \
          end {align}$



          $4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.



          For $dfrac {1}{37} dots$



          $begin {align}
          1operatorname {mod} 37 equiv 1 \
          10operatorname {mod} 37 equiv 10 \
          100operatorname {mod} 37 equiv 26 \
          260operatorname {mod} 37 equiv 1 \
          end {align}$



          $1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.



          It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.






          share|cite|improve this answer























          • But then what about the example. 1 is subtracted, not 3.
            – xAly
            Nov 3 at 19:01










          • Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
            – bjcolby15
            Nov 3 at 20:20

















          up vote
          1
          down vote



          accepted










          For $dfrac {1}{6} dots$



          $begin {align}
          1operatorname {mod} 6 equiv 1 \
          10operatorname {mod} 6 equiv 4 \
          40operatorname {mod} 6 equiv 4 \
          end {align}$



          $4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.



          For $dfrac {1}{37} dots$



          $begin {align}
          1operatorname {mod} 37 equiv 1 \
          10operatorname {mod} 37 equiv 10 \
          100operatorname {mod} 37 equiv 26 \
          260operatorname {mod} 37 equiv 1 \
          end {align}$



          $1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.



          It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.






          share|cite|improve this answer























          • But then what about the example. 1 is subtracted, not 3.
            – xAly
            Nov 3 at 19:01










          • Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
            – bjcolby15
            Nov 3 at 20:20















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          For $dfrac {1}{6} dots$



          $begin {align}
          1operatorname {mod} 6 equiv 1 \
          10operatorname {mod} 6 equiv 4 \
          40operatorname {mod} 6 equiv 4 \
          end {align}$



          $4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.



          For $dfrac {1}{37} dots$



          $begin {align}
          1operatorname {mod} 37 equiv 1 \
          10operatorname {mod} 37 equiv 10 \
          100operatorname {mod} 37 equiv 26 \
          260operatorname {mod} 37 equiv 1 \
          end {align}$



          $1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.



          It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.






          share|cite|improve this answer














          For $dfrac {1}{6} dots$



          $begin {align}
          1operatorname {mod} 6 equiv 1 \
          10operatorname {mod} 6 equiv 4 \
          40operatorname {mod} 6 equiv 4 \
          end {align}$



          $4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.



          For $dfrac {1}{37} dots$



          $begin {align}
          1operatorname {mod} 37 equiv 1 \
          10operatorname {mod} 37 equiv 10 \
          100operatorname {mod} 37 equiv 26 \
          260operatorname {mod} 37 equiv 1 \
          end {align}$



          $1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.



          It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 13 at 1:10

























          answered Nov 3 at 18:07









          bjcolby15

          1,0961916




          1,0961916












          • But then what about the example. 1 is subtracted, not 3.
            – xAly
            Nov 3 at 19:01










          • Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
            – bjcolby15
            Nov 3 at 20:20




















          • But then what about the example. 1 is subtracted, not 3.
            – xAly
            Nov 3 at 19:01










          • Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
            – bjcolby15
            Nov 3 at 20:20


















          But then what about the example. 1 is subtracted, not 3.
          – xAly
          Nov 3 at 19:01




          But then what about the example. 1 is subtracted, not 3.
          – xAly
          Nov 3 at 19:01












          Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
          – bjcolby15
          Nov 3 at 20:20






          Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
          – bjcolby15
          Nov 3 at 20:20




















           

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