Understanding fractions in modular arithmeticRing of fractions of $mathbb Z_6$Interesting problem on “neighbor fractions”Proof on linear congruence solutionsAddition on an Elliptic Curve and Modular Arithmetic involving fractionsModular fractions: $5 big| 3- frac 12$Prove that there are no solutions to the modular equation 9x + 10 = 6y - 1 (mod 15)Is this a valid proof? Modular arithmetic with fractionsdivision in modular arithmeticCan you make sense of roots of unity in modular arithmetic?Is there a name for this exponential analog to modular arithmetic? (octave equivalency)Modular inverse of a fraction vs just the numerator

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Understanding fractions in modular arithmetic


Ring of fractions of $mathbb Z_6$Interesting problem on “neighbor fractions”Proof on linear congruence solutionsAddition on an Elliptic Curve and Modular Arithmetic involving fractionsModular fractions: $5 big| 3- frac 12$Prove that there are no solutions to the modular equation 9x + 10 = 6y - 1 (mod 15)Is this a valid proof? Modular arithmetic with fractionsdivision in modular arithmeticCan you make sense of roots of unity in modular arithmetic?Is there a name for this exponential analog to modular arithmetic? (octave equivalency)Modular inverse of a fraction vs just the numerator













3












$begingroup$


I would like confirmation that my understanding of a congruence of two fractions is correct.



Suppose we have



$$frac23 equiv frac46mod 5$$



What happened was multiplying both the numerator and denominator of the left fraction by $2$.



So $frac23$ is essentially the same as $3xequiv 2 mod 5$ and we can multiply both sides by $2$ since that is valid in modular arithmetic we get $6x equiv 4mod 5$ and rewriting that as a fraction we get $frac46$.



A fraction really represents that $x$ and since that $x$ remains unchanged via such manipulations then those fractions are essentially congruent.



Of course I am aware this is true only for valid fractions, i.e. where the denominator is invertible mod $5$.



Is my understanding of this correct?










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    I would like confirmation that my understanding of a congruence of two fractions is correct.



    Suppose we have



    $$frac23 equiv frac46mod 5$$



    What happened was multiplying both the numerator and denominator of the left fraction by $2$.



    So $frac23$ is essentially the same as $3xequiv 2 mod 5$ and we can multiply both sides by $2$ since that is valid in modular arithmetic we get $6x equiv 4mod 5$ and rewriting that as a fraction we get $frac46$.



    A fraction really represents that $x$ and since that $x$ remains unchanged via such manipulations then those fractions are essentially congruent.



    Of course I am aware this is true only for valid fractions, i.e. where the denominator is invertible mod $5$.



    Is my understanding of this correct?










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I would like confirmation that my understanding of a congruence of two fractions is correct.



      Suppose we have



      $$frac23 equiv frac46mod 5$$



      What happened was multiplying both the numerator and denominator of the left fraction by $2$.



      So $frac23$ is essentially the same as $3xequiv 2 mod 5$ and we can multiply both sides by $2$ since that is valid in modular arithmetic we get $6x equiv 4mod 5$ and rewriting that as a fraction we get $frac46$.



      A fraction really represents that $x$ and since that $x$ remains unchanged via such manipulations then those fractions are essentially congruent.



      Of course I am aware this is true only for valid fractions, i.e. where the denominator is invertible mod $5$.



      Is my understanding of this correct?










      share|cite|improve this question











      $endgroup$




      I would like confirmation that my understanding of a congruence of two fractions is correct.



      Suppose we have



      $$frac23 equiv frac46mod 5$$



      What happened was multiplying both the numerator and denominator of the left fraction by $2$.



      So $frac23$ is essentially the same as $3xequiv 2 mod 5$ and we can multiply both sides by $2$ since that is valid in modular arithmetic we get $6x equiv 4mod 5$ and rewriting that as a fraction we get $frac46$.



      A fraction really represents that $x$ and since that $x$ remains unchanged via such manipulations then those fractions are essentially congruent.



      Of course I am aware this is true only for valid fractions, i.e. where the denominator is invertible mod $5$.



      Is my understanding of this correct?







      elementary-number-theory modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 7 at 9:10









      mrtaurho

      6,07771641




      6,07771641










      asked Mar 7 at 9:06









      Michael MuntaMichael Munta

      99111




      99111




















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$

          Fractions work fine in modular arithmetic. They're not very commonly used, though.



          If you want to use fractions, you have to stop thinking about $frac23$ as $0.666ldots$ and instead think of it as $4$ (at least in modulo $5$). Also, there is no simple way to just look at a modular fraction and know more or less what its value is, the way you can just know that $frac1234561000001$ is very close to $0.123456$. In other words, small change in the denominator gives large changes in fraction value in modular arithmetic.



          So there are a few mental roadblocks on the way to using fractions in modular arithmetic, and I suspect that that's why it isn't done more. That being said, if you do use fractions, the same rules that you learned in elementary school for regular (rational) fractions still all apply: How to expand / simplify fractions, how to add or multiply them, or how to check that they are equal. The technique known as "cross multiplication" still works: $frac abequiv frac cdiff ad = bc$ (this does require that only invertible elements are allowed as denominators, though).



          You can allow non-invertible elements as denominators, but then more care is needed. See, for instance, my answer here on allowing fractions with odd denominators (including $3$) in modulo $6$.



          If you only allow invertible elements as denominators, then because they're already invertible, you're not really changing anything, as all fractions are elements that already exist as whole numbers. If you allow non-invertible elements, you basically get a smaller ring as a result (modulo $6$ with $3$ allowed as a denominator gives you what is essentially modulo $2$). Sadly, you can't get something "bigger" (the way $Bbb Q$ is bigger than $Bbb Z$), beause that requires you to have denominators which are neither invertible nor zero-divisors.



          Edit: Since I apparently missed the real point of the question. Yes, those manipulations are fine. Whatever $frac23$ is, give it the name $x$. By the definition of fractions, $3x = 2$. Multiplying both sides by $2$ can't make them not-equal, so $6x = 4$. Since $6$ is a valid denominator, by definition of fractions we get $x = frac46$. What you have ultimately done is to expand the fraction $frac23$ by $2$ and shown that this doesn't change the value of the fraction.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
            $endgroup$
            – Michael Munta
            Mar 7 at 9:41






          • 1




            $begingroup$
            @MichaelMunta You're right. I added a bit at the end.
            $endgroup$
            – Arthur
            Mar 7 at 9:51


















          1












          $begingroup$

          Yes. If $a,b,c,d$ are all co-prime to $n$ and $ad=bc mod n$ then we know that $bd$ is co-prime to $n$ so it is invertible in $mathbbZ_n$, and so we can divide by $bd$ to get



          $fracab=fraccd mod n$






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            Yes, your understanding is correct, There are a few reasons that they typically aren't used though:



            • They aren't always defined ex. $frac52 bmod 6$

            • Because you can give a second name mod anything, you can get multiple answers when not coprime to the modulus ex. $frac33equivfrac3-3bmod 6 $ at least in the sense of $3equiv -3 bmod 6$ but These give back two answers, The first is $1 bmod 6$ if taken literally, the second is $-1equiv 5 bmod 6$, You can also get 3 mod 6, because $frac6x+33=2x+1$ which lands on all remainders corresponding to odd numbers mod 6.

            So using them recklessly, comes with some perils.






            share|cite|improve this answer









            $endgroup$




















              -1












              $begingroup$

              Yes, and just like you can do modular arithmetic to simplify ordinary arithmetic using integers, one can also do this for fractions. This looks impossible because of the multiple ways one can write a fraction modulo an integer representing a given number. But if it is known that the answer one is interested in, has a numerator between $0$ and $N$ and a denominator between $0$ and $D$ then a result modulo a number larger than $2 N D$ will allow one to reconstruct this rational number. As explained here, finding this fraction involves just the Euclidean algorithm.






              share|cite|improve this answer











              $endgroup$












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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4












                $begingroup$

                Fractions work fine in modular arithmetic. They're not very commonly used, though.



                If you want to use fractions, you have to stop thinking about $frac23$ as $0.666ldots$ and instead think of it as $4$ (at least in modulo $5$). Also, there is no simple way to just look at a modular fraction and know more or less what its value is, the way you can just know that $frac1234561000001$ is very close to $0.123456$. In other words, small change in the denominator gives large changes in fraction value in modular arithmetic.



                So there are a few mental roadblocks on the way to using fractions in modular arithmetic, and I suspect that that's why it isn't done more. That being said, if you do use fractions, the same rules that you learned in elementary school for regular (rational) fractions still all apply: How to expand / simplify fractions, how to add or multiply them, or how to check that they are equal. The technique known as "cross multiplication" still works: $frac abequiv frac cdiff ad = bc$ (this does require that only invertible elements are allowed as denominators, though).



                You can allow non-invertible elements as denominators, but then more care is needed. See, for instance, my answer here on allowing fractions with odd denominators (including $3$) in modulo $6$.



                If you only allow invertible elements as denominators, then because they're already invertible, you're not really changing anything, as all fractions are elements that already exist as whole numbers. If you allow non-invertible elements, you basically get a smaller ring as a result (modulo $6$ with $3$ allowed as a denominator gives you what is essentially modulo $2$). Sadly, you can't get something "bigger" (the way $Bbb Q$ is bigger than $Bbb Z$), beause that requires you to have denominators which are neither invertible nor zero-divisors.



                Edit: Since I apparently missed the real point of the question. Yes, those manipulations are fine. Whatever $frac23$ is, give it the name $x$. By the definition of fractions, $3x = 2$. Multiplying both sides by $2$ can't make them not-equal, so $6x = 4$. Since $6$ is a valid denominator, by definition of fractions we get $x = frac46$. What you have ultimately done is to expand the fraction $frac23$ by $2$ and shown that this doesn't change the value of the fraction.






                share|cite|improve this answer











                $endgroup$












                • $begingroup$
                  I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
                  $endgroup$
                  – Michael Munta
                  Mar 7 at 9:41






                • 1




                  $begingroup$
                  @MichaelMunta You're right. I added a bit at the end.
                  $endgroup$
                  – Arthur
                  Mar 7 at 9:51















                4












                $begingroup$

                Fractions work fine in modular arithmetic. They're not very commonly used, though.



                If you want to use fractions, you have to stop thinking about $frac23$ as $0.666ldots$ and instead think of it as $4$ (at least in modulo $5$). Also, there is no simple way to just look at a modular fraction and know more or less what its value is, the way you can just know that $frac1234561000001$ is very close to $0.123456$. In other words, small change in the denominator gives large changes in fraction value in modular arithmetic.



                So there are a few mental roadblocks on the way to using fractions in modular arithmetic, and I suspect that that's why it isn't done more. That being said, if you do use fractions, the same rules that you learned in elementary school for regular (rational) fractions still all apply: How to expand / simplify fractions, how to add or multiply them, or how to check that they are equal. The technique known as "cross multiplication" still works: $frac abequiv frac cdiff ad = bc$ (this does require that only invertible elements are allowed as denominators, though).



                You can allow non-invertible elements as denominators, but then more care is needed. See, for instance, my answer here on allowing fractions with odd denominators (including $3$) in modulo $6$.



                If you only allow invertible elements as denominators, then because they're already invertible, you're not really changing anything, as all fractions are elements that already exist as whole numbers. If you allow non-invertible elements, you basically get a smaller ring as a result (modulo $6$ with $3$ allowed as a denominator gives you what is essentially modulo $2$). Sadly, you can't get something "bigger" (the way $Bbb Q$ is bigger than $Bbb Z$), beause that requires you to have denominators which are neither invertible nor zero-divisors.



                Edit: Since I apparently missed the real point of the question. Yes, those manipulations are fine. Whatever $frac23$ is, give it the name $x$. By the definition of fractions, $3x = 2$. Multiplying both sides by $2$ can't make them not-equal, so $6x = 4$. Since $6$ is a valid denominator, by definition of fractions we get $x = frac46$. What you have ultimately done is to expand the fraction $frac23$ by $2$ and shown that this doesn't change the value of the fraction.






                share|cite|improve this answer











                $endgroup$












                • $begingroup$
                  I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
                  $endgroup$
                  – Michael Munta
                  Mar 7 at 9:41






                • 1




                  $begingroup$
                  @MichaelMunta You're right. I added a bit at the end.
                  $endgroup$
                  – Arthur
                  Mar 7 at 9:51













                4












                4








                4





                $begingroup$

                Fractions work fine in modular arithmetic. They're not very commonly used, though.



                If you want to use fractions, you have to stop thinking about $frac23$ as $0.666ldots$ and instead think of it as $4$ (at least in modulo $5$). Also, there is no simple way to just look at a modular fraction and know more or less what its value is, the way you can just know that $frac1234561000001$ is very close to $0.123456$. In other words, small change in the denominator gives large changes in fraction value in modular arithmetic.



                So there are a few mental roadblocks on the way to using fractions in modular arithmetic, and I suspect that that's why it isn't done more. That being said, if you do use fractions, the same rules that you learned in elementary school for regular (rational) fractions still all apply: How to expand / simplify fractions, how to add or multiply them, or how to check that they are equal. The technique known as "cross multiplication" still works: $frac abequiv frac cdiff ad = bc$ (this does require that only invertible elements are allowed as denominators, though).



                You can allow non-invertible elements as denominators, but then more care is needed. See, for instance, my answer here on allowing fractions with odd denominators (including $3$) in modulo $6$.



                If you only allow invertible elements as denominators, then because they're already invertible, you're not really changing anything, as all fractions are elements that already exist as whole numbers. If you allow non-invertible elements, you basically get a smaller ring as a result (modulo $6$ with $3$ allowed as a denominator gives you what is essentially modulo $2$). Sadly, you can't get something "bigger" (the way $Bbb Q$ is bigger than $Bbb Z$), beause that requires you to have denominators which are neither invertible nor zero-divisors.



                Edit: Since I apparently missed the real point of the question. Yes, those manipulations are fine. Whatever $frac23$ is, give it the name $x$. By the definition of fractions, $3x = 2$. Multiplying both sides by $2$ can't make them not-equal, so $6x = 4$. Since $6$ is a valid denominator, by definition of fractions we get $x = frac46$. What you have ultimately done is to expand the fraction $frac23$ by $2$ and shown that this doesn't change the value of the fraction.






                share|cite|improve this answer











                $endgroup$



                Fractions work fine in modular arithmetic. They're not very commonly used, though.



                If you want to use fractions, you have to stop thinking about $frac23$ as $0.666ldots$ and instead think of it as $4$ (at least in modulo $5$). Also, there is no simple way to just look at a modular fraction and know more or less what its value is, the way you can just know that $frac1234561000001$ is very close to $0.123456$. In other words, small change in the denominator gives large changes in fraction value in modular arithmetic.



                So there are a few mental roadblocks on the way to using fractions in modular arithmetic, and I suspect that that's why it isn't done more. That being said, if you do use fractions, the same rules that you learned in elementary school for regular (rational) fractions still all apply: How to expand / simplify fractions, how to add or multiply them, or how to check that they are equal. The technique known as "cross multiplication" still works: $frac abequiv frac cdiff ad = bc$ (this does require that only invertible elements are allowed as denominators, though).



                You can allow non-invertible elements as denominators, but then more care is needed. See, for instance, my answer here on allowing fractions with odd denominators (including $3$) in modulo $6$.



                If you only allow invertible elements as denominators, then because they're already invertible, you're not really changing anything, as all fractions are elements that already exist as whole numbers. If you allow non-invertible elements, you basically get a smaller ring as a result (modulo $6$ with $3$ allowed as a denominator gives you what is essentially modulo $2$). Sadly, you can't get something "bigger" (the way $Bbb Q$ is bigger than $Bbb Z$), beause that requires you to have denominators which are neither invertible nor zero-divisors.



                Edit: Since I apparently missed the real point of the question. Yes, those manipulations are fine. Whatever $frac23$ is, give it the name $x$. By the definition of fractions, $3x = 2$. Multiplying both sides by $2$ can't make them not-equal, so $6x = 4$. Since $6$ is a valid denominator, by definition of fractions we get $x = frac46$. What you have ultimately done is to expand the fraction $frac23$ by $2$ and shown that this doesn't change the value of the fraction.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 17 at 23:46









                Bill Dubuque

                213k29195654




                213k29195654










                answered Mar 7 at 9:20









                ArthurArthur

                120k7120204




                120k7120204











                • $begingroup$
                  I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
                  $endgroup$
                  – Michael Munta
                  Mar 7 at 9:41






                • 1




                  $begingroup$
                  @MichaelMunta You're right. I added a bit at the end.
                  $endgroup$
                  – Arthur
                  Mar 7 at 9:51
















                • $begingroup$
                  I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
                  $endgroup$
                  – Michael Munta
                  Mar 7 at 9:41






                • 1




                  $begingroup$
                  @MichaelMunta You're right. I added a bit at the end.
                  $endgroup$
                  – Arthur
                  Mar 7 at 9:51















                $begingroup$
                I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
                $endgroup$
                – Michael Munta
                Mar 7 at 9:41




                $begingroup$
                I see what you are saying, but it was not really the point of my question. I am wondering if my way of reaching the conclusion in the question is correct.
                $endgroup$
                – Michael Munta
                Mar 7 at 9:41




                1




                1




                $begingroup$
                @MichaelMunta You're right. I added a bit at the end.
                $endgroup$
                – Arthur
                Mar 7 at 9:51




                $begingroup$
                @MichaelMunta You're right. I added a bit at the end.
                $endgroup$
                – Arthur
                Mar 7 at 9:51











                1












                $begingroup$

                Yes. If $a,b,c,d$ are all co-prime to $n$ and $ad=bc mod n$ then we know that $bd$ is co-prime to $n$ so it is invertible in $mathbbZ_n$, and so we can divide by $bd$ to get



                $fracab=fraccd mod n$






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  Yes. If $a,b,c,d$ are all co-prime to $n$ and $ad=bc mod n$ then we know that $bd$ is co-prime to $n$ so it is invertible in $mathbbZ_n$, and so we can divide by $bd$ to get



                  $fracab=fraccd mod n$






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    Yes. If $a,b,c,d$ are all co-prime to $n$ and $ad=bc mod n$ then we know that $bd$ is co-prime to $n$ so it is invertible in $mathbbZ_n$, and so we can divide by $bd$ to get



                    $fracab=fraccd mod n$






                    share|cite|improve this answer









                    $endgroup$



                    Yes. If $a,b,c,d$ are all co-prime to $n$ and $ad=bc mod n$ then we know that $bd$ is co-prime to $n$ so it is invertible in $mathbbZ_n$, and so we can divide by $bd$ to get



                    $fracab=fraccd mod n$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 7 at 10:40









                    gandalf61gandalf61

                    9,164825




                    9,164825





















                        1












                        $begingroup$

                        Yes, your understanding is correct, There are a few reasons that they typically aren't used though:



                        • They aren't always defined ex. $frac52 bmod 6$

                        • Because you can give a second name mod anything, you can get multiple answers when not coprime to the modulus ex. $frac33equivfrac3-3bmod 6 $ at least in the sense of $3equiv -3 bmod 6$ but These give back two answers, The first is $1 bmod 6$ if taken literally, the second is $-1equiv 5 bmod 6$, You can also get 3 mod 6, because $frac6x+33=2x+1$ which lands on all remainders corresponding to odd numbers mod 6.

                        So using them recklessly, comes with some perils.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          Yes, your understanding is correct, There are a few reasons that they typically aren't used though:



                          • They aren't always defined ex. $frac52 bmod 6$

                          • Because you can give a second name mod anything, you can get multiple answers when not coprime to the modulus ex. $frac33equivfrac3-3bmod 6 $ at least in the sense of $3equiv -3 bmod 6$ but These give back two answers, The first is $1 bmod 6$ if taken literally, the second is $-1equiv 5 bmod 6$, You can also get 3 mod 6, because $frac6x+33=2x+1$ which lands on all remainders corresponding to odd numbers mod 6.

                          So using them recklessly, comes with some perils.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            Yes, your understanding is correct, There are a few reasons that they typically aren't used though:



                            • They aren't always defined ex. $frac52 bmod 6$

                            • Because you can give a second name mod anything, you can get multiple answers when not coprime to the modulus ex. $frac33equivfrac3-3bmod 6 $ at least in the sense of $3equiv -3 bmod 6$ but These give back two answers, The first is $1 bmod 6$ if taken literally, the second is $-1equiv 5 bmod 6$, You can also get 3 mod 6, because $frac6x+33=2x+1$ which lands on all remainders corresponding to odd numbers mod 6.

                            So using them recklessly, comes with some perils.






                            share|cite|improve this answer









                            $endgroup$



                            Yes, your understanding is correct, There are a few reasons that they typically aren't used though:



                            • They aren't always defined ex. $frac52 bmod 6$

                            • Because you can give a second name mod anything, you can get multiple answers when not coprime to the modulus ex. $frac33equivfrac3-3bmod 6 $ at least in the sense of $3equiv -3 bmod 6$ but These give back two answers, The first is $1 bmod 6$ if taken literally, the second is $-1equiv 5 bmod 6$, You can also get 3 mod 6, because $frac6x+33=2x+1$ which lands on all remainders corresponding to odd numbers mod 6.

                            So using them recklessly, comes with some perils.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 7 at 13:18









                            Roddy MacPheeRoddy MacPhee

                            529118




                            529118





















                                -1












                                $begingroup$

                                Yes, and just like you can do modular arithmetic to simplify ordinary arithmetic using integers, one can also do this for fractions. This looks impossible because of the multiple ways one can write a fraction modulo an integer representing a given number. But if it is known that the answer one is interested in, has a numerator between $0$ and $N$ and a denominator between $0$ and $D$ then a result modulo a number larger than $2 N D$ will allow one to reconstruct this rational number. As explained here, finding this fraction involves just the Euclidean algorithm.






                                share|cite|improve this answer











                                $endgroup$

















                                  -1












                                  $begingroup$

                                  Yes, and just like you can do modular arithmetic to simplify ordinary arithmetic using integers, one can also do this for fractions. This looks impossible because of the multiple ways one can write a fraction modulo an integer representing a given number. But if it is known that the answer one is interested in, has a numerator between $0$ and $N$ and a denominator between $0$ and $D$ then a result modulo a number larger than $2 N D$ will allow one to reconstruct this rational number. As explained here, finding this fraction involves just the Euclidean algorithm.






                                  share|cite|improve this answer











                                  $endgroup$















                                    -1












                                    -1








                                    -1





                                    $begingroup$

                                    Yes, and just like you can do modular arithmetic to simplify ordinary arithmetic using integers, one can also do this for fractions. This looks impossible because of the multiple ways one can write a fraction modulo an integer representing a given number. But if it is known that the answer one is interested in, has a numerator between $0$ and $N$ and a denominator between $0$ and $D$ then a result modulo a number larger than $2 N D$ will allow one to reconstruct this rational number. As explained here, finding this fraction involves just the Euclidean algorithm.






                                    share|cite|improve this answer











                                    $endgroup$



                                    Yes, and just like you can do modular arithmetic to simplify ordinary arithmetic using integers, one can also do this for fractions. This looks impossible because of the multiple ways one can write a fraction modulo an integer representing a given number. But if it is known that the answer one is interested in, has a numerator between $0$ and $N$ and a denominator between $0$ and $D$ then a result modulo a number larger than $2 N D$ will allow one to reconstruct this rational number. As explained here, finding this fraction involves just the Euclidean algorithm.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Mar 17 at 22:26

























                                    answered Mar 17 at 21:52









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