Smallest positive element of $ {ax + by: x,y in mathbb{Z}}$ is $gcd(a,b)$ [duplicate]












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  • Proving that $ gcd(a,b) = as + bt $, i.e., $ gcd $ is a linear combination. [duplicate]

    4 answers




Let $S = {ax + by: x,y in mathbb{Z}}$ and let $e > 0$ be the smallest element in $S$. Prove that $e mid a$, and hence prove that $e = gcd(a,b)$



I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.



I suppose that to show $emid a$ I would have to show that $eg = a$ for some $g in mathbb{Z}$. But I have no idea how I would do this.










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marked as duplicate by Jonas Meyer, colormegone, M. Vinay, user91500, Claude Leibovici Jun 22 '16 at 5:06


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.























    4












    $begingroup$



    This question already has an answer here:




    • Proving that $ gcd(a,b) = as + bt $, i.e., $ gcd $ is a linear combination. [duplicate]

      4 answers




    Let $S = {ax + by: x,y in mathbb{Z}}$ and let $e > 0$ be the smallest element in $S$. Prove that $e mid a$, and hence prove that $e = gcd(a,b)$



    I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.



    I suppose that to show $emid a$ I would have to show that $eg = a$ for some $g in mathbb{Z}$. But I have no idea how I would do this.










    share|cite|improve this question











    $endgroup$



    marked as duplicate by Jonas Meyer, colormegone, M. Vinay, user91500, Claude Leibovici Jun 22 '16 at 5:06


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





















      4












      4








      4


      0



      $begingroup$



      This question already has an answer here:




      • Proving that $ gcd(a,b) = as + bt $, i.e., $ gcd $ is a linear combination. [duplicate]

        4 answers




      Let $S = {ax + by: x,y in mathbb{Z}}$ and let $e > 0$ be the smallest element in $S$. Prove that $e mid a$, and hence prove that $e = gcd(a,b)$



      I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.



      I suppose that to show $emid a$ I would have to show that $eg = a$ for some $g in mathbb{Z}$. But I have no idea how I would do this.










      share|cite|improve this question











      $endgroup$





      This question already has an answer here:




      • Proving that $ gcd(a,b) = as + bt $, i.e., $ gcd $ is a linear combination. [duplicate]

        4 answers




      Let $S = {ax + by: x,y in mathbb{Z}}$ and let $e > 0$ be the smallest element in $S$. Prove that $e mid a$, and hence prove that $e = gcd(a,b)$



      I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.



      I suppose that to show $emid a$ I would have to show that $eg = a$ for some $g in mathbb{Z}$. But I have no idea how I would do this.





      This question already has an answer here:




      • Proving that $ gcd(a,b) = as + bt $, i.e., $ gcd $ is a linear combination. [duplicate]

        4 answers








      elementary-number-theory greatest-common-divisor






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      edited Jun 22 '16 at 6:22









      Martin Sleziak

      44.8k10118272




      44.8k10118272










      asked Jun 22 '16 at 3:35









      leanneleanne

      655




      655




      marked as duplicate by Jonas Meyer, colormegone, M. Vinay, user91500, Claude Leibovici Jun 22 '16 at 5:06


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Jonas Meyer, colormegone, M. Vinay, user91500, Claude Leibovici Jun 22 '16 at 5:06


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          3 Answers
          3






          active

          oldest

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          3












          $begingroup$

          When I first saw this problem, I was pretty surprised at how the proof went, so I'll just get you started on it.



          By the Division Theorem, we have:
          $$a=qe+r text{ for } 0 leq r < e$$
          Now, we need to show $r=0$. To do this, we can use the fact that $e=ax+by$ for some $x,y in Bbb{Z}$ and that $a=1a+0b$, so:
          $$1a+0b=q(ax+by)+r implies a(1-q)-bqy=r$$
          Now, combine the fact that $r=au+bv$ for $u,v in Bbb{Z}$ and that $0 leq r < e$ to prove $r=0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
            $endgroup$
            – leanne
            Jun 22 '16 at 3:48






          • 1




            $begingroup$
            @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
            $endgroup$
            – Noble Mushtak
            Jun 22 '16 at 12:17






          • 1




            $begingroup$
            @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
            $endgroup$
            – Bill Dubuque
            Jun 22 '16 at 15:13












          • $begingroup$
            @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
            $endgroup$
            – Noble Mushtak
            Jun 22 '16 at 15:15






          • 1




            $begingroup$
            @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
            $endgroup$
            – Bill Dubuque
            Jun 22 '16 at 15:51





















          1












          $begingroup$

          Here is one conceptual way to present a proof of Bezout's Identity for the gcd. The set $rm,S,$ of integers of the form $rm,a x + b y, x,yin mathbb Z,,$ is closed under subtraction hence, by the Lemma, every positive $rm,kin S,$ is divisible by $rm,d = $ least positive $rmin S.,$ Therefore $rm,a,bin S$ $,Rightarrow,$ $rm dmid a,b,,$ i.e. $rm,d,$ is a common divisor of $rm,a,b,,$ the greatest by $rm cmid a,b,$ $Rightarrow$ $rm,cmid d = ax+by$ $Rightarrow$ $rm,cle d$



          Lemma $ $ Let $,rm Sneemptyset ,$ be a set of integers $>0,$ closed under subtraction $> 0,,$ i.e. for all $rm,n,min S, ,$ $rm n > m Rightarrow n-m, in, S.,$ Then the least $rm:ellin S,$ divides every element of $,rm S.$



          Proof ${bf 1}, $ If not there is a least nonmultiple $rm,nin S,,$ contra $rm,n-ell in S,$ is a nonmultiple of $rm,ell.$



          Proof ${bf 2},rm, S,$ closed under subtraction $rm,Rightarrow,S,$ closed under remainder (mod), when it is $ne 0,$ since mod is simply repeated subtraction, i.e. $rm a mod b, =, a - k b, =, a!-!b!-!b!-cdots! -!b.,$ Thus $rm,nin S,$ $Rightarrow$ $rm, (n mod ell) = 0,,$ else it is in $,rm S,$ and smaller than $rm,ell,,$ contra minimality of $rm,ell.$



          Remark $ $ In a nutshell, two applications of induction yield the following inferences



          $ rmbegin{eqnarray} S closed under {bf subtraction} &:Rightarrow:&rm S closed under {bf mod} = remainder = repeated subtraction \
          &:Rightarrow:&rm S closed under {bf gcd} = repeated mod (Euclid's algorithm) end{eqnarray}$



          Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.



          The conceptual structure will be clarified when one studies ideals of rings, where the above proof generalizes to show that Euclidean domains are PIDs.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            One must define $e$ as the smallest positive element of $S$.



            Since $ein {ax + by: x,y in mathbb{Z}}$, there are some numbers $x_1,y_1$ such that $e = ax_1+by_1$.



            Suppose some positive member $f$ of $S = {ax + by: x,y in mathbb{Z}}$ were not a multiple of $e$. Then we have some numbers $x_2,y_2$ such that $f = ax_2+by_2$.



            Since $e$ is the smallest, we must have $e<f$. Let $r$ be the remainder when $f$ is divided by $e$. Then using the relations
            begin{align}
            f & = ax_2 + by_2 \
            e & = ax_1 + by_1
            end{align}
            you can show that $r in S$. But $0<r$ since $f$ is not a multiple of $e$ and $r<e$ since $r$ was a remainder upon division by $e$.



            That means there is a positive member of $S$, namely $r$ that is smaller than $e$. That can't happen since $e$ is the smallest.



            Hence every member of $S$ is a multiple of $e$. Therefore $e$ is a common divisor of $ain S$ and $bin S$.






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              When I first saw this problem, I was pretty surprised at how the proof went, so I'll just get you started on it.



              By the Division Theorem, we have:
              $$a=qe+r text{ for } 0 leq r < e$$
              Now, we need to show $r=0$. To do this, we can use the fact that $e=ax+by$ for some $x,y in Bbb{Z}$ and that $a=1a+0b$, so:
              $$1a+0b=q(ax+by)+r implies a(1-q)-bqy=r$$
              Now, combine the fact that $r=au+bv$ for $u,v in Bbb{Z}$ and that $0 leq r < e$ to prove $r=0$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
                $endgroup$
                – leanne
                Jun 22 '16 at 3:48






              • 1




                $begingroup$
                @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 12:17






              • 1




                $begingroup$
                @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:13












              • $begingroup$
                @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 15:15






              • 1




                $begingroup$
                @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:51


















              3












              $begingroup$

              When I first saw this problem, I was pretty surprised at how the proof went, so I'll just get you started on it.



              By the Division Theorem, we have:
              $$a=qe+r text{ for } 0 leq r < e$$
              Now, we need to show $r=0$. To do this, we can use the fact that $e=ax+by$ for some $x,y in Bbb{Z}$ and that $a=1a+0b$, so:
              $$1a+0b=q(ax+by)+r implies a(1-q)-bqy=r$$
              Now, combine the fact that $r=au+bv$ for $u,v in Bbb{Z}$ and that $0 leq r < e$ to prove $r=0$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
                $endgroup$
                – leanne
                Jun 22 '16 at 3:48






              • 1




                $begingroup$
                @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 12:17






              • 1




                $begingroup$
                @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:13












              • $begingroup$
                @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 15:15






              • 1




                $begingroup$
                @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:51
















              3












              3








              3





              $begingroup$

              When I first saw this problem, I was pretty surprised at how the proof went, so I'll just get you started on it.



              By the Division Theorem, we have:
              $$a=qe+r text{ for } 0 leq r < e$$
              Now, we need to show $r=0$. To do this, we can use the fact that $e=ax+by$ for some $x,y in Bbb{Z}$ and that $a=1a+0b$, so:
              $$1a+0b=q(ax+by)+r implies a(1-q)-bqy=r$$
              Now, combine the fact that $r=au+bv$ for $u,v in Bbb{Z}$ and that $0 leq r < e$ to prove $r=0$.






              share|cite|improve this answer











              $endgroup$



              When I first saw this problem, I was pretty surprised at how the proof went, so I'll just get you started on it.



              By the Division Theorem, we have:
              $$a=qe+r text{ for } 0 leq r < e$$
              Now, we need to show $r=0$. To do this, we can use the fact that $e=ax+by$ for some $x,y in Bbb{Z}$ and that $a=1a+0b$, so:
              $$1a+0b=q(ax+by)+r implies a(1-q)-bqy=r$$
              Now, combine the fact that $r=au+bv$ for $u,v in Bbb{Z}$ and that $0 leq r < e$ to prove $r=0$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jun 22 '16 at 12:16

























              answered Jun 22 '16 at 3:44









              Noble MushtakNoble Mushtak

              15.3k1835




              15.3k1835












              • $begingroup$
                Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
                $endgroup$
                – leanne
                Jun 22 '16 at 3:48






              • 1




                $begingroup$
                @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 12:17






              • 1




                $begingroup$
                @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:13












              • $begingroup$
                @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 15:15






              • 1




                $begingroup$
                @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:51




















              • $begingroup$
                Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
                $endgroup$
                – leanne
                Jun 22 '16 at 3:48






              • 1




                $begingroup$
                @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 12:17






              • 1




                $begingroup$
                @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:13












              • $begingroup$
                @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
                $endgroup$
                – Noble Mushtak
                Jun 22 '16 at 15:15






              • 1




                $begingroup$
                @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
                $endgroup$
                – Bill Dubuque
                Jun 22 '16 at 15:51


















              $begingroup$
              Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
              $endgroup$
              – leanne
              Jun 22 '16 at 3:48




              $begingroup$
              Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$?
              $endgroup$
              – leanne
              Jun 22 '16 at 3:48




              1




              1




              $begingroup$
              @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
              $endgroup$
              – Noble Mushtak
              Jun 22 '16 at 12:17




              $begingroup$
              @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q in Bbb{Z}$ and $v=-qy in Bbb{Z}$ so that $r=au+bv$.
              $endgroup$
              – Noble Mushtak
              Jun 22 '16 at 12:17




              1




              1




              $begingroup$
              @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
              $endgroup$
              – Bill Dubuque
              Jun 22 '16 at 15:13






              $begingroup$
              @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore.
              $endgroup$
              – Bill Dubuque
              Jun 22 '16 at 15:13














              $begingroup$
              @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
              $endgroup$
              – Noble Mushtak
              Jun 22 '16 at 15:15




              $begingroup$
              @BillDubuque "Closed under remainders" means that the Division Theorem is true for $Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure.
              $endgroup$
              – Noble Mushtak
              Jun 22 '16 at 15:15




              1




              1




              $begingroup$
              @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
              $endgroup$
              – Bill Dubuque
              Jun 22 '16 at 15:51






              $begingroup$
              @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so.
              $endgroup$
              – Bill Dubuque
              Jun 22 '16 at 15:51













              1












              $begingroup$

              Here is one conceptual way to present a proof of Bezout's Identity for the gcd. The set $rm,S,$ of integers of the form $rm,a x + b y, x,yin mathbb Z,,$ is closed under subtraction hence, by the Lemma, every positive $rm,kin S,$ is divisible by $rm,d = $ least positive $rmin S.,$ Therefore $rm,a,bin S$ $,Rightarrow,$ $rm dmid a,b,,$ i.e. $rm,d,$ is a common divisor of $rm,a,b,,$ the greatest by $rm cmid a,b,$ $Rightarrow$ $rm,cmid d = ax+by$ $Rightarrow$ $rm,cle d$



              Lemma $ $ Let $,rm Sneemptyset ,$ be a set of integers $>0,$ closed under subtraction $> 0,,$ i.e. for all $rm,n,min S, ,$ $rm n > m Rightarrow n-m, in, S.,$ Then the least $rm:ellin S,$ divides every element of $,rm S.$



              Proof ${bf 1}, $ If not there is a least nonmultiple $rm,nin S,,$ contra $rm,n-ell in S,$ is a nonmultiple of $rm,ell.$



              Proof ${bf 2},rm, S,$ closed under subtraction $rm,Rightarrow,S,$ closed under remainder (mod), when it is $ne 0,$ since mod is simply repeated subtraction, i.e. $rm a mod b, =, a - k b, =, a!-!b!-!b!-cdots! -!b.,$ Thus $rm,nin S,$ $Rightarrow$ $rm, (n mod ell) = 0,,$ else it is in $,rm S,$ and smaller than $rm,ell,,$ contra minimality of $rm,ell.$



              Remark $ $ In a nutshell, two applications of induction yield the following inferences



              $ rmbegin{eqnarray} S closed under {bf subtraction} &:Rightarrow:&rm S closed under {bf mod} = remainder = repeated subtraction \
              &:Rightarrow:&rm S closed under {bf gcd} = repeated mod (Euclid's algorithm) end{eqnarray}$



              Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.



              The conceptual structure will be clarified when one studies ideals of rings, where the above proof generalizes to show that Euclidean domains are PIDs.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Here is one conceptual way to present a proof of Bezout's Identity for the gcd. The set $rm,S,$ of integers of the form $rm,a x + b y, x,yin mathbb Z,,$ is closed under subtraction hence, by the Lemma, every positive $rm,kin S,$ is divisible by $rm,d = $ least positive $rmin S.,$ Therefore $rm,a,bin S$ $,Rightarrow,$ $rm dmid a,b,,$ i.e. $rm,d,$ is a common divisor of $rm,a,b,,$ the greatest by $rm cmid a,b,$ $Rightarrow$ $rm,cmid d = ax+by$ $Rightarrow$ $rm,cle d$



                Lemma $ $ Let $,rm Sneemptyset ,$ be a set of integers $>0,$ closed under subtraction $> 0,,$ i.e. for all $rm,n,min S, ,$ $rm n > m Rightarrow n-m, in, S.,$ Then the least $rm:ellin S,$ divides every element of $,rm S.$



                Proof ${bf 1}, $ If not there is a least nonmultiple $rm,nin S,,$ contra $rm,n-ell in S,$ is a nonmultiple of $rm,ell.$



                Proof ${bf 2},rm, S,$ closed under subtraction $rm,Rightarrow,S,$ closed under remainder (mod), when it is $ne 0,$ since mod is simply repeated subtraction, i.e. $rm a mod b, =, a - k b, =, a!-!b!-!b!-cdots! -!b.,$ Thus $rm,nin S,$ $Rightarrow$ $rm, (n mod ell) = 0,,$ else it is in $,rm S,$ and smaller than $rm,ell,,$ contra minimality of $rm,ell.$



                Remark $ $ In a nutshell, two applications of induction yield the following inferences



                $ rmbegin{eqnarray} S closed under {bf subtraction} &:Rightarrow:&rm S closed under {bf mod} = remainder = repeated subtraction \
                &:Rightarrow:&rm S closed under {bf gcd} = repeated mod (Euclid's algorithm) end{eqnarray}$



                Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.



                The conceptual structure will be clarified when one studies ideals of rings, where the above proof generalizes to show that Euclidean domains are PIDs.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Here is one conceptual way to present a proof of Bezout's Identity for the gcd. The set $rm,S,$ of integers of the form $rm,a x + b y, x,yin mathbb Z,,$ is closed under subtraction hence, by the Lemma, every positive $rm,kin S,$ is divisible by $rm,d = $ least positive $rmin S.,$ Therefore $rm,a,bin S$ $,Rightarrow,$ $rm dmid a,b,,$ i.e. $rm,d,$ is a common divisor of $rm,a,b,,$ the greatest by $rm cmid a,b,$ $Rightarrow$ $rm,cmid d = ax+by$ $Rightarrow$ $rm,cle d$



                  Lemma $ $ Let $,rm Sneemptyset ,$ be a set of integers $>0,$ closed under subtraction $> 0,,$ i.e. for all $rm,n,min S, ,$ $rm n > m Rightarrow n-m, in, S.,$ Then the least $rm:ellin S,$ divides every element of $,rm S.$



                  Proof ${bf 1}, $ If not there is a least nonmultiple $rm,nin S,,$ contra $rm,n-ell in S,$ is a nonmultiple of $rm,ell.$



                  Proof ${bf 2},rm, S,$ closed under subtraction $rm,Rightarrow,S,$ closed under remainder (mod), when it is $ne 0,$ since mod is simply repeated subtraction, i.e. $rm a mod b, =, a - k b, =, a!-!b!-!b!-cdots! -!b.,$ Thus $rm,nin S,$ $Rightarrow$ $rm, (n mod ell) = 0,,$ else it is in $,rm S,$ and smaller than $rm,ell,,$ contra minimality of $rm,ell.$



                  Remark $ $ In a nutshell, two applications of induction yield the following inferences



                  $ rmbegin{eqnarray} S closed under {bf subtraction} &:Rightarrow:&rm S closed under {bf mod} = remainder = repeated subtraction \
                  &:Rightarrow:&rm S closed under {bf gcd} = repeated mod (Euclid's algorithm) end{eqnarray}$



                  Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.



                  The conceptual structure will be clarified when one studies ideals of rings, where the above proof generalizes to show that Euclidean domains are PIDs.






                  share|cite|improve this answer











                  $endgroup$



                  Here is one conceptual way to present a proof of Bezout's Identity for the gcd. The set $rm,S,$ of integers of the form $rm,a x + b y, x,yin mathbb Z,,$ is closed under subtraction hence, by the Lemma, every positive $rm,kin S,$ is divisible by $rm,d = $ least positive $rmin S.,$ Therefore $rm,a,bin S$ $,Rightarrow,$ $rm dmid a,b,,$ i.e. $rm,d,$ is a common divisor of $rm,a,b,,$ the greatest by $rm cmid a,b,$ $Rightarrow$ $rm,cmid d = ax+by$ $Rightarrow$ $rm,cle d$



                  Lemma $ $ Let $,rm Sneemptyset ,$ be a set of integers $>0,$ closed under subtraction $> 0,,$ i.e. for all $rm,n,min S, ,$ $rm n > m Rightarrow n-m, in, S.,$ Then the least $rm:ellin S,$ divides every element of $,rm S.$



                  Proof ${bf 1}, $ If not there is a least nonmultiple $rm,nin S,,$ contra $rm,n-ell in S,$ is a nonmultiple of $rm,ell.$



                  Proof ${bf 2},rm, S,$ closed under subtraction $rm,Rightarrow,S,$ closed under remainder (mod), when it is $ne 0,$ since mod is simply repeated subtraction, i.e. $rm a mod b, =, a - k b, =, a!-!b!-!b!-cdots! -!b.,$ Thus $rm,nin S,$ $Rightarrow$ $rm, (n mod ell) = 0,,$ else it is in $,rm S,$ and smaller than $rm,ell,,$ contra minimality of $rm,ell.$



                  Remark $ $ In a nutshell, two applications of induction yield the following inferences



                  $ rmbegin{eqnarray} S closed under {bf subtraction} &:Rightarrow:&rm S closed under {bf mod} = remainder = repeated subtraction \
                  &:Rightarrow:&rm S closed under {bf gcd} = repeated mod (Euclid's algorithm) end{eqnarray}$



                  Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.



                  The conceptual structure will be clarified when one studies ideals of rings, where the above proof generalizes to show that Euclidean domains are PIDs.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 13 '17 at 12:21









                  Community

                  1




                  1










                  answered Jun 22 '16 at 3:50









                  Bill DubuqueBill Dubuque

                  210k29192645




                  210k29192645























                      0












                      $begingroup$

                      One must define $e$ as the smallest positive element of $S$.



                      Since $ein {ax + by: x,y in mathbb{Z}}$, there are some numbers $x_1,y_1$ such that $e = ax_1+by_1$.



                      Suppose some positive member $f$ of $S = {ax + by: x,y in mathbb{Z}}$ were not a multiple of $e$. Then we have some numbers $x_2,y_2$ such that $f = ax_2+by_2$.



                      Since $e$ is the smallest, we must have $e<f$. Let $r$ be the remainder when $f$ is divided by $e$. Then using the relations
                      begin{align}
                      f & = ax_2 + by_2 \
                      e & = ax_1 + by_1
                      end{align}
                      you can show that $r in S$. But $0<r$ since $f$ is not a multiple of $e$ and $r<e$ since $r$ was a remainder upon division by $e$.



                      That means there is a positive member of $S$, namely $r$ that is smaller than $e$. That can't happen since $e$ is the smallest.



                      Hence every member of $S$ is a multiple of $e$. Therefore $e$ is a common divisor of $ain S$ and $bin S$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        One must define $e$ as the smallest positive element of $S$.



                        Since $ein {ax + by: x,y in mathbb{Z}}$, there are some numbers $x_1,y_1$ such that $e = ax_1+by_1$.



                        Suppose some positive member $f$ of $S = {ax + by: x,y in mathbb{Z}}$ were not a multiple of $e$. Then we have some numbers $x_2,y_2$ such that $f = ax_2+by_2$.



                        Since $e$ is the smallest, we must have $e<f$. Let $r$ be the remainder when $f$ is divided by $e$. Then using the relations
                        begin{align}
                        f & = ax_2 + by_2 \
                        e & = ax_1 + by_1
                        end{align}
                        you can show that $r in S$. But $0<r$ since $f$ is not a multiple of $e$ and $r<e$ since $r$ was a remainder upon division by $e$.



                        That means there is a positive member of $S$, namely $r$ that is smaller than $e$. That can't happen since $e$ is the smallest.



                        Hence every member of $S$ is a multiple of $e$. Therefore $e$ is a common divisor of $ain S$ and $bin S$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          One must define $e$ as the smallest positive element of $S$.



                          Since $ein {ax + by: x,y in mathbb{Z}}$, there are some numbers $x_1,y_1$ such that $e = ax_1+by_1$.



                          Suppose some positive member $f$ of $S = {ax + by: x,y in mathbb{Z}}$ were not a multiple of $e$. Then we have some numbers $x_2,y_2$ such that $f = ax_2+by_2$.



                          Since $e$ is the smallest, we must have $e<f$. Let $r$ be the remainder when $f$ is divided by $e$. Then using the relations
                          begin{align}
                          f & = ax_2 + by_2 \
                          e & = ax_1 + by_1
                          end{align}
                          you can show that $r in S$. But $0<r$ since $f$ is not a multiple of $e$ and $r<e$ since $r$ was a remainder upon division by $e$.



                          That means there is a positive member of $S$, namely $r$ that is smaller than $e$. That can't happen since $e$ is the smallest.



                          Hence every member of $S$ is a multiple of $e$. Therefore $e$ is a common divisor of $ain S$ and $bin S$.






                          share|cite|improve this answer









                          $endgroup$



                          One must define $e$ as the smallest positive element of $S$.



                          Since $ein {ax + by: x,y in mathbb{Z}}$, there are some numbers $x_1,y_1$ such that $e = ax_1+by_1$.



                          Suppose some positive member $f$ of $S = {ax + by: x,y in mathbb{Z}}$ were not a multiple of $e$. Then we have some numbers $x_2,y_2$ such that $f = ax_2+by_2$.



                          Since $e$ is the smallest, we must have $e<f$. Let $r$ be the remainder when $f$ is divided by $e$. Then using the relations
                          begin{align}
                          f & = ax_2 + by_2 \
                          e & = ax_1 + by_1
                          end{align}
                          you can show that $r in S$. But $0<r$ since $f$ is not a multiple of $e$ and $r<e$ since $r$ was a remainder upon division by $e$.



                          That means there is a positive member of $S$, namely $r$ that is smaller than $e$. That can't happen since $e$ is the smallest.



                          Hence every member of $S$ is a multiple of $e$. Therefore $e$ is a common divisor of $ain S$ and $bin S$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jun 22 '16 at 4:00









                          Michael HardyMichael Hardy

                          1




                          1















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