Smallest positive element of $ {ax + by: x,y in mathbb{Z}}$ is $gcd(a,b)$ [duplicate]
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.
elementary-number-theory greatest-common-divisor
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.
add a comment |
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.
elementary-number-theory greatest-common-divisor
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.
add a comment |
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.
elementary-number-theory greatest-common-divisor
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
elementary-number-theory greatest-common-divisor
edited Jun 22 '16 at 6:22
Martin Sleziak
44.7k7115270
44.7k7115270
asked Jun 22 '16 at 3:35
leanne
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.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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 $r=0$.
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
1
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
|
show 2 more comments
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.
add a comment |
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 $bin S$.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $r=0$.
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
1
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
|
show 2 more comments
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 $r=0$.
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
1
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
|
show 2 more comments
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 $r=0$.
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 $r=0$.
edited Jun 22 '16 at 12:16
answered Jun 22 '16 at 3:44
Noble Mushtak
14.6k1734
14.6k1734
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
1
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
|
show 2 more comments
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
1
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
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 $r = au + bv$?
– leanne
Jun 22 '16 at 3:48
1
1
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
@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 $r=au+bv$.
– Noble Mushtak
Jun 22 '16 at 12:17
1
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Bill Dubuque
Jun 22 '16 at 15:13
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
@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.
– Noble Mushtak
Jun 22 '16 at 15:15
1
1
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
@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.
– Bill Dubuque
Jun 22 '16 at 15:51
|
show 2 more comments
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.
add a comment |
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.
add a comment |
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.
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.
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jun 22 '16 at 3:50
Bill Dubuque
208k29190628
208k29190628
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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 $bin S$.
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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 $bin S$.
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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 $bin S$.
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 $bin S$.
answered Jun 22 '16 at 4:00
Michael Hardy
1
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