Prove that $gcd(a, b) = gcd(b, a-b)$
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I can understand the concept that $gcd(a, b) = gcd(b, r)$, where $a = bq + r$, which is grounded from the fact that $gcd(a, b) = gcd(b, a-b)$, but I have no intuition for the latter.
elementary-number-theory divisibility greatest-common-divisor
$endgroup$
add a comment |
$begingroup$
I can understand the concept that $gcd(a, b) = gcd(b, r)$, where $a = bq + r$, which is grounded from the fact that $gcd(a, b) = gcd(b, a-b)$, but I have no intuition for the latter.
elementary-number-theory divisibility greatest-common-divisor
$endgroup$
add a comment |
$begingroup$
I can understand the concept that $gcd(a, b) = gcd(b, r)$, where $a = bq + r$, which is grounded from the fact that $gcd(a, b) = gcd(b, a-b)$, but I have no intuition for the latter.
elementary-number-theory divisibility greatest-common-divisor
$endgroup$
I can understand the concept that $gcd(a, b) = gcd(b, r)$, where $a = bq + r$, which is grounded from the fact that $gcd(a, b) = gcd(b, a-b)$, but I have no intuition for the latter.
elementary-number-theory divisibility greatest-common-divisor
elementary-number-theory divisibility greatest-common-divisor
edited Oct 18 '17 at 12:02
Martin Sleziak
44.8k10119272
44.8k10119272
asked Oct 24 '14 at 5:58
Joel ChristophelJoel Christophel
324519
324519
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7 Answers
7
active
oldest
votes
$begingroup$
Lemma: $gcd(x,y)=d$ iff $d$ is the smallest natural number that can be expressed as a linear combination of $x$ and $y$ (in $mathbb{Z}$)
We have $gcd(a,b)=d$
$d=ap+bq$ for some $p,qin mathbb{Z}$
$d=ap-bp+bp+bq=(a-b)p+b(p+q)=(a-b)p+bq'$
where $q'=p+qin mathbb{Z}$
Thus,
$gcd(a-b,b)=d$
$endgroup$
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
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Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
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yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
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Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
add a comment |
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Let $d = text{gcd}(a,b)$, then $d|a$, and $d|b$, so $d|(a-b)$, and $d|mbox {gcd}(b,a-b)= d'$. Also $d'|b$, and $d'|(a-b)$, so $d'|(b+(a-b)) = a$, and $d'|mbox {gcd}(a,b) = d$. So $d = d'$.
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$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
add a comment |
$begingroup$
Suppose $gcd(a, b) = q$ it means at least that $q | a, q|b$ then $q|(a - b)$. This is true for every divisor so that for $gcd$ also.
One user asked converse but it is also obvious. Denote $s = a - b$, then $gcd(b, s) = gcd(b, s + b) = gcd(b, a)$.
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$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
1
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
add a comment |
$begingroup$
By definition
$$gcd(a,b)=giff g|a,bland (forall h:h|a,bimplies hle g).$$
Then
$$begin{align}gcd(a,b)=g&implies g|a,bland (forall h:h|a,bimplies hle g)\&implies g|b,(a-b)land (forall h:h|b,(a-b)implies h|a,bimplies hle g)\&implies gcd(b,a-b)=g.end{align}$$
$endgroup$
add a comment |
$begingroup$
Every common divisor of $a$ and $b$ is also a common divisor of $b$ and $a-b$.
Every common divisor of $b$ and $a-b$ is also a common divisor of $a$ and $b$, as $a=(a-b)+b$.
$endgroup$
$begingroup$
This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
$endgroup$
– Yves Daoust
Oct 26 '14 at 21:00
add a comment |
$begingroup$
$dmid a-b$ means $dfrac{a-b}d=dfrac ad-dfrac bd$ is an integer. It's clear that if one of these fractions is an integer, then so is the other, i.e., $dmid aiff dmid b$.
$dmidgcd(b,a-b)implies dmid b$, thus $dmid a$.
Therefore a divisor of both $a-b$ and $b$ is also a divisor of $a$.
Conversely, $kmid a$ means $dfrac ak=dfrac{a-b}k+dfrac bk$ is an integer, so $kmid a-biff kmid b$.
$kmidgcd(a,b)implies kmid b$, thus $kmid a-b$.
Therefore a divisor of both $a$ and $b$ is also a divisor of $a-b$. QED
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add a comment |
$begingroup$
Write $u=b$ and $v=a-b$. Then $a=u+v$ and $b=u$.
These two sets of equations imply that $d$ divides $u,v$ iff $d$ divides $a,b$. Therefore, $gcd(u,v)=gcd(a,b)$.
In algebra terms, the equations imply that ${a,b}$ and ${u,v}$ generate the same subgroup of $mathbb Z$. It's well known that $langle x,y rangle = gcd(x,y)mathbb Z$, hence the result.
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Lemma: $gcd(x,y)=d$ iff $d$ is the smallest natural number that can be expressed as a linear combination of $x$ and $y$ (in $mathbb{Z}$)
We have $gcd(a,b)=d$
$d=ap+bq$ for some $p,qin mathbb{Z}$
$d=ap-bp+bp+bq=(a-b)p+b(p+q)=(a-b)p+bq'$
where $q'=p+qin mathbb{Z}$
Thus,
$gcd(a-b,b)=d$
$endgroup$
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
$begingroup$
Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
$begingroup$
yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
$begingroup$
Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
add a comment |
$begingroup$
Lemma: $gcd(x,y)=d$ iff $d$ is the smallest natural number that can be expressed as a linear combination of $x$ and $y$ (in $mathbb{Z}$)
We have $gcd(a,b)=d$
$d=ap+bq$ for some $p,qin mathbb{Z}$
$d=ap-bp+bp+bq=(a-b)p+b(p+q)=(a-b)p+bq'$
where $q'=p+qin mathbb{Z}$
Thus,
$gcd(a-b,b)=d$
$endgroup$
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
$begingroup$
Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
$begingroup$
yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
$begingroup$
Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
add a comment |
$begingroup$
Lemma: $gcd(x,y)=d$ iff $d$ is the smallest natural number that can be expressed as a linear combination of $x$ and $y$ (in $mathbb{Z}$)
We have $gcd(a,b)=d$
$d=ap+bq$ for some $p,qin mathbb{Z}$
$d=ap-bp+bp+bq=(a-b)p+b(p+q)=(a-b)p+bq'$
where $q'=p+qin mathbb{Z}$
Thus,
$gcd(a-b,b)=d$
$endgroup$
Lemma: $gcd(x,y)=d$ iff $d$ is the smallest natural number that can be expressed as a linear combination of $x$ and $y$ (in $mathbb{Z}$)
We have $gcd(a,b)=d$
$d=ap+bq$ for some $p,qin mathbb{Z}$
$d=ap-bp+bp+bq=(a-b)p+b(p+q)=(a-b)p+bq'$
where $q'=p+qin mathbb{Z}$
Thus,
$gcd(a-b,b)=d$
edited Oct 22 '17 at 18:31
answered Oct 24 '14 at 6:12
Swapnil TripathiSwapnil Tripathi
3,08621936
3,08621936
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
$begingroup$
Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
$begingroup$
yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
$begingroup$
Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
add a comment |
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
$begingroup$
Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
$begingroup$
yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
$begingroup$
Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
$begingroup$
That lemma is only true in the forward direction. For example, $1cdot5+1cdot4=9$ but $gcd(5,4)=1neq9$.
$endgroup$
– Harry
Oct 18 '17 at 12:14
$begingroup$
Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
$begingroup$
Ah, I see. Would the converse be true if the condition of d being the smallest such positive natural number is added?
$endgroup$
– Swapnil Tripathi
Oct 21 '17 at 1:17
$begingroup$
yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
$begingroup$
yes, it would. The $gcd$ of $x,y$ is the smallest positive natural number that can be expressed as an integral linear combination of $x,y$.
$endgroup$
– Harry
Oct 21 '17 at 2:02
$begingroup$
Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
$begingroup$
Edited, thank you! @Harry
$endgroup$
– Swapnil Tripathi
Oct 22 '17 at 18:32
add a comment |
$begingroup$
Let $d = text{gcd}(a,b)$, then $d|a$, and $d|b$, so $d|(a-b)$, and $d|mbox {gcd}(b,a-b)= d'$. Also $d'|b$, and $d'|(a-b)$, so $d'|(b+(a-b)) = a$, and $d'|mbox {gcd}(a,b) = d$. So $d = d'$.
$endgroup$
$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
add a comment |
$begingroup$
Let $d = text{gcd}(a,b)$, then $d|a$, and $d|b$, so $d|(a-b)$, and $d|mbox {gcd}(b,a-b)= d'$. Also $d'|b$, and $d'|(a-b)$, so $d'|(b+(a-b)) = a$, and $d'|mbox {gcd}(a,b) = d$. So $d = d'$.
$endgroup$
$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
add a comment |
$begingroup$
Let $d = text{gcd}(a,b)$, then $d|a$, and $d|b$, so $d|(a-b)$, and $d|mbox {gcd}(b,a-b)= d'$. Also $d'|b$, and $d'|(a-b)$, so $d'|(b+(a-b)) = a$, and $d'|mbox {gcd}(a,b) = d$. So $d = d'$.
$endgroup$
Let $d = text{gcd}(a,b)$, then $d|a$, and $d|b$, so $d|(a-b)$, and $d|mbox {gcd}(b,a-b)= d'$. Also $d'|b$, and $d'|(a-b)$, so $d'|(b+(a-b)) = a$, and $d'|mbox {gcd}(a,b) = d$. So $d = d'$.
edited Jan 1 at 14:46
user376343
3,8733829
3,8733829
answered Oct 24 '14 at 6:07
DeepSeaDeepSea
71.3k54487
71.3k54487
$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
add a comment |
$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
$begingroup$
How do you prove that $$d mid gcd(b, a-b)=d'$$
$endgroup$
– Dr. Mathva
Jan 1 at 14:23
add a comment |
$begingroup$
Suppose $gcd(a, b) = q$ it means at least that $q | a, q|b$ then $q|(a - b)$. This is true for every divisor so that for $gcd$ also.
One user asked converse but it is also obvious. Denote $s = a - b$, then $gcd(b, s) = gcd(b, s + b) = gcd(b, a)$.
$endgroup$
$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
1
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
add a comment |
$begingroup$
Suppose $gcd(a, b) = q$ it means at least that $q | a, q|b$ then $q|(a - b)$. This is true for every divisor so that for $gcd$ also.
One user asked converse but it is also obvious. Denote $s = a - b$, then $gcd(b, s) = gcd(b, s + b) = gcd(b, a)$.
$endgroup$
$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
1
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
add a comment |
$begingroup$
Suppose $gcd(a, b) = q$ it means at least that $q | a, q|b$ then $q|(a - b)$. This is true for every divisor so that for $gcd$ also.
One user asked converse but it is also obvious. Denote $s = a - b$, then $gcd(b, s) = gcd(b, s + b) = gcd(b, a)$.
$endgroup$
Suppose $gcd(a, b) = q$ it means at least that $q | a, q|b$ then $q|(a - b)$. This is true for every divisor so that for $gcd$ also.
One user asked converse but it is also obvious. Denote $s = a - b$, then $gcd(b, s) = gcd(b, s + b) = gcd(b, a)$.
edited Oct 24 '14 at 6:08
answered Oct 24 '14 at 6:01
JihadJihad
38417
38417
$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
1
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
add a comment |
$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
1
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
$begingroup$
You need to prove also the converse, if $q'|b$ and $q'|a-b$ then $q'|a$, but +1 anyway.
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:02
1
1
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
$begingroup$
It's just that your first statement only proved that $gcd(a,b)|gcd(b,a-b)$, not equality. ;-)
$endgroup$
– Jean-Claude Arbaut
Oct 24 '14 at 6:15
add a comment |
$begingroup$
By definition
$$gcd(a,b)=giff g|a,bland (forall h:h|a,bimplies hle g).$$
Then
$$begin{align}gcd(a,b)=g&implies g|a,bland (forall h:h|a,bimplies hle g)\&implies g|b,(a-b)land (forall h:h|b,(a-b)implies h|a,bimplies hle g)\&implies gcd(b,a-b)=g.end{align}$$
$endgroup$
add a comment |
$begingroup$
By definition
$$gcd(a,b)=giff g|a,bland (forall h:h|a,bimplies hle g).$$
Then
$$begin{align}gcd(a,b)=g&implies g|a,bland (forall h:h|a,bimplies hle g)\&implies g|b,(a-b)land (forall h:h|b,(a-b)implies h|a,bimplies hle g)\&implies gcd(b,a-b)=g.end{align}$$
$endgroup$
add a comment |
$begingroup$
By definition
$$gcd(a,b)=giff g|a,bland (forall h:h|a,bimplies hle g).$$
Then
$$begin{align}gcd(a,b)=g&implies g|a,bland (forall h:h|a,bimplies hle g)\&implies g|b,(a-b)land (forall h:h|b,(a-b)implies h|a,bimplies hle g)\&implies gcd(b,a-b)=g.end{align}$$
$endgroup$
By definition
$$gcd(a,b)=giff g|a,bland (forall h:h|a,bimplies hle g).$$
Then
$$begin{align}gcd(a,b)=g&implies g|a,bland (forall h:h|a,bimplies hle g)\&implies g|b,(a-b)land (forall h:h|b,(a-b)implies h|a,bimplies hle g)\&implies gcd(b,a-b)=g.end{align}$$
answered Oct 24 '14 at 7:35
Yves DaoustYves Daoust
129k675227
129k675227
add a comment |
add a comment |
$begingroup$
Every common divisor of $a$ and $b$ is also a common divisor of $b$ and $a-b$.
Every common divisor of $b$ and $a-b$ is also a common divisor of $a$ and $b$, as $a=(a-b)+b$.
$endgroup$
$begingroup$
This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
$endgroup$
– Yves Daoust
Oct 26 '14 at 21:00
add a comment |
$begingroup$
Every common divisor of $a$ and $b$ is also a common divisor of $b$ and $a-b$.
Every common divisor of $b$ and $a-b$ is also a common divisor of $a$ and $b$, as $a=(a-b)+b$.
$endgroup$
$begingroup$
This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
$endgroup$
– Yves Daoust
Oct 26 '14 at 21:00
add a comment |
$begingroup$
Every common divisor of $a$ and $b$ is also a common divisor of $b$ and $a-b$.
Every common divisor of $b$ and $a-b$ is also a common divisor of $a$ and $b$, as $a=(a-b)+b$.
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Every common divisor of $a$ and $b$ is also a common divisor of $b$ and $a-b$.
Every common divisor of $b$ and $a-b$ is also a common divisor of $a$ and $b$, as $a=(a-b)+b$.
answered Oct 24 '14 at 7:48
Yves DaoustYves Daoust
129k675227
129k675227
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This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
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– Yves Daoust
Oct 26 '14 at 21:00
add a comment |
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This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
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– Yves Daoust
Oct 26 '14 at 21:00
$begingroup$
This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
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– Yves Daoust
Oct 26 '14 at 21:00
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This answer moves the problem from $gcd$ to divisors. $d$ divides $a$ and $b$ implies $d$ divides $apm b$ is a rather trivial property. ($a=md,b=ndimplies apm b=(mpm n)d$.)
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– Yves Daoust
Oct 26 '14 at 21:00
add a comment |
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$dmid a-b$ means $dfrac{a-b}d=dfrac ad-dfrac bd$ is an integer. It's clear that if one of these fractions is an integer, then so is the other, i.e., $dmid aiff dmid b$.
$dmidgcd(b,a-b)implies dmid b$, thus $dmid a$.
Therefore a divisor of both $a-b$ and $b$ is also a divisor of $a$.
Conversely, $kmid a$ means $dfrac ak=dfrac{a-b}k+dfrac bk$ is an integer, so $kmid a-biff kmid b$.
$kmidgcd(a,b)implies kmid b$, thus $kmid a-b$.
Therefore a divisor of both $a$ and $b$ is also a divisor of $a-b$. QED
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add a comment |
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$dmid a-b$ means $dfrac{a-b}d=dfrac ad-dfrac bd$ is an integer. It's clear that if one of these fractions is an integer, then so is the other, i.e., $dmid aiff dmid b$.
$dmidgcd(b,a-b)implies dmid b$, thus $dmid a$.
Therefore a divisor of both $a-b$ and $b$ is also a divisor of $a$.
Conversely, $kmid a$ means $dfrac ak=dfrac{a-b}k+dfrac bk$ is an integer, so $kmid a-biff kmid b$.
$kmidgcd(a,b)implies kmid b$, thus $kmid a-b$.
Therefore a divisor of both $a$ and $b$ is also a divisor of $a-b$. QED
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add a comment |
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$dmid a-b$ means $dfrac{a-b}d=dfrac ad-dfrac bd$ is an integer. It's clear that if one of these fractions is an integer, then so is the other, i.e., $dmid aiff dmid b$.
$dmidgcd(b,a-b)implies dmid b$, thus $dmid a$.
Therefore a divisor of both $a-b$ and $b$ is also a divisor of $a$.
Conversely, $kmid a$ means $dfrac ak=dfrac{a-b}k+dfrac bk$ is an integer, so $kmid a-biff kmid b$.
$kmidgcd(a,b)implies kmid b$, thus $kmid a-b$.
Therefore a divisor of both $a$ and $b$ is also a divisor of $a-b$. QED
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$dmid a-b$ means $dfrac{a-b}d=dfrac ad-dfrac bd$ is an integer. It's clear that if one of these fractions is an integer, then so is the other, i.e., $dmid aiff dmid b$.
$dmidgcd(b,a-b)implies dmid b$, thus $dmid a$.
Therefore a divisor of both $a-b$ and $b$ is also a divisor of $a$.
Conversely, $kmid a$ means $dfrac ak=dfrac{a-b}k+dfrac bk$ is an integer, so $kmid a-biff kmid b$.
$kmidgcd(a,b)implies kmid b$, thus $kmid a-b$.
Therefore a divisor of both $a$ and $b$ is also a divisor of $a-b$. QED
edited Jan 29 '15 at 9:58
answered Oct 25 '14 at 22:32
Jaycob ColemanJaycob Coleman
413524
413524
add a comment |
add a comment |
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Write $u=b$ and $v=a-b$. Then $a=u+v$ and $b=u$.
These two sets of equations imply that $d$ divides $u,v$ iff $d$ divides $a,b$. Therefore, $gcd(u,v)=gcd(a,b)$.
In algebra terms, the equations imply that ${a,b}$ and ${u,v}$ generate the same subgroup of $mathbb Z$. It's well known that $langle x,y rangle = gcd(x,y)mathbb Z$, hence the result.
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add a comment |
$begingroup$
Write $u=b$ and $v=a-b$. Then $a=u+v$ and $b=u$.
These two sets of equations imply that $d$ divides $u,v$ iff $d$ divides $a,b$. Therefore, $gcd(u,v)=gcd(a,b)$.
In algebra terms, the equations imply that ${a,b}$ and ${u,v}$ generate the same subgroup of $mathbb Z$. It's well known that $langle x,y rangle = gcd(x,y)mathbb Z$, hence the result.
$endgroup$
add a comment |
$begingroup$
Write $u=b$ and $v=a-b$. Then $a=u+v$ and $b=u$.
These two sets of equations imply that $d$ divides $u,v$ iff $d$ divides $a,b$. Therefore, $gcd(u,v)=gcd(a,b)$.
In algebra terms, the equations imply that ${a,b}$ and ${u,v}$ generate the same subgroup of $mathbb Z$. It's well known that $langle x,y rangle = gcd(x,y)mathbb Z$, hence the result.
$endgroup$
Write $u=b$ and $v=a-b$. Then $a=u+v$ and $b=u$.
These two sets of equations imply that $d$ divides $u,v$ iff $d$ divides $a,b$. Therefore, $gcd(u,v)=gcd(a,b)$.
In algebra terms, the equations imply that ${a,b}$ and ${u,v}$ generate the same subgroup of $mathbb Z$. It's well known that $langle x,y rangle = gcd(x,y)mathbb Z$, hence the result.
answered Jan 29 '15 at 10:38
lhflhf
166k10171396
166k10171396
add a comment |
add a comment |
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