Prove that $gcd(a,b) = gcd(b,a)$

2

$begingroup$

I want to prove $gcd(a,b)=gcd(b,a)$. I tried using the euclidean algorithm but that didn't help me much.

number-theory greatest-common-divisor

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edited Dec 14 '15 at 15:37

user98602

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

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• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What definition of $gcd(a,b)$ do you operate with? Mine is obviously symmetric.
  $endgroup$
  – Abstraction
  Dec 14 '15 at 15:31
  

  
  
  
  

add a comment | 

2

$begingroup$

I want to prove $gcd(a,b)=gcd(b,a)$. I tried using the euclidean algorithm but that didn't help me much.

number-theory greatest-common-divisor

share|cite|improve this question

edited Dec 14 '15 at 15:37

user98602

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What definition of $gcd(a,b)$ do you operate with? Mine is obviously symmetric.
  $endgroup$
  – Abstraction
  Dec 14 '15 at 15:31
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

I want to prove $gcd(a,b)=gcd(b,a)$. I tried using the euclidean algorithm but that didn't help me much.

number-theory greatest-common-divisor

share|cite|improve this question

edited Dec 14 '15 at 15:37

user98602

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

$endgroup$

I want to prove $gcd(a,b)=gcd(b,a)$. I tried using the euclidean algorithm but that didn't help me much.

number-theory greatest-common-divisor

number-theory greatest-common-divisor

share|cite|improve this question

edited Dec 14 '15 at 15:37

user98602

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

share|cite|improve this question

edited Dec 14 '15 at 15:37

user98602

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

share|cite|improve this question

share|cite|improve this question

edited Dec 14 '15 at 15:37

user98602

edited Dec 14 '15 at 15:37

user98602

edited Dec 14 '15 at 15:37

user98602

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

asked Dec 14 '15 at 15:28

GeoffGeoff

221212

221212

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What definition of $gcd(a,b)$ do you operate with? Mine is obviously symmetric.
  $endgroup$
  – Abstraction
  Dec 14 '15 at 15:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What definition of $gcd(a,b)$ do you operate with? Mine is obviously symmetric.
  $endgroup$
  – Abstraction
  Dec 14 '15 at 15:31
  

  
  
  
  

3

3

$begingroup$
What definition of $gcd(a,b)$ do you operate with? Mine is obviously symmetric.
$endgroup$
– Abstraction
Dec 14 '15 at 15:31

$begingroup$
What definition of $gcd(a,b)$ do you operate with? Mine is obviously symmetric.
$endgroup$
– Abstraction
Dec 14 '15 at 15:31

add a comment | 

2 Answers
2

active

oldest

votes

4

$begingroup$

If $gcd(a,b)=x$ then $x mid a$ and $xmid b$ this implies $x mid gcd(b,a)$. By similar argument $gcd(b,a)midgcd(a,b).$ i.e., $gcd(a,b) = gcd(b,a). $

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
  $endgroup$
  – Geoff
  Dec 14 '15 at 15:36
  

  
  
  
  

add a comment | 

2

$begingroup$

A divisor of $a$ and $b$ is a number $c$ such that $cmid a$
and $cmid b$. Note that a divisor of $a$ and $b$ is trivially a divisor
of $b$ and $a$.

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

4

$begingroup$

If $gcd(a,b)=x$ then $x mid a$ and $xmid b$ this implies $x mid gcd(b,a)$. By similar argument $gcd(b,a)midgcd(a,b).$ i.e., $gcd(a,b) = gcd(b,a). $

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
  $endgroup$
  – Geoff
  Dec 14 '15 at 15:36
  

  
  
  
  

add a comment | 

4

$begingroup$

If $gcd(a,b)=x$ then $x mid a$ and $xmid b$ this implies $x mid gcd(b,a)$. By similar argument $gcd(b,a)midgcd(a,b).$ i.e., $gcd(a,b) = gcd(b,a). $

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
  $endgroup$
  – Geoff
  Dec 14 '15 at 15:36
  

  
  
  
  

add a comment | 

4

4

4

$begingroup$

If $gcd(a,b)=x$ then $x mid a$ and $xmid b$ this implies $x mid gcd(b,a)$. By similar argument $gcd(b,a)midgcd(a,b).$ i.e., $gcd(a,b) = gcd(b,a). $

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

$endgroup$

If $gcd(a,b)=x$ then $x mid a$ and $xmid b$ this implies $x mid gcd(b,a)$. By similar argument $gcd(b,a)midgcd(a,b).$ i.e., $gcd(a,b) = gcd(b,a). $

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

share|cite|improve this answer

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

edited Dec 14 '15 at 15:46

Michael Hardy

1

edited Dec 14 '15 at 15:46

Michael Hardy

1

1

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

answered Dec 14 '15 at 15:33

SuhailSuhail

1,791514

1,791514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
  $endgroup$
  – Geoff
  Dec 14 '15 at 15:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
  $endgroup$
  – Geoff
  Dec 14 '15 at 15:36
  

  
  
  
  

$begingroup$
ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
$endgroup$
– Geoff
Dec 14 '15 at 15:36

$begingroup$
ah thank you, i got very close to that argument but i couldn't quite put the last pieces together. I often struggle with these proofs.
$endgroup$
– Geoff
Dec 14 '15 at 15:36

add a comment | 

2

$begingroup$

A divisor of $a$ and $b$ is a number $c$ such that $cmid a$
and $cmid b$. Note that a divisor of $a$ and $b$ is trivially a divisor
of $b$ and $a$.

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

$endgroup$

add a comment | 

2

$begingroup$

A divisor of $a$ and $b$ is a number $c$ such that $cmid a$
and $cmid b$. Note that a divisor of $a$ and $b$ is trivially a divisor
of $b$ and $a$.

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

$endgroup$

add a comment | 

2

2

2

$begingroup$

A divisor of $a$ and $b$ is a number $c$ such that $cmid a$
and $cmid b$. Note that a divisor of $a$ and $b$ is trivially a divisor
of $b$ and $a$.

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

$endgroup$

A divisor of $a$ and $b$ is a number $c$ such that $cmid a$
and $cmid b$. Note that a divisor of $a$ and $b$ is trivially a divisor
of $b$ and $a$.

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

share|cite|improve this answer

share|cite|improve this answer

edited Dec 14 '15 at 15:46

Michael Hardy

1

edited Dec 14 '15 at 15:46

Michael Hardy

1

edited Dec 14 '15 at 15:46

Michael Hardy

1

1

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

answered Dec 14 '15 at 15:33

parsiadparsiad

18.6k32453

18.6k32453

add a comment | 

add a comment | 

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