If $2$ numbers are co-prime, does it imply that their difference is also prime to those numbers?

Clash Royale CLAN TAG#URR8PPP

up vote
6
down vote

favorite

1

Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.

If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?

elementary-number-theory coprime

share|cite|improve this question

edited Aug 7 at 0:21

Daniel Buck

2,3341624

asked Aug 6 at 18:15

Dean Yang

1388

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
  – DanielWainfleet
  Aug 6 at 20:27
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

1

Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.

If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?

elementary-number-theory coprime

share|cite|improve this question

edited Aug 7 at 0:21

Daniel Buck

2,3341624

asked Aug 6 at 18:15

Dean Yang

1388

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
  – DanielWainfleet
  Aug 6 at 20:27
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

1

up vote
6
down vote

favorite

1

1

Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.

If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?

elementary-number-theory coprime

share|cite|improve this question

edited Aug 7 at 0:21

Daniel Buck

2,3341624

asked Aug 6 at 18:15

Dean Yang

1388

Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.

If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?

elementary-number-theory coprime

share|cite|improve this question

edited Aug 7 at 0:21

Daniel Buck

2,3341624

asked Aug 6 at 18:15

Dean Yang

1388

share|cite|improve this question

share|cite|improve this question

edited Aug 7 at 0:21

Daniel Buck

2,3341624

edited Aug 7 at 0:21

Daniel Buck

2,3341624

edited Aug 7 at 0:21

Daniel Buck

2,3341624

2,3341624

asked Aug 6 at 18:15

Dean Yang

1388

asked Aug 6 at 18:15

Dean Yang

1388

asked Aug 6 at 18:15

Dean Yang

1388

1388

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
  – DanielWainfleet
  Aug 6 at 20:27
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
  – DanielWainfleet
  Aug 6 at 20:27
  
  

  
  
  
  

1

1

Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
– DanielWainfleet
Aug 6 at 20:27

Yes but perhaps you should explain why: If $d/q$ $could $ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q' $ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, pm (pn+q)$ are co-prime for any $nin Bbb Z.$
– DanielWainfleet
Aug 6 at 20:27

add a comment | 

2 Answers
2

active

oldest

votes

up vote
16
down vote

Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.

share|cite|improve this answer

answered Aug 6 at 18:19

Ross Millikan

276k21187352

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, thanks :)
  – Dean Yang
  Aug 6 at 18:20
  

  
  
  
  

add a comment | 

up vote
4
down vote

Of course it is. If $q$ and $d$ have a common factor $f$:

$q = Qf$

$d = Df$

then

$p = q + d = (Q +D)f$

and more generally

$kqpm md = (kQ pm mD)f$

will have that factor, too.

share|cite|improve this answer

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,93311044

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874176%2fif-2-numbers-are-co-prime-does-it-imply-that-their-difference-is-also-prime-t%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
16
down vote

Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.

share|cite|improve this answer

answered Aug 6 at 18:19

Ross Millikan

276k21187352

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, thanks :)
  – Dean Yang
  Aug 6 at 18:20
  

  
  
  
  

add a comment | 

up vote
16
down vote

Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.

share|cite|improve this answer

answered Aug 6 at 18:19

Ross Millikan

276k21187352

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, thanks :)
  – Dean Yang
  Aug 6 at 18:20
  

  
  
  
  

add a comment | 

up vote
16
down vote

up vote
16
down vote

Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.

share|cite|improve this answer

answered Aug 6 at 18:19

Ross Millikan

276k21187352

Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.

share|cite|improve this answer

answered Aug 6 at 18:19

Ross Millikan

276k21187352

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 18:19

Ross Millikan

276k21187352

answered Aug 6 at 18:19

Ross Millikan

276k21187352

answered Aug 6 at 18:19

Ross Millikan

276k21187352

276k21187352

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, thanks :)
  – Dean Yang
  Aug 6 at 18:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, thanks :)
  – Dean Yang
  Aug 6 at 18:20
  

  
  
  
  

Alright, thanks :)
– Dean Yang
Aug 6 at 18:20

Alright, thanks :)
– Dean Yang
Aug 6 at 18:20

add a comment | 

up vote
4
down vote

Of course it is. If $q$ and $d$ have a common factor $f$:

$q = Qf$

$d = Df$

then

$p = q + d = (Q +D)f$

and more generally

$kqpm md = (kQ pm mD)f$

will have that factor, too.

share|cite|improve this answer

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,93311044

add a comment | 

up vote
4
down vote

Of course it is. If $q$ and $d$ have a common factor $f$:

$q = Qf$

$d = Df$

then

$p = q + d = (Q +D)f$

and more generally

$kqpm md = (kQ pm mD)f$

will have that factor, too.

share|cite|improve this answer

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,93311044

add a comment | 

up vote
4
down vote

up vote
4
down vote

Of course it is. If $q$ and $d$ have a common factor $f$:

$q = Qf$

$d = Df$

then

$p = q + d = (Q +D)f$

and more generally

$kqpm md = (kQ pm mD)f$

will have that factor, too.

share|cite|improve this answer

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,93311044

Of course it is. If $q$ and $d$ have a common factor $f$:

$q = Qf$

$d = Df$

then

$p = q + d = (Q +D)f$

and more generally

$kqpm md = (kQ pm mD)f$

will have that factor, too.

share|cite|improve this answer

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,93311044

share|cite|improve this answer

share|cite|improve this answer

edited Aug 8 at 10:25

edited Aug 8 at 10:25

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,93311044

answered Aug 6 at 18:21

CiaPan

9,93311044

answered Aug 6 at 18:21

CiaPan

9,93311044

9,93311044

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874176%2fif-2-numbers-are-co-prime-does-it-imply-that-their-difference-is-also-prime-t%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email