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

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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?

edited Aug 7 at 0:21

Daniel Buck

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asked Aug 6 at 18:15

Dean Yang

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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 $nin Bbb Z.$
â€“Â DanielWainfleet
Aug 6 at 20:27

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up vote
6
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Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.

edited Aug 7 at 0:21

Daniel Buck

2,3041623

asked Aug 6 at 18:15

Dean Yang

1115

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 $nin Bbb Z.$
â€“Â DanielWainfleet
Aug 6 at 20:27

add a commentÂ |Â

up vote
6
down vote

favorite

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

edited Aug 7 at 0:21

Daniel Buck

2,3041623

asked Aug 6 at 18:15

Dean Yang

1115

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

edited Aug 7 at 0:21

Daniel Buck

2,3041623

asked Aug 6 at 18:15

Dean Yang

1115

edited Aug 7 at 0:21

Daniel Buck

2,3041623

edited Aug 7 at 0:21

Daniel Buck

2,3041623

edited Aug 7 at 0:21

Daniel Buck

2,3041623

asked Aug 6 at 18:15

Dean Yang

1115

asked Aug 6 at 18:15

Dean Yang

1115

asked Aug 6 at 18:15

Dean Yang

1115

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 $nin Bbb Z.$
â€“Â DanielWainfleet
Aug 6 at 20:27

add a commentÂ |Â

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 $nin Bbb Z.$
â€“Â 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 $nin Bbb Z.$
â€“Â DanielWainfleet
Aug 6 at 20:27

add a commentÂ |Â

2 Answers
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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.

answered Aug 6 at 18:19

Ross Millikan

275k21186351

Alright, thanks :)
â€“Â Dean Yang
Aug 6 at 18:20

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4
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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.

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,85311044

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2 Answers
2

active

oldest

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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.

answered Aug 6 at 18:19

Ross Millikan

275k21186351

Alright, thanks :)
â€“Â Dean Yang
Aug 6 at 18:20

add a commentÂ |Â

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.

answered Aug 6 at 18:19

Ross Millikan

275k21186351

Alright, thanks :)
â€“Â Dean Yang
Aug 6 at 18:20

add a commentÂ |Â

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.

answered Aug 6 at 18:19

Ross Millikan

275k21186351

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.

answered Aug 6 at 18:19

Ross Millikan

275k21186351

answered Aug 6 at 18:19

Ross Millikan

275k21186351

answered Aug 6 at 18:19

Ross Millikan

275k21186351

answered Aug 6 at 18:19

Ross Millikan

275k21186351

Alright, thanks :)
â€“Â Dean Yang
Aug 6 at 18:20

add a commentÂ |Â

Alright, thanks :)
â€“Â Dean Yang
Aug 6 at 18:20

Alright, thanks :)
â€“Â Dean Yang
Aug 6 at 18:20

add a commentÂ |Â

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.

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,85311044

add a commentÂ |Â

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.

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,85311044

add a commentÂ |Â

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.

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,85311044

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.

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,85311044

edited Aug 8 at 10:25

answered Aug 6 at 18:21

CiaPan

9,85311044

answered Aug 6 at 18:21

CiaPan

9,85311044

answered Aug 6 at 18:21

CiaPan

9,85311044

add a commentÂ |Â

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