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?
elementary-number-theory coprime
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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?
elementary-number-theory coprime
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 DanielWainfleet
Aug 6 at 20:27
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up vote
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up vote
6
down vote
favorite
Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.
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
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
edited Aug 7 at 0:21
Daniel Buck
2,3041623
2,3041623
asked Aug 6 at 18:15
Dean Yang
1115
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 DanielWainfleet
Aug 6 at 20:27
add a comment |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 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 $nin Bbb Z.$
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 DanielWainfleet
Aug 6 at 20:27
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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.
Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
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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.
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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.
Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
add a comment |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.
Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
add a comment |Â
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.
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
275k21186351
Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
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Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
Alright, thanks :)
â Dean Yang
Aug 6 at 18:20
add a comment |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.
add a comment |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.
add a comment |Â
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.
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
9,85311044
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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 DanielWainfleet
Aug 6 at 20:27