If $c | b(x, y, z,…)$, does $c | b$?
$begingroup$
If $c$ divides something like $bxcd + bhdwou + bn$, does $ c | b$?
I'm confused because say that's true. Then let $c = 5$ and $b = 6$.
Then $c | b(10)$, but $5$ does not divide $6$.
So it this false?
If it is false, how come the solution to this, $c|abx + cby$ is true in my textbook?
edit: My bad, I was missing some info from the question, I'll delete soon
elementary-number-theory
edited Dec 9 '18 at 5:58
Shaun
9,010113682
asked Dec 9 '18 at 5:47
mingming
3365
$endgroup$
add a comment |
$begingroup$
If $c$ divides something like $bxcd + bhdwou + bn$, does $ c | b$?
I'm confused because say that's true. Then let $c = 5$ and $b = 6$.
Then $c | b(10)$, but $5$ does not divide $6$.
So it this false?
If it is false, how come the solution to this, $c|abx + cby$ is true in my textbook?
edit: My bad, I was missing some info from the question, I'll delete soon
elementary-number-theory
edited Dec 9 '18 at 5:58
Shaun
9,010113682
asked Dec 9 '18 at 5:47
mingming
3365
$endgroup$
add a comment |
$begingroup$
If $c$ divides something like $bxcd + bhdwou + bn$, does $ c | b$?
I'm confused because say that's true. Then let $c = 5$ and $b = 6$.
Then $c | b(10)$, but $5$ does not divide $6$.
So it this false?
If it is false, how come the solution to this, $c|abx + cby$ is true in my textbook?
edit: My bad, I was missing some info from the question, I'll delete soon
elementary-number-theory
edited Dec 9 '18 at 5:58
Shaun
9,010113682
asked Dec 9 '18 at 5:47
mingming
3365
$endgroup$
If $c$ divides something like $bxcd + bhdwou + bn$, does $ c | b$?
I'm confused because say that's true. Then let $c = 5$ and $b = 6$.
Then $c | b(10)$, but $5$ does not divide $6$.
So it this false?
If it is false, how come the solution to this, $c|abx + cby$ is true in my textbook?
edit: My bad, I was missing some info from the question, I'll delete soon
elementary-number-theory
elementary-number-theory
edited Dec 9 '18 at 5:58
Shaun
9,010113682
asked Dec 9 '18 at 5:47
mingming
3365
edited Dec 9 '18 at 5:58
Shaun
9,010113682
asked Dec 9 '18 at 5:47
mingming
3365
edited Dec 9 '18 at 5:58
Shaun
9,010113682
edited Dec 9 '18 at 5:58
Shaun
9,010113682
edited Dec 9 '18 at 5:58
Shaun
9,010113682
9,010113682
asked Dec 9 '18 at 5:47
mingming
3365
asked Dec 9 '18 at 5:47
mingming
3365
asked Dec 9 '18 at 5:47
mingming
3365
3365
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Let $pinBbb N$. Then $p$ is prime if and only if for all $a,binBbb N$, whenever $pmid ab$, we have either $pmid a$ or $pmid b$.
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
$endgroup$
add a comment |
$begingroup$
If $(xcd + hdwou + n,c)=1$,it's true.
More precisely, if $a|bc$ and $(a,c)=1$,then $a|b $.
I can't comment about the textbook problem, as context is missing.
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $pinBbb N$. Then $p$ is prime if and only if for all $a,binBbb N$, whenever $pmid ab$, we have either $pmid a$ or $pmid b$.
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
$endgroup$
add a comment |
$begingroup$
Hint: Let $pinBbb N$. Then $p$ is prime if and only if for all $a,binBbb N$, whenever $pmid ab$, we have either $pmid a$ or $pmid b$.
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
$endgroup$
add a comment |
$begingroup$
Hint: Let $pinBbb N$. Then $p$ is prime if and only if for all $a,binBbb N$, whenever $pmid ab$, we have either $pmid a$ or $pmid b$.
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
$endgroup$
Hint: Let $pinBbb N$. Then $p$ is prime if and only if for all $a,binBbb N$, whenever $pmid ab$, we have either $pmid a$ or $pmid b$.
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
answered Dec 9 '18 at 5:55
ShaunShaun
9,010113682
9,010113682
add a comment |
add a comment |
$begingroup$
If $(xcd + hdwou + n,c)=1$,it's true.
More precisely, if $a|bc$ and $(a,c)=1$,then $a|b $.
I can't comment about the textbook problem, as context is missing.
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
add a comment |
$begingroup$
If $(xcd + hdwou + n,c)=1$,it's true.
More precisely, if $a|bc$ and $(a,c)=1$,then $a|b $.
I can't comment about the textbook problem, as context is missing.
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
add a comment |
$begingroup$
If $(xcd + hdwou + n,c)=1$,it's true.
More precisely, if $a|bc$ and $(a,c)=1$,then $a|b $.
I can't comment about the textbook problem, as context is missing.
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
If $(xcd + hdwou + n,c)=1$,it's true.
More precisely, if $a|bc$ and $(a,c)=1$,then $a|b $.
I can't comment about the textbook problem, as context is missing.
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
edited Dec 9 '18 at 6:06
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
answered Dec 9 '18 at 5:56
Thomas ShelbyThomas Shelby
2,710421
2,710421
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
add a comment |
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
$begingroup$
So the question was suppose $c|ab$ and $gcd(a, c) = 1$. Then $c = b$. So the solution did what I described up there but I'm actually still unsure why they said $c|abx + cby$ rather than $c|abx + cy$ like DIC actually says.
$endgroup$
– ming
Dec 9 '18 at 6:53
add a comment |
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