Prove this function is injective and determine its image
$begingroup$
Fix three distinct primes p, q, r,
prove that the map
$Z_{pqr} → Z_{pq} × Z_{qr} × Z_{pr}$ by $[x]_{pqr}$ → ($[x]_{pq}$, $[x]_{qr}$, $[x]_{pr}$)
is injective and determine its image.
My attempt:
To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x-y)$
So, one of $p$ or $q$ must divide $(x-y)$.
Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x-y$ as well. Therefore, one of p,q,r must divide $x-y$.
This implies $pqr|(x-y)$,
So, $[x]_{pqr}=[y]_{pqr}$
This proves that the function is injective (if I'm correct in my implications).
Now, how do I determine its image?
elementary-number-theory modular-arithmetic
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
$endgroup$
add a comment |
$begingroup$
Fix three distinct primes p, q, r,
prove that the map
$Z_{pqr} → Z_{pq} × Z_{qr} × Z_{pr}$ by $[x]_{pqr}$ → ($[x]_{pq}$, $[x]_{qr}$, $[x]_{pr}$)
is injective and determine its image.
My attempt:
To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x-y)$
So, one of $p$ or $q$ must divide $(x-y)$.
Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x-y$ as well. Therefore, one of p,q,r must divide $x-y$.
This implies $pqr|(x-y)$,
So, $[x]_{pqr}=[y]_{pqr}$
This proves that the function is injective (if I'm correct in my implications).
Now, how do I determine its image?
elementary-number-theory modular-arithmetic
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
$endgroup$
$begingroup$
From the fact that one of $p,q,r$ divide $x-y$ you can't conclude $pqr$ divide $x-y$
$endgroup$
– jjagmath
Dec 14 '18 at 4:51
add a comment |
$begingroup$
Fix three distinct primes p, q, r,
prove that the map
$Z_{pqr} → Z_{pq} × Z_{qr} × Z_{pr}$ by $[x]_{pqr}$ → ($[x]_{pq}$, $[x]_{qr}$, $[x]_{pr}$)
is injective and determine its image.
My attempt:
To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x-y)$
So, one of $p$ or $q$ must divide $(x-y)$.
Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x-y$ as well. Therefore, one of p,q,r must divide $x-y$.
This implies $pqr|(x-y)$,
So, $[x]_{pqr}=[y]_{pqr}$
This proves that the function is injective (if I'm correct in my implications).
Now, how do I determine its image?
elementary-number-theory modular-arithmetic
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
$endgroup$
Fix three distinct primes p, q, r,
prove that the map
$Z_{pqr} → Z_{pq} × Z_{qr} × Z_{pr}$ by $[x]_{pqr}$ → ($[x]_{pq}$, $[x]_{qr}$, $[x]_{pr}$)
is injective and determine its image.
My attempt:
To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x-y)$
So, one of $p$ or $q$ must divide $(x-y)$.
Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x-y$ as well. Therefore, one of p,q,r must divide $x-y$.
This implies $pqr|(x-y)$,
So, $[x]_{pqr}=[y]_{pqr}$
This proves that the function is injective (if I'm correct in my implications).
Now, how do I determine its image?
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
asked Dec 12 '18 at 16:00
childishsadbinochildishsadbino
1148
1148
$begingroup$
From the fact that one of $p,q,r$ divide $x-y$ you can't conclude $pqr$ divide $x-y$
$endgroup$
– jjagmath
Dec 14 '18 at 4:51
add a comment |
$begingroup$
From the fact that one of $p,q,r$ divide $x-y$ you can't conclude $pqr$ divide $x-y$
$endgroup$
– jjagmath
Dec 14 '18 at 4:51
$begingroup$
From the fact that one of $p,q,r$ divide $x-y$ you can't conclude $pqr$ divide $x-y$
$endgroup$
– jjagmath
Dec 14 '18 at 4:51
$begingroup$
From the fact that one of $p,q,r$ divide $x-y$ you can't conclude $pqr$ divide $x-y$
$endgroup$
– jjagmath
Dec 14 '18 at 4:51
add a comment |
1 Answer
1
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$begingroup$
$pq mid x-y$, so we have $p mid x-y$ AND $qmid x-y$
$qr mid x-y$, so ALSO $rmid x-y$
As $p$, $q$ AND $r$ divide $x-y$ and are distinct primes, we have $pqr mid x-y$
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$pq mid x-y$, so we have $p mid x-y$ AND $qmid x-y$
$qr mid x-y$, so ALSO $rmid x-y$
As $p$, $q$ AND $r$ divide $x-y$ and are distinct primes, we have $pqr mid x-y$
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
$endgroup$
add a comment |
$begingroup$
$pq mid x-y$, so we have $p mid x-y$ AND $qmid x-y$
$qr mid x-y$, so ALSO $rmid x-y$
As $p$, $q$ AND $r$ divide $x-y$ and are distinct primes, we have $pqr mid x-y$
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
$endgroup$
add a comment |
$begingroup$
$pq mid x-y$, so we have $p mid x-y$ AND $qmid x-y$
$qr mid x-y$, so ALSO $rmid x-y$
As $p$, $q$ AND $r$ divide $x-y$ and are distinct primes, we have $pqr mid x-y$
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
$endgroup$
$pq mid x-y$, so we have $p mid x-y$ AND $qmid x-y$
$qr mid x-y$, so ALSO $rmid x-y$
As $p$, $q$ AND $r$ divide $x-y$ and are distinct primes, we have $pqr mid x-y$
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
answered Dec 14 '18 at 5:07
jjagmathjjagmath
3387
3387
add a comment |
add a comment |
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$begingroup$
From the fact that one of $p,q,r$ divide $x-y$ you can't conclude $pqr$ divide $x-y$
$endgroup$
– jjagmath
Dec 14 '18 at 4:51