Proof that $mathbb Z[sqrt{3}]$ is a Euclidean Domain
$begingroup$
Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
$$omega =
begin{cases}
sqrt{d}, & text{if } quad d not equiv 1mod 4 \
frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
end{cases}$$
It has been proven that $R_d$ is Euclidean for several positive values of $d$.
Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?
Thank you.
abstract-algebra reference-request ring-theory
$endgroup$
add a comment |
$begingroup$
Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
$$omega =
begin{cases}
sqrt{d}, & text{if } quad d not equiv 1mod 4 \
frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
end{cases}$$
It has been proven that $R_d$ is Euclidean for several positive values of $d$.
Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?
Thank you.
abstract-algebra reference-request ring-theory
$endgroup$
add a comment |
$begingroup$
Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
$$omega =
begin{cases}
sqrt{d}, & text{if } quad d not equiv 1mod 4 \
frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
end{cases}$$
It has been proven that $R_d$ is Euclidean for several positive values of $d$.
Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?
Thank you.
abstract-algebra reference-request ring-theory
$endgroup$
Let $R_d$ be the ring defined as $R_d=left { x+yomega : x,yin mathbb{Z} right}$, where
$$omega =
begin{cases}
sqrt{d}, & text{if } quad d not equiv 1mod 4 \
frac{1+sqrt{d}}{2}, & text{if } quad dequiv 1mod 4.
end{cases}$$
It has been proven that $R_d$ is Euclidean for several positive values of $d$.
Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?
Thank you.
abstract-algebra reference-request ring-theory
abstract-algebra reference-request ring-theory
edited Feb 18 '15 at 4:45
Robert Cardona
5,392234102
5,392234102
asked Oct 7 '13 at 3:57
SarahSarah
10719
10719
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.
Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.
Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.
Notice that
begin{align*}
fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
&= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
&= r + ssqrt 3
end{align*}
where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.
We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)
Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
begin{align*}
gamma &= beta cdot theta\
&= beta ( (r - p) + (s - q)sqrt 3)\
&= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
&= beta cdotfracalphabeta - beta (p + qsqrt 3) \
&= alpha - beta (p + qsqrt 3)
end{align*}
Hence we have $alpha = beta(p + qsqrt 3) + gamma$.
Finally notice that
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
&leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
& leqfrac34 N(beta)\
&< N(beta)
end{align*}
The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.
$endgroup$
1
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
2
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.
Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.
Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.
Notice that
begin{align*}
fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
&= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
&= r + ssqrt 3
end{align*}
where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.
We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)
Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
begin{align*}
gamma &= beta cdot theta\
&= beta ( (r - p) + (s - q)sqrt 3)\
&= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
&= beta cdotfracalphabeta - beta (p + qsqrt 3) \
&= alpha - beta (p + qsqrt 3)
end{align*}
Hence we have $alpha = beta(p + qsqrt 3) + gamma$.
Finally notice that
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
&leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
& leqfrac34 N(beta)\
&< N(beta)
end{align*}
The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.
$endgroup$
1
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
2
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
add a comment |
$begingroup$
Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.
Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.
Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.
Notice that
begin{align*}
fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
&= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
&= r + ssqrt 3
end{align*}
where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.
We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)
Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
begin{align*}
gamma &= beta cdot theta\
&= beta ( (r - p) + (s - q)sqrt 3)\
&= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
&= beta cdotfracalphabeta - beta (p + qsqrt 3) \
&= alpha - beta (p + qsqrt 3)
end{align*}
Hence we have $alpha = beta(p + qsqrt 3) + gamma$.
Finally notice that
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
&leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
& leqfrac34 N(beta)\
&< N(beta)
end{align*}
The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.
$endgroup$
1
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
2
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
add a comment |
$begingroup$
Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.
Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.
Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.
Notice that
begin{align*}
fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
&= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
&= r + ssqrt 3
end{align*}
where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.
We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)
Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
begin{align*}
gamma &= beta cdot theta\
&= beta ( (r - p) + (s - q)sqrt 3)\
&= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
&= beta cdotfracalphabeta - beta (p + qsqrt 3) \
&= alpha - beta (p + qsqrt 3)
end{align*}
Hence we have $alpha = beta(p + qsqrt 3) + gamma$.
Finally notice that
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
&leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
& leqfrac34 N(beta)\
&< N(beta)
end{align*}
The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.
$endgroup$
Define the norm on $mathbb Z[sqrt 3]$ to be $N(a + b sqrt 3) = vert a^2 - 3 b^2 vert$.
Let $alpha, beta in mathbb Z[sqrt 3]$ with $beta neq 0$.
Say $alpha = a + b sqrt 3$ and $beta = c + d sqrt 3$.
Notice that
begin{align*}
fracalphabeta &= frac{a + b sqrt 3}{c + d sqrt 3} cdot frac{c - d sqrt 3}{c - d sqrt 3} \
&= frac{ac - 3bd}{c^2 - 3d^2} + frac{-ad + bc}{c^2 - 3d^2} sqrt 3 \
&= r + ssqrt 3
end{align*}
where $r = displaystyle frac{ac - 3bd}{c^2 - 3d^2}$ and $s = displaystyle frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $vert r - p vert leq 1/2$ and $vert s - q vert leq 1/2$.
We want to show that $alpha = (p + qsqrt 3) beta + gamma$ for some $gamma in mathbb Z[sqrt 3]$ such that either $gamma = 0$ or $N(gamma) < N(beta)$. (We'll show the latter holds always.)
Define $theta := (r - p) + (s - q)sqrt 3$ and define $gamma = beta cdot theta in mathbb Z[sqrt 3]$ and observe that
begin{align*}
gamma &= beta cdot theta\
&= beta ( (r - p) + (s - q)sqrt 3)\
&= beta (r + ssqrt 3) - beta(p + qsqrt 3) \
&= beta cdotfracalphabeta - beta (p + qsqrt 3) \
&= alpha - beta (p + qsqrt 3)
end{align*}
Hence we have $alpha = beta(p + qsqrt 3) + gamma$.
Finally notice that
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (r - p)^2 - 3 (s - q)^2 vert \
&leq N(beta) cdot max{ (r - p)^2, 3(s - q)^2} \
& leqfrac34 N(beta)\
&< N(beta)
end{align*}
The key here was that $vert (r - p)^2 - 3 (s - q)^2 vert leq max{ (r - p)^2, 3(s - q)^2}$ since $(r - p)^2, 3(s - q)^2 geq 0$ and then we use that $(r - p)^2 leq 1/4$ and $3(s - q)^2 leq 3/4$.
edited Jan 16 at 7:10
user549397
1,6611418
1,6611418
answered Feb 18 '15 at 4:43
Robert CardonaRobert Cardona
5,392234102
5,392234102
1
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
2
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
add a comment |
1
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
2
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
1
1
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
$begingroup$
When you're proving that $N(gamma) < N(beta)$, you say that $N(theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $Bbb{Z}[sqrt{3}]$. You showed that $(beta)(theta) in Bbb{Z}[sqrt{3}]$. But how does this imply that $theta in Bbb{Z}[sqrt{3}]$?
$endgroup$
– Artus
Mar 12 '15 at 14:05
2
2
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
That's a very good question! We can just think of the norm $N$ on $mathbb Z[sqrt 3]$ as being the norm on $mathbb Q[sqrt 3]$, restricted to $mathbb Z[sqrt 3]$.
$endgroup$
– Robert Cardona
Mar 12 '15 at 15:13
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
why is |p - r |< 1/2 and |q- s| < 1/2
$endgroup$
– user10024395
Apr 23 '15 at 9:33
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = frac14$, the closest integer is $0$ and the difference is less than $frac12$, if $p = frac34$, then the closest integer is $1$. If $p = frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to).
$endgroup$
– Robert Cardona
Apr 23 '15 at 14:21
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
$begingroup$
@RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2.
$endgroup$
– user10024395
Apr 24 '15 at 1:45
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