Choice of a square root in $mathbb{Q}_7$
While studying a calculation in Koblitz's P-adic Numbers, P-adic Analysis, and Zeta-Functions, he remarks the following:
...we were sloppy when we wrote $4/3 = (1 + 7/9)^{1/2}$. In both $mathbb{R}$ and $mathbb{Q}_7$ the number $16/9$ has two square roots $pm 4/3$. In $mathbb{R}$, the series for $(1 + 7/9)^{1/2}$
converges to $4/3$, i.e., the positive value is favored. But in $mathbb{Q}_7$ the square root congruent to $1 text{ mod } 7$, i.e., $-4/3 = 1 - 7/3$, is favored.
So, my question is why is, in $mathbb{Q}_7$, the root $-4/3$ "favoured"?
p-adic-number-theory
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
add a comment |
While studying a calculation in Koblitz's P-adic Numbers, P-adic Analysis, and Zeta-Functions, he remarks the following:
...we were sloppy when we wrote $4/3 = (1 + 7/9)^{1/2}$. In both $mathbb{R}$ and $mathbb{Q}_7$ the number $16/9$ has two square roots $pm 4/3$. In $mathbb{R}$, the series for $(1 + 7/9)^{1/2}$
converges to $4/3$, i.e., the positive value is favored. But in $mathbb{Q}_7$ the square root congruent to $1 text{ mod } 7$, i.e., $-4/3 = 1 - 7/3$, is favored.
So, my question is why is, in $mathbb{Q}_7$, the root $-4/3$ "favoured"?
p-adic-number-theory
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
3
It's nothing more than noting that $-4/3equiv 1pmod{7}$, while $4/3 equiv -1pmod{7}$. Unlike $mathbb{R}$, the space $mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over.
– anomaly
Nov 12 '18 at 0:33
2
"favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$.
– reuns
Nov 12 '18 at 0:51
Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too.
– Lubin
Nov 24 '18 at 21:39
add a comment |
While studying a calculation in Koblitz's P-adic Numbers, P-adic Analysis, and Zeta-Functions, he remarks the following:
...we were sloppy when we wrote $4/3 = (1 + 7/9)^{1/2}$. In both $mathbb{R}$ and $mathbb{Q}_7$ the number $16/9$ has two square roots $pm 4/3$. In $mathbb{R}$, the series for $(1 + 7/9)^{1/2}$
converges to $4/3$, i.e., the positive value is favored. But in $mathbb{Q}_7$ the square root congruent to $1 text{ mod } 7$, i.e., $-4/3 = 1 - 7/3$, is favored.
So, my question is why is, in $mathbb{Q}_7$, the root $-4/3$ "favoured"?
p-adic-number-theory
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
While studying a calculation in Koblitz's P-adic Numbers, P-adic Analysis, and Zeta-Functions, he remarks the following:
...we were sloppy when we wrote $4/3 = (1 + 7/9)^{1/2}$. In both $mathbb{R}$ and $mathbb{Q}_7$ the number $16/9$ has two square roots $pm 4/3$. In $mathbb{R}$, the series for $(1 + 7/9)^{1/2}$
converges to $4/3$, i.e., the positive value is favored. But in $mathbb{Q}_7$ the square root congruent to $1 text{ mod } 7$, i.e., $-4/3 = 1 - 7/3$, is favored.
So, my question is why is, in $mathbb{Q}_7$, the root $-4/3$ "favoured"?
p-adic-number-theory
p-adic-number-theory
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
asked Nov 12 '18 at 0:27
Naweed G. Seldon
1,309419
1,309419
3
It's nothing more than noting that $-4/3equiv 1pmod{7}$, while $4/3 equiv -1pmod{7}$. Unlike $mathbb{R}$, the space $mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over.
– anomaly
Nov 12 '18 at 0:33
2
"favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$.
– reuns
Nov 12 '18 at 0:51
Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too.
– Lubin
Nov 24 '18 at 21:39
add a comment |
3
It's nothing more than noting that $-4/3equiv 1pmod{7}$, while $4/3 equiv -1pmod{7}$. Unlike $mathbb{R}$, the space $mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over.
– anomaly
Nov 12 '18 at 0:33
2
"favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$.
– reuns
Nov 12 '18 at 0:51
Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too.
– Lubin
Nov 24 '18 at 21:39
3
3
It's nothing more than noting that $-4/3equiv 1pmod{7}$, while $4/3 equiv -1pmod{7}$. Unlike $mathbb{R}$, the space $mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over.
– anomaly
Nov 12 '18 at 0:33
It's nothing more than noting that $-4/3equiv 1pmod{7}$, while $4/3 equiv -1pmod{7}$. Unlike $mathbb{R}$, the space $mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over.
– anomaly
Nov 12 '18 at 0:33
2
2
"favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$.
– reuns
Nov 12 '18 at 0:51
"favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$.
– reuns
Nov 12 '18 at 0:51
Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too.
– Lubin
Nov 24 '18 at 21:39
Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too.
– Lubin
Nov 24 '18 at 21:39
add a comment |
1 Answer
1
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Consider the polynomial $1+X$ in the ring $A=mathbb Q[[X]]$ of power series over $mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $mathbb Q[X]$ but it
has two solutions over the domain $A$. Mainly, $T = pmleft( 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dotsright)$
The "positive" or "favoured" one is
$$
B_{1/2}(X)= 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dots
$$
This is the Taylor expansion of $sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series.
It is easy to check that $B_{1/2}in mathbb Z[1/2][[X]] subset mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function
$$
B_{1/2} : pmathbb Z_p rightarrow 1+ pmathbb Z_p
$$
and also
$$
B_{1/2} : (-1,1)subset mathbb Rrightarrow [0,infty) subsetmathbb R.
$$
Koblitz points out that $7/9$ is in both $(-1,1)$ and $7mathbb Z_7$. Hence
$B_{1/2}(7/9) = 4/3$ as function of $(-1,1)subset mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7mathbb Z_7$.
answered Dec 2 '18 at 23:23
eduard
38417
add a comment |
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Consider the polynomial $1+X$ in the ring $A=mathbb Q[[X]]$ of power series over $mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $mathbb Q[X]$ but it
has two solutions over the domain $A$. Mainly, $T = pmleft( 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dotsright)$
The "positive" or "favoured" one is
$$
B_{1/2}(X)= 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dots
$$
This is the Taylor expansion of $sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series.
It is easy to check that $B_{1/2}in mathbb Z[1/2][[X]] subset mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function
$$
B_{1/2} : pmathbb Z_p rightarrow 1+ pmathbb Z_p
$$
and also
$$
B_{1/2} : (-1,1)subset mathbb Rrightarrow [0,infty) subsetmathbb R.
$$
Koblitz points out that $7/9$ is in both $(-1,1)$ and $7mathbb Z_7$. Hence
$B_{1/2}(7/9) = 4/3$ as function of $(-1,1)subset mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7mathbb Z_7$.
answered Dec 2 '18 at 23:23
eduard
38417
add a comment |
Consider the polynomial $1+X$ in the ring $A=mathbb Q[[X]]$ of power series over $mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $mathbb Q[X]$ but it
has two solutions over the domain $A$. Mainly, $T = pmleft( 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dotsright)$
The "positive" or "favoured" one is
$$
B_{1/2}(X)= 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dots
$$
This is the Taylor expansion of $sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series.
It is easy to check that $B_{1/2}in mathbb Z[1/2][[X]] subset mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function
$$
B_{1/2} : pmathbb Z_p rightarrow 1+ pmathbb Z_p
$$
and also
$$
B_{1/2} : (-1,1)subset mathbb Rrightarrow [0,infty) subsetmathbb R.
$$
Koblitz points out that $7/9$ is in both $(-1,1)$ and $7mathbb Z_7$. Hence
$B_{1/2}(7/9) = 4/3$ as function of $(-1,1)subset mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7mathbb Z_7$.
answered Dec 2 '18 at 23:23
eduard
38417
add a comment |
Consider the polynomial $1+X$ in the ring $A=mathbb Q[[X]]$ of power series over $mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $mathbb Q[X]$ but it
has two solutions over the domain $A$. Mainly, $T = pmleft( 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dotsright)$
The "positive" or "favoured" one is
$$
B_{1/2}(X)= 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dots
$$
This is the Taylor expansion of $sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series.
It is easy to check that $B_{1/2}in mathbb Z[1/2][[X]] subset mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function
$$
B_{1/2} : pmathbb Z_p rightarrow 1+ pmathbb Z_p
$$
and also
$$
B_{1/2} : (-1,1)subset mathbb Rrightarrow [0,infty) subsetmathbb R.
$$
Koblitz points out that $7/9$ is in both $(-1,1)$ and $7mathbb Z_7$. Hence
$B_{1/2}(7/9) = 4/3$ as function of $(-1,1)subset mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7mathbb Z_7$.
answered Dec 2 '18 at 23:23
eduard
38417
Consider the polynomial $1+X$ in the ring $A=mathbb Q[[X]]$ of power series over $mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $mathbb Q[X]$ but it
has two solutions over the domain $A$. Mainly, $T = pmleft( 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dotsright)$
The "positive" or "favoured" one is
$$
B_{1/2}(X)= 1+frac{X}{2}-frac{X^2}{8}+frac{X^3}{16}-frac{5 X^4}{128}+frac{7 X^5}{256}-frac{21 X^6}{1024}+frac{33 X^7}{2048}-frac{429 X^8}{32768}+dots
$$
This is the Taylor expansion of $sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series.
It is easy to check that $B_{1/2}in mathbb Z[1/2][[X]] subset mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function
$$
B_{1/2} : pmathbb Z_p rightarrow 1+ pmathbb Z_p
$$
and also
$$
B_{1/2} : (-1,1)subset mathbb Rrightarrow [0,infty) subsetmathbb R.
$$
Koblitz points out that $7/9$ is in both $(-1,1)$ and $7mathbb Z_7$. Hence
$B_{1/2}(7/9) = 4/3$ as function of $(-1,1)subset mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7mathbb Z_7$.
answered Dec 2 '18 at 23:23
eduard
38417
answered Dec 2 '18 at 23:23
eduard
38417
answered Dec 2 '18 at 23:23
eduard
38417
answered Dec 2 '18 at 23:23
eduard
38417
38417
add a comment |
add a comment |
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3
It's nothing more than noting that $-4/3equiv 1pmod{7}$, while $4/3 equiv -1pmod{7}$. Unlike $mathbb{R}$, the space $mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over.
– anomaly
Nov 12 '18 at 0:33
2
"favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$.
– reuns
Nov 12 '18 at 0:51
Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too.
– Lubin
Nov 24 '18 at 21:39