What is the meaning of modular fractions?
up vote
0
down vote
favorite
What is the meaning of
$$
\frac{m}{n}equiv kpmod l
$$
and
$$
\frac{m}{n} mod l
$$
$m, n, k, l inmathbb N$ and $gcd(m, n)=1$.
number-theory elementary-number-theory
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
asked Nov 17 at 0:26
Samvel Safaryan
38211
add a comment |
up vote
0
down vote
favorite
What is the meaning of
$$
\frac{m}{n}equiv kpmod l
$$
and
$$
\frac{m}{n} mod l
$$
$m, n, k, l inmathbb N$ and $gcd(m, n)=1$.
number-theory elementary-number-theory
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
asked Nov 17 at 0:26
Samvel Safaryan
38211
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the meaning of
$$
\frac{m}{n}equiv kpmod l
$$
and
$$
\frac{m}{n} mod l
$$
$m, n, k, l inmathbb N$ and $gcd(m, n)=1$.
number-theory elementary-number-theory
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
asked Nov 17 at 0:26
Samvel Safaryan
38211
What is the meaning of
$$
\frac{m}{n}equiv kpmod l
$$
and
$$
\frac{m}{n} mod l
$$
$m, n, k, l inmathbb N$ and $gcd(m, n)=1$.
number-theory elementary-number-theory
number-theory elementary-number-theory
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
asked Nov 17 at 0:26
Samvel Safaryan
38211
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
asked Nov 17 at 0:26
Samvel Safaryan
38211
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
edited Nov 18 at 17:44
Bill Dubuque
206k29189621
206k29189621
asked Nov 17 at 0:26
Samvel Safaryan
38211
asked Nov 17 at 0:26
Samvel Safaryan
38211
asked Nov 17 at 0:26
Samvel Safaryan
38211
38211
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Recall $b$ is invertible mod $niff b$ is coprime to $n $ [by Bezout's identity for $,gcd(b,n)=1$]
When so we define $ a/b := ab^{-1},,$ which is the unique solution of $,bxequiv apmod{!n}$
Hence $ a/b = ab^{-1}equiv c iff aequiv bcpmod{!n}. $ Operationally $, a/bbmod n, :=, ab^{-1}bmod n$
We can do modular fraction arithmetic on all rationals writable with denominator coprime to $n$ (which comprise a subring of the rationals, i.e. are closed under addition and multiplication). The well-known formulas for fraction addition and multiplication remain true for modular fractions, e.g. the addition law $,a/b+c/d = (ad+bc)/(bd),$ holds true because $$,(ad+bc)color{#0a0}{(bd)^{-1}}!equiv ad,color{#0a0}{d^{-1}b^{-1}}+ color{#0a0}{d^{-1}b^{-1}}bcequiv ab^{-1}+d^{-1}b $$
See here for many examples of modular fraction arithmetic.
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Remark $ $ You may already be familiar with this idea from calculus. For example, consider polynomials with real coefficients. Every real $rneq 0$ is invertible so we can work with fractional expressions $f(x)/r := r^{-1} f(x),$ just as above. Then $,f/r+g/s = (sf!+!rg)/(rs),,$ and $,(f/r)(g/s) = fg/(rs).,$ More generally the same idea works for invertible elements in any commutative domain (e.g. we can work with "compound fractions" of rationals, reals, or elements of any field). These ideas are algebraically reified in university algebra when one studies fraction fields and localizations of rings and modules which - unlike the degenerate cases above - typically enlarge the domain by adjoining inverses of non-invertible elements.
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
add a comment |
up vote
1
down vote
The first one means
$$
kn equiv m pmod{l} .
$$
The second is a name for the solution to the congruence
$$
n times ? equiv m pmod{l} .
$$
That congruence has a unique solution when $l$ and $n$ are relatively prime.
answered Nov 17 at 0:30
Ethan Bolker
39k543102
add a comment |
up vote
1
down vote
I am guessing the meaning of your poorly worded question. The first one is some kind of equality (in congruences) between two objects. The second one simply refers to an object (of some congruence calculation).
The following two things give another example of similar nature: $A=5+3$, and $ 5+3$
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall $b$ is invertible mod $niff b$ is coprime to $n $ [by Bezout's identity for $,gcd(b,n)=1$]
When so we define $ a/b := ab^{-1},,$ which is the unique solution of $,bxequiv apmod{!n}$
Hence $ a/b = ab^{-1}equiv c iff aequiv bcpmod{!n}. $ Operationally $, a/bbmod n, :=, ab^{-1}bmod n$
We can do modular fraction arithmetic on all rationals writable with denominator coprime to $n$ (which comprise a subring of the rationals, i.e. are closed under addition and multiplication). The well-known formulas for fraction addition and multiplication remain true for modular fractions, e.g. the addition law $,a/b+c/d = (ad+bc)/(bd),$ holds true because $$,(ad+bc)color{#0a0}{(bd)^{-1}}!equiv ad,color{#0a0}{d^{-1}b^{-1}}+ color{#0a0}{d^{-1}b^{-1}}bcequiv ab^{-1}+d^{-1}b $$
See here for many examples of modular fraction arithmetic.
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Remark $ $ You may already be familiar with this idea from calculus. For example, consider polynomials with real coefficients. Every real $rneq 0$ is invertible so we can work with fractional expressions $f(x)/r := r^{-1} f(x),$ just as above. Then $,f/r+g/s = (sf!+!rg)/(rs),,$ and $,(f/r)(g/s) = fg/(rs).,$ More generally the same idea works for invertible elements in any commutative domain (e.g. we can work with "compound fractions" of rationals, reals, or elements of any field). These ideas are algebraically reified in university algebra when one studies fraction fields and localizations of rings and modules which - unlike the degenerate cases above - typically enlarge the domain by adjoining inverses of non-invertible elements.
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
add a comment |
up vote
2
down vote
accepted
Recall $b$ is invertible mod $niff b$ is coprime to $n $ [by Bezout's identity for $,gcd(b,n)=1$]
When so we define $ a/b := ab^{-1},,$ which is the unique solution of $,bxequiv apmod{!n}$
Hence $ a/b = ab^{-1}equiv c iff aequiv bcpmod{!n}. $ Operationally $, a/bbmod n, :=, ab^{-1}bmod n$
We can do modular fraction arithmetic on all rationals writable with denominator coprime to $n$ (which comprise a subring of the rationals, i.e. are closed under addition and multiplication). The well-known formulas for fraction addition and multiplication remain true for modular fractions, e.g. the addition law $,a/b+c/d = (ad+bc)/(bd),$ holds true because $$,(ad+bc)color{#0a0}{(bd)^{-1}}!equiv ad,color{#0a0}{d^{-1}b^{-1}}+ color{#0a0}{d^{-1}b^{-1}}bcequiv ab^{-1}+d^{-1}b $$
See here for many examples of modular fraction arithmetic.
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Remark $ $ You may already be familiar with this idea from calculus. For example, consider polynomials with real coefficients. Every real $rneq 0$ is invertible so we can work with fractional expressions $f(x)/r := r^{-1} f(x),$ just as above. Then $,f/r+g/s = (sf!+!rg)/(rs),,$ and $,(f/r)(g/s) = fg/(rs).,$ More generally the same idea works for invertible elements in any commutative domain (e.g. we can work with "compound fractions" of rationals, reals, or elements of any field). These ideas are algebraically reified in university algebra when one studies fraction fields and localizations of rings and modules which - unlike the degenerate cases above - typically enlarge the domain by adjoining inverses of non-invertible elements.
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall $b$ is invertible mod $niff b$ is coprime to $n $ [by Bezout's identity for $,gcd(b,n)=1$]
When so we define $ a/b := ab^{-1},,$ which is the unique solution of $,bxequiv apmod{!n}$
Hence $ a/b = ab^{-1}equiv c iff aequiv bcpmod{!n}. $ Operationally $, a/bbmod n, :=, ab^{-1}bmod n$
We can do modular fraction arithmetic on all rationals writable with denominator coprime to $n$ (which comprise a subring of the rationals, i.e. are closed under addition and multiplication). The well-known formulas for fraction addition and multiplication remain true for modular fractions, e.g. the addition law $,a/b+c/d = (ad+bc)/(bd),$ holds true because $$,(ad+bc)color{#0a0}{(bd)^{-1}}!equiv ad,color{#0a0}{d^{-1}b^{-1}}+ color{#0a0}{d^{-1}b^{-1}}bcequiv ab^{-1}+d^{-1}b $$
See here for many examples of modular fraction arithmetic.
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Remark $ $ You may already be familiar with this idea from calculus. For example, consider polynomials with real coefficients. Every real $rneq 0$ is invertible so we can work with fractional expressions $f(x)/r := r^{-1} f(x),$ just as above. Then $,f/r+g/s = (sf!+!rg)/(rs),,$ and $,(f/r)(g/s) = fg/(rs).,$ More generally the same idea works for invertible elements in any commutative domain (e.g. we can work with "compound fractions" of rationals, reals, or elements of any field). These ideas are algebraically reified in university algebra when one studies fraction fields and localizations of rings and modules which - unlike the degenerate cases above - typically enlarge the domain by adjoining inverses of non-invertible elements.
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
Recall $b$ is invertible mod $niff b$ is coprime to $n $ [by Bezout's identity for $,gcd(b,n)=1$]
When so we define $ a/b := ab^{-1},,$ which is the unique solution of $,bxequiv apmod{!n}$
Hence $ a/b = ab^{-1}equiv c iff aequiv bcpmod{!n}. $ Operationally $, a/bbmod n, :=, ab^{-1}bmod n$
We can do modular fraction arithmetic on all rationals writable with denominator coprime to $n$ (which comprise a subring of the rationals, i.e. are closed under addition and multiplication). The well-known formulas for fraction addition and multiplication remain true for modular fractions, e.g. the addition law $,a/b+c/d = (ad+bc)/(bd),$ holds true because $$,(ad+bc)color{#0a0}{(bd)^{-1}}!equiv ad,color{#0a0}{d^{-1}b^{-1}}+ color{#0a0}{d^{-1}b^{-1}}bcequiv ab^{-1}+d^{-1}b $$
See here for many examples of modular fraction arithmetic.
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Remark $ $ You may already be familiar with this idea from calculus. For example, consider polynomials with real coefficients. Every real $rneq 0$ is invertible so we can work with fractional expressions $f(x)/r := r^{-1} f(x),$ just as above. Then $,f/r+g/s = (sf!+!rg)/(rs),,$ and $,(f/r)(g/s) = fg/(rs).,$ More generally the same idea works for invertible elements in any commutative domain (e.g. we can work with "compound fractions" of rationals, reals, or elements of any field). These ideas are algebraically reified in university algebra when one studies fraction fields and localizations of rings and modules which - unlike the degenerate cases above - typically enlarge the domain by adjoining inverses of non-invertible elements.
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
edited Nov 18 at 17:44
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
answered Nov 17 at 1:43
Bill Dubuque
206k29189621
206k29189621
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
add a comment |
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
+1 for the careful explicit connection to "ordinary" arithmetic with fractions. But is the reference to fraction fields right? In $mathbb{Z}_n$ whatever fractions exist are already present.
– Ethan Bolker
Nov 17 at 12:16
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
@Ethan Yes, it is a degenerate case of the fraction field / localization construction: if the denominators are all units (invertibles) then we get nothing new, i.e. the construction yields an isomorphic ring, e.g if we apply the fraction field construction to a fraction field then it yields an isomorphic ring of "compound fractions". But such universality often proves very handy, e.g. $,1/2 - 1/3 = 1/6,$ can be mapped into any ring (or $Bbb Z$-module) where $6$ is invertible (e.g. in a group if $b$ is a square-root of $a$ and $c$ is a cube-root then $b/c$ is a sixth-root).
– Bill Dubuque
Nov 17 at 14:33
add a comment |
up vote
1
down vote
The first one means
$$
kn equiv m pmod{l} .
$$
The second is a name for the solution to the congruence
$$
n times ? equiv m pmod{l} .
$$
That congruence has a unique solution when $l$ and $n$ are relatively prime.
answered Nov 17 at 0:30
Ethan Bolker
39k543102
add a comment |
up vote
1
down vote
The first one means
$$
kn equiv m pmod{l} .
$$
The second is a name for the solution to the congruence
$$
n times ? equiv m pmod{l} .
$$
That congruence has a unique solution when $l$ and $n$ are relatively prime.
answered Nov 17 at 0:30
Ethan Bolker
39k543102
add a comment |
up vote
1
down vote
up vote
1
down vote
The first one means
$$
kn equiv m pmod{l} .
$$
The second is a name for the solution to the congruence
$$
n times ? equiv m pmod{l} .
$$
That congruence has a unique solution when $l$ and $n$ are relatively prime.
answered Nov 17 at 0:30
Ethan Bolker
39k543102
The first one means
$$
kn equiv m pmod{l} .
$$
The second is a name for the solution to the congruence
$$
n times ? equiv m pmod{l} .
$$
That congruence has a unique solution when $l$ and $n$ are relatively prime.
answered Nov 17 at 0:30
Ethan Bolker
39k543102
answered Nov 17 at 0:30
Ethan Bolker
39k543102
answered Nov 17 at 0:30
Ethan Bolker
39k543102
answered Nov 17 at 0:30
Ethan Bolker
39k543102
39k543102
add a comment |
add a comment |
up vote
1
down vote
I am guessing the meaning of your poorly worded question. The first one is some kind of equality (in congruences) between two objects. The second one simply refers to an object (of some congruence calculation).
The following two things give another example of similar nature: $A=5+3$, and $ 5+3$
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
add a comment |
up vote
1
down vote
I am guessing the meaning of your poorly worded question. The first one is some kind of equality (in congruences) between two objects. The second one simply refers to an object (of some congruence calculation).
The following two things give another example of similar nature: $A=5+3$, and $ 5+3$
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
add a comment |
up vote
1
down vote
up vote
1
down vote
I am guessing the meaning of your poorly worded question. The first one is some kind of equality (in congruences) between two objects. The second one simply refers to an object (of some congruence calculation).
The following two things give another example of similar nature: $A=5+3$, and $ 5+3$
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
I am guessing the meaning of your poorly worded question. The first one is some kind of equality (in congruences) between two objects. The second one simply refers to an object (of some congruence calculation).
The following two things give another example of similar nature: $A=5+3$, and $ 5+3$
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
answered Nov 17 at 1:26
P Vanchinathan
14.6k12036
14.6k12036
add a comment |
add a comment |
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