Trying to extend distributive property of modulo operation to real numbers
$begingroup$
Here Wikipedia states that modulo operation is distributive:
$$a cdot b mod n = (a mod n)cdot (b mod n) mod n$$
Which is true for every natural number. Unfortunately it is not for rational ones:
given$q in Bbb Q, n in Bbb Z, qcdot n notin Bbb Z $.
$$qcdot n mod 1 ne 0 $$
$$ left{ begin{array}{c}q mod 1ne0\n mod 1 = 0 end{array}right.$$
Therefore $$(q mod 1)cdot(n mod 1)=0 $$
I was wondering if you guys have any idea on how to extend this very property to real numbers (or rational ones).
modular-arithmetic real-numbers irrational-numbers integers
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
$endgroup$
add a comment |
$begingroup$
Here Wikipedia states that modulo operation is distributive:
$$a cdot b mod n = (a mod n)cdot (b mod n) mod n$$
Which is true for every natural number. Unfortunately it is not for rational ones:
given$q in Bbb Q, n in Bbb Z, qcdot n notin Bbb Z $.
$$qcdot n mod 1 ne 0 $$
$$ left{ begin{array}{c}q mod 1ne0\n mod 1 = 0 end{array}right.$$
Therefore $$(q mod 1)cdot(n mod 1)=0 $$
I was wondering if you guys have any idea on how to extend this very property to real numbers (or rational ones).
modular-arithmetic real-numbers irrational-numbers integers
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
$endgroup$
$begingroup$
you can't, not if you mean real distributivity. The reason for that is that the modulo construction on integers is actually taking the quotient by "nice" subgroups which are closed under arbitrary multiplication (ideals). sadly however $mathbb{Q}$ is a field, and hence has no nontrivial prime ideals. (this corresponds precisely to your problem above, that as soon as you have a number vanishing in modulo, $1$ has to vanish as well)
$endgroup$
– Enkidu
Dec 18 '18 at 14:28
add a comment |
$begingroup$
Here Wikipedia states that modulo operation is distributive:
$$a cdot b mod n = (a mod n)cdot (b mod n) mod n$$
Which is true for every natural number. Unfortunately it is not for rational ones:
given$q in Bbb Q, n in Bbb Z, qcdot n notin Bbb Z $.
$$qcdot n mod 1 ne 0 $$
$$ left{ begin{array}{c}q mod 1ne0\n mod 1 = 0 end{array}right.$$
Therefore $$(q mod 1)cdot(n mod 1)=0 $$
I was wondering if you guys have any idea on how to extend this very property to real numbers (or rational ones).
modular-arithmetic real-numbers irrational-numbers integers
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
$endgroup$
Here Wikipedia states that modulo operation is distributive:
$$a cdot b mod n = (a mod n)cdot (b mod n) mod n$$
Which is true for every natural number. Unfortunately it is not for rational ones:
given$q in Bbb Q, n in Bbb Z, qcdot n notin Bbb Z $.
$$qcdot n mod 1 ne 0 $$
$$ left{ begin{array}{c}q mod 1ne0\n mod 1 = 0 end{array}right.$$
Therefore $$(q mod 1)cdot(n mod 1)=0 $$
I was wondering if you guys have any idea on how to extend this very property to real numbers (or rational ones).
modular-arithmetic real-numbers irrational-numbers integers
modular-arithmetic real-numbers irrational-numbers integers
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
asked Dec 18 '18 at 12:19
Lyn CassidyLyn Cassidy
436
436
$begingroup$
you can't, not if you mean real distributivity. The reason for that is that the modulo construction on integers is actually taking the quotient by "nice" subgroups which are closed under arbitrary multiplication (ideals). sadly however $mathbb{Q}$ is a field, and hence has no nontrivial prime ideals. (this corresponds precisely to your problem above, that as soon as you have a number vanishing in modulo, $1$ has to vanish as well)
$endgroup$
– Enkidu
Dec 18 '18 at 14:28
add a comment |
$begingroup$
you can't, not if you mean real distributivity. The reason for that is that the modulo construction on integers is actually taking the quotient by "nice" subgroups which are closed under arbitrary multiplication (ideals). sadly however $mathbb{Q}$ is a field, and hence has no nontrivial prime ideals. (this corresponds precisely to your problem above, that as soon as you have a number vanishing in modulo, $1$ has to vanish as well)
$endgroup$
– Enkidu
Dec 18 '18 at 14:28
$begingroup$
you can't, not if you mean real distributivity. The reason for that is that the modulo construction on integers is actually taking the quotient by "nice" subgroups which are closed under arbitrary multiplication (ideals). sadly however $mathbb{Q}$ is a field, and hence has no nontrivial prime ideals. (this corresponds precisely to your problem above, that as soon as you have a number vanishing in modulo, $1$ has to vanish as well)
$endgroup$
– Enkidu
Dec 18 '18 at 14:28
$begingroup$
you can't, not if you mean real distributivity. The reason for that is that the modulo construction on integers is actually taking the quotient by "nice" subgroups which are closed under arbitrary multiplication (ideals). sadly however $mathbb{Q}$ is a field, and hence has no nontrivial prime ideals. (this corresponds precisely to your problem above, that as soon as you have a number vanishing in modulo, $1$ has to vanish as well)
$endgroup$
– Enkidu
Dec 18 '18 at 14:28
add a comment |
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$begingroup$
you can't, not if you mean real distributivity. The reason for that is that the modulo construction on integers is actually taking the quotient by "nice" subgroups which are closed under arbitrary multiplication (ideals). sadly however $mathbb{Q}$ is a field, and hence has no nontrivial prime ideals. (this corresponds precisely to your problem above, that as soon as you have a number vanishing in modulo, $1$ has to vanish as well)
$endgroup$
– Enkidu
Dec 18 '18 at 14:28