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Distributivity of a dot product-like operation

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2

0

$begingroup$

Let $b_1, ldots, b_n in mathbb{N}$. For $x, y in mathbb{Z}^n$, define $x cdot y$ as $newcommand{lcm}{operatorname{lcm}}$ $$x cdot y = left(sum_{i=1}^n (x_i text{ mod } b_i)(y_i text{ mod } b_i) text{ mod } b_iright) text{ mod } lcm(b_1, ldots, b_n)$$

Informally, each multiplication $x_iy_i$ is carried over $mathbb{Z}/b_imathbb{Z}$ and the sum is carried over $mathbb{Z}/ lcm(b_1, ldots, b_n)mathbb{Z}$.

My question is the following: does $x cdot (y + z) equiv x cdot y + x cdot z quad (text{mod} lcm(b_1, ldots, b_n))$ hold?

elementary-number-theory

share|cite|improve this question

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have a sum on $i$ with $i$ not appearing in the summand. If that's not what you intended, please edit.
  $endgroup$
  – Gerry Myerson
  Sep 14 '11 at 22:26
  

  
  
  
  

add a comment | 

2

0

$begingroup$

Let $b_1, ldots, b_n in mathbb{N}$. For $x, y in mathbb{Z}^n$, define $x cdot y$ as $newcommand{lcm}{operatorname{lcm}}$ $$x cdot y = left(sum_{i=1}^n (x_i text{ mod } b_i)(y_i text{ mod } b_i) text{ mod } b_iright) text{ mod } lcm(b_1, ldots, b_n)$$

Informally, each multiplication $x_iy_i$ is carried over $mathbb{Z}/b_imathbb{Z}$ and the sum is carried over $mathbb{Z}/ lcm(b_1, ldots, b_n)mathbb{Z}$.

My question is the following: does $x cdot (y + z) equiv x cdot y + x cdot z quad (text{mod} lcm(b_1, ldots, b_n))$ hold?

elementary-number-theory

share|cite|improve this question

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have a sum on $i$ with $i$ not appearing in the summand. If that's not what you intended, please edit.
  $endgroup$
  – Gerry Myerson
  Sep 14 '11 at 22:26
  

  
  
  
  

add a comment | 

$begingroup$

Let $b_1, ldots, b_n in mathbb{N}$. For $x, y in mathbb{Z}^n$, define $x cdot y$ as $newcommand{lcm}{operatorname{lcm}}$ $$x cdot y = left(sum_{i=1}^n (x_i text{ mod } b_i)(y_i text{ mod } b_i) text{ mod } b_iright) text{ mod } lcm(b_1, ldots, b_n)$$

Informally, each multiplication $x_iy_i$ is carried over $mathbb{Z}/b_imathbb{Z}$ and the sum is carried over $mathbb{Z}/ lcm(b_1, ldots, b_n)mathbb{Z}$.

My question is the following: does $x cdot (y + z) equiv x cdot y + x cdot z quad (text{mod} lcm(b_1, ldots, b_n))$ hold?

elementary-number-theory

share|cite|improve this question

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

$endgroup$

share|cite|improve this question

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

share|cite|improve this question

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

edited Sep 14 '11 at 22:33

Srivatsan

21k371126

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

asked Sep 14 '11 at 22:21

Li-thiLi-thi

12916

1 Answer
1

active

oldest

votes

1

$begingroup$

For $n>1$, no. The individual $mathbb{Z}/b_imathbb{Z}$'s will cycle around to $0$ under addition before the same occurs in the broader group of $mathbb{Z}/operatorname{lcm}(b_1,dots,b_n)mathbb{Z}$. Namely, let $e_1=(1,0,0,dots)$. Then

$$e_1cdot e_1equiv 1modell $$
but
$$(e_1+cdots+e_1)cdot e_1equiv(b_1e_1)cdot e_1equiv0cdot e_1equiv0modell $$
while $e_1cdot e_1+cdots+e_1cdot e_1$ with $b_1$ terms is congruent to $b_1(e_1cdot e_1)equiv b_1notequiv 0 modell$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214

$endgroup$

add a comment | 

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1

$begingroup$

For $n>1$, no. The individual $mathbb{Z}/b_imathbb{Z}$'s will cycle around to $0$ under addition before the same occurs in the broader group of $mathbb{Z}/operatorname{lcm}(b_1,dots,b_n)mathbb{Z}$. Namely, let $e_1=(1,0,0,dots)$. Then

$$e_1cdot e_1equiv 1modell $$
but
$$(e_1+cdots+e_1)cdot e_1equiv(b_1e_1)cdot e_1equiv0cdot e_1equiv0modell $$
while $e_1cdot e_1+cdots+e_1cdot e_1$ with $b_1$ terms is congruent to $b_1(e_1cdot e_1)equiv b_1notequiv 0 modell$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214

$endgroup$

add a comment | 

1

$begingroup$

For $n>1$, no. The individual $mathbb{Z}/b_imathbb{Z}$'s will cycle around to $0$ under addition before the same occurs in the broader group of $mathbb{Z}/operatorname{lcm}(b_1,dots,b_n)mathbb{Z}$. Namely, let $e_1=(1,0,0,dots)$. Then

$$e_1cdot e_1equiv 1modell $$
but
$$(e_1+cdots+e_1)cdot e_1equiv(b_1e_1)cdot e_1equiv0cdot e_1equiv0modell $$
while $e_1cdot e_1+cdots+e_1cdot e_1$ with $b_1$ terms is congruent to $b_1(e_1cdot e_1)equiv b_1notequiv 0 modell$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214

$endgroup$

add a comment | 

$begingroup$

For $n>1$, no. The individual $mathbb{Z}/b_imathbb{Z}$'s will cycle around to $0$ under addition before the same occurs in the broader group of $mathbb{Z}/operatorname{lcm}(b_1,dots,b_n)mathbb{Z}$. Namely, let $e_1=(1,0,0,dots)$. Then

$$e_1cdot e_1equiv 1modell $$
but
$$(e_1+cdots+e_1)cdot e_1equiv(b_1e_1)cdot e_1equiv0cdot e_1equiv0modell $$
while $e_1cdot e_1+cdots+e_1cdot e_1$ with $b_1$ terms is congruent to $b_1(e_1cdot e_1)equiv b_1notequiv 0 modell$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214

$endgroup$

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214

edited yesterday

Martin Sleziak

44.8k10119273

edited yesterday

Martin Sleziak

44.8k10119273

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214

answered Sep 14 '11 at 22:38

anonanon

72.4k5111214