If the sum and the product of two sets are equal are the sets?
I have the following lemma that I can't seem to disprove or prove but which I am assuming someone here will make short work of. I will first write a simple case and then a more general case.
Simple case:
Let $mathbb{F}_q$ be a finite field of prime order $q$.
$forall a, b, c, d in mathbb{F}_q$ where $a$ and $b$ are non zero
if $a + b = c + d$ and $a*b=c*d$
then $(a=c land b=d) lor (a=d land b=c)$
General case:
Let $mathbb{F}_q^n$ be a vector space of dimension $n$ over a finite field of prime order $q$.
$forall a,b in mathbb{F}_q^n$, where all elements in $a$ are non zero
if $sum_{i=1}^n a_i = sum_{i=1}^n b_i$ and $prod_{i=1}^n a_i = prod_{i=1}^n b_i$
then there exists a permutation matrix $M$ such that $aM=b$.
Thanks in advance,
abstract-algebra
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
asked Nov 27 '18 at 16:31
hainest
82
|
show 5 more comments
I have the following lemma that I can't seem to disprove or prove but which I am assuming someone here will make short work of. I will first write a simple case and then a more general case.
Simple case:
Let $mathbb{F}_q$ be a finite field of prime order $q$.
$forall a, b, c, d in mathbb{F}_q$ where $a$ and $b$ are non zero
if $a + b = c + d$ and $a*b=c*d$
then $(a=c land b=d) lor (a=d land b=c)$
General case:
Let $mathbb{F}_q^n$ be a vector space of dimension $n$ over a finite field of prime order $q$.
$forall a,b in mathbb{F}_q^n$, where all elements in $a$ are non zero
if $sum_{i=1}^n a_i = sum_{i=1}^n b_i$ and $prod_{i=1}^n a_i = prod_{i=1}^n b_i$
then there exists a permutation matrix $M$ such that $aM=b$.
Thanks in advance,
abstract-algebra
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
asked Nov 27 '18 at 16:31
hainest
82
Little tip [irrelevant to the question]: the tags should be correct. So i assume it should be "abstract-algebra"
– xbh
Nov 27 '18 at 16:36
The first one is true. Write $a+b=x=c+d,$ then substitute $b=x-a,d=x-c$ into $ab=cd,$ and forge ahead.
– saulspatz
Nov 27 '18 at 17:14
1
If $mathbb F$ is an $m$-element field, then there are $binom{m-1}3$ sets ${a_1,a_2,a_3}$ of three distinct nonzero elements, while there are only $m(m-1)$ possible pairs $(a_1+a_2+a_3,a_1a_2a_3)$.
– bof
Nov 27 '18 at 17:55
2
In $mathbb F_{11}$, $1+4+8+9=2+3+7+10$ and $1cdot4cdot8cdot9=2cdot3cdot7cdot10$.
– bof
Nov 27 '18 at 19:00
1
"Are $n$ nonzero numbers determined by their sum and product?" It doesn't seem too likely that $n$ numbers could be determined by only $2$, does it?
– bof
Nov 27 '18 at 19:03
|
show 5 more comments
I have the following lemma that I can't seem to disprove or prove but which I am assuming someone here will make short work of. I will first write a simple case and then a more general case.
Simple case:
Let $mathbb{F}_q$ be a finite field of prime order $q$.
$forall a, b, c, d in mathbb{F}_q$ where $a$ and $b$ are non zero
if $a + b = c + d$ and $a*b=c*d$
then $(a=c land b=d) lor (a=d land b=c)$
General case:
Let $mathbb{F}_q^n$ be a vector space of dimension $n$ over a finite field of prime order $q$.
$forall a,b in mathbb{F}_q^n$, where all elements in $a$ are non zero
if $sum_{i=1}^n a_i = sum_{i=1}^n b_i$ and $prod_{i=1}^n a_i = prod_{i=1}^n b_i$
then there exists a permutation matrix $M$ such that $aM=b$.
Thanks in advance,
abstract-algebra
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
asked Nov 27 '18 at 16:31
hainest
82
I have the following lemma that I can't seem to disprove or prove but which I am assuming someone here will make short work of. I will first write a simple case and then a more general case.
Simple case:
Let $mathbb{F}_q$ be a finite field of prime order $q$.
$forall a, b, c, d in mathbb{F}_q$ where $a$ and $b$ are non zero
if $a + b = c + d$ and $a*b=c*d$
then $(a=c land b=d) lor (a=d land b=c)$
General case:
Let $mathbb{F}_q^n$ be a vector space of dimension $n$ over a finite field of prime order $q$.
$forall a,b in mathbb{F}_q^n$, where all elements in $a$ are non zero
if $sum_{i=1}^n a_i = sum_{i=1}^n b_i$ and $prod_{i=1}^n a_i = prod_{i=1}^n b_i$
then there exists a permutation matrix $M$ such that $aM=b$.
Thanks in advance,
abstract-algebra
abstract-algebra
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
asked Nov 27 '18 at 16:31
hainest
82
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
asked Nov 27 '18 at 16:31
hainest
82
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
edited Nov 27 '18 at 16:38
Tianlalu
3,09621038
3,09621038
asked Nov 27 '18 at 16:31
hainest
82
asked Nov 27 '18 at 16:31
hainest
82
asked Nov 27 '18 at 16:31
hainest
82
82
Little tip [irrelevant to the question]: the tags should be correct. So i assume it should be "abstract-algebra"
– xbh
Nov 27 '18 at 16:36
The first one is true. Write $a+b=x=c+d,$ then substitute $b=x-a,d=x-c$ into $ab=cd,$ and forge ahead.
– saulspatz
Nov 27 '18 at 17:14
1
If $mathbb F$ is an $m$-element field, then there are $binom{m-1}3$ sets ${a_1,a_2,a_3}$ of three distinct nonzero elements, while there are only $m(m-1)$ possible pairs $(a_1+a_2+a_3,a_1a_2a_3)$.
– bof
Nov 27 '18 at 17:55
2
In $mathbb F_{11}$, $1+4+8+9=2+3+7+10$ and $1cdot4cdot8cdot9=2cdot3cdot7cdot10$.
– bof
Nov 27 '18 at 19:00
1
"Are $n$ nonzero numbers determined by their sum and product?" It doesn't seem too likely that $n$ numbers could be determined by only $2$, does it?
– bof
Nov 27 '18 at 19:03
|
show 5 more comments
Little tip [irrelevant to the question]: the tags should be correct. So i assume it should be "abstract-algebra"
– xbh
Nov 27 '18 at 16:36
The first one is true. Write $a+b=x=c+d,$ then substitute $b=x-a,d=x-c$ into $ab=cd,$ and forge ahead.
– saulspatz
Nov 27 '18 at 17:14
1
If $mathbb F$ is an $m$-element field, then there are $binom{m-1}3$ sets ${a_1,a_2,a_3}$ of three distinct nonzero elements, while there are only $m(m-1)$ possible pairs $(a_1+a_2+a_3,a_1a_2a_3)$.
– bof
Nov 27 '18 at 17:55
2
In $mathbb F_{11}$, $1+4+8+9=2+3+7+10$ and $1cdot4cdot8cdot9=2cdot3cdot7cdot10$.
– bof
Nov 27 '18 at 19:00
1
"Are $n$ nonzero numbers determined by their sum and product?" It doesn't seem too likely that $n$ numbers could be determined by only $2$, does it?
– bof
Nov 27 '18 at 19:03
Little tip [irrelevant to the question]: the tags should be correct. So i assume it should be "abstract-algebra"
– xbh
Nov 27 '18 at 16:36
Little tip [irrelevant to the question]: the tags should be correct. So i assume it should be "abstract-algebra"
– xbh
Nov 27 '18 at 16:36
The first one is true. Write $a+b=x=c+d,$ then substitute $b=x-a,d=x-c$ into $ab=cd,$ and forge ahead.
– saulspatz
Nov 27 '18 at 17:14
The first one is true. Write $a+b=x=c+d,$ then substitute $b=x-a,d=x-c$ into $ab=cd,$ and forge ahead.
– saulspatz
Nov 27 '18 at 17:14
1
1
If $mathbb F$ is an $m$-element field, then there are $binom{m-1}3$ sets ${a_1,a_2,a_3}$ of three distinct nonzero elements, while there are only $m(m-1)$ possible pairs $(a_1+a_2+a_3,a_1a_2a_3)$.
– bof
Nov 27 '18 at 17:55
If $mathbb F$ is an $m$-element field, then there are $binom{m-1}3$ sets ${a_1,a_2,a_3}$ of three distinct nonzero elements, while there are only $m(m-1)$ possible pairs $(a_1+a_2+a_3,a_1a_2a_3)$.
– bof
Nov 27 '18 at 17:55
2
2
In $mathbb F_{11}$, $1+4+8+9=2+3+7+10$ and $1cdot4cdot8cdot9=2cdot3cdot7cdot10$.
– bof
Nov 27 '18 at 19:00
In $mathbb F_{11}$, $1+4+8+9=2+3+7+10$ and $1cdot4cdot8cdot9=2cdot3cdot7cdot10$.
– bof
Nov 27 '18 at 19:00
1
1
"Are $n$ nonzero numbers determined by their sum and product?" It doesn't seem too likely that $n$ numbers could be determined by only $2$, does it?
– bof
Nov 27 '18 at 19:03
"Are $n$ nonzero numbers determined by their sum and product?" It doesn't seem too likely that $n$ numbers could be determined by only $2$, does it?
– bof
Nov 27 '18 at 19:03
|
show 5 more comments
1 Answer
1
active
oldest
votes
It's true for two elements in any field. We have $a+b=x=c+d,$ so
$b=x-a, d=x-c.$ Then $$ab=cd iff a(x-a)=c(x-c)iff x(a-c)=(a-c)(a+c)$$ Either $a-c = 0,$ and then $b=d,$ or $$x=a+ciff a+b=a+ciff b=c.$$
As @bof has indicated in a comment, the statement is false for $n>2.$
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
add a comment |
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1 Answer
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It's true for two elements in any field. We have $a+b=x=c+d,$ so
$b=x-a, d=x-c.$ Then $$ab=cd iff a(x-a)=c(x-c)iff x(a-c)=(a-c)(a+c)$$ Either $a-c = 0,$ and then $b=d,$ or $$x=a+ciff a+b=a+ciff b=c.$$
As @bof has indicated in a comment, the statement is false for $n>2.$
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
add a comment |
It's true for two elements in any field. We have $a+b=x=c+d,$ so
$b=x-a, d=x-c.$ Then $$ab=cd iff a(x-a)=c(x-c)iff x(a-c)=(a-c)(a+c)$$ Either $a-c = 0,$ and then $b=d,$ or $$x=a+ciff a+b=a+ciff b=c.$$
As @bof has indicated in a comment, the statement is false for $n>2.$
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
add a comment |
It's true for two elements in any field. We have $a+b=x=c+d,$ so
$b=x-a, d=x-c.$ Then $$ab=cd iff a(x-a)=c(x-c)iff x(a-c)=(a-c)(a+c)$$ Either $a-c = 0,$ and then $b=d,$ or $$x=a+ciff a+b=a+ciff b=c.$$
As @bof has indicated in a comment, the statement is false for $n>2.$
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
It's true for two elements in any field. We have $a+b=x=c+d,$ so
$b=x-a, d=x-c.$ Then $$ab=cd iff a(x-a)=c(x-c)iff x(a-c)=(a-c)(a+c)$$ Either $a-c = 0,$ and then $b=d,$ or $$x=a+ciff a+b=a+ciff b=c.$$
As @bof has indicated in a comment, the statement is false for $n>2.$
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
edited Nov 29 '18 at 14:51
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
answered Nov 28 '18 at 15:08
saulspatz
13.9k21329
13.9k21329
add a comment |
add a comment |
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Little tip [irrelevant to the question]: the tags should be correct. So i assume it should be "abstract-algebra"
– xbh
Nov 27 '18 at 16:36
The first one is true. Write $a+b=x=c+d,$ then substitute $b=x-a,d=x-c$ into $ab=cd,$ and forge ahead.
– saulspatz
Nov 27 '18 at 17:14
1
If $mathbb F$ is an $m$-element field, then there are $binom{m-1}3$ sets ${a_1,a_2,a_3}$ of three distinct nonzero elements, while there are only $m(m-1)$ possible pairs $(a_1+a_2+a_3,a_1a_2a_3)$.
– bof
Nov 27 '18 at 17:55
2
In $mathbb F_{11}$, $1+4+8+9=2+3+7+10$ and $1cdot4cdot8cdot9=2cdot3cdot7cdot10$.
– bof
Nov 27 '18 at 19:00
1
"Are $n$ nonzero numbers determined by their sum and product?" It doesn't seem too likely that $n$ numbers could be determined by only $2$, does it?
– bof
Nov 27 '18 at 19:03