Does the system of equations a + b, a * b provide a one-to-one correspondence between ordered and unordered...
$begingroup$
Let's write ordered sets as $[a,b,c,...]$ and ${a,b,c,...}$ for unordered sets.
Now let $f(a,b)=[a+b,a b]$. Note $f(a,b)=f(b,a).$
Consider $g([x,y])={a,b}$ where $a+b=x,a b=y$.
Mathematica's Solve function came up with $left{frac{1}{2} left(x-sqrt{x^2-4 y}right),frac{1}{2} left(sqrt{x^2-4 y}+xright)right}$ as the value of ${a,b}$ given $[x,y]$, and in my testing, it worked for everything I threw at it. (On a philosophical note it is interesting that just from the definitions provided there's no implied order for $a$ and $b$ given $[x,y]$.)
Is this a correct one-to-one correspondence from ordered sets of two elements to unordered sets of two elements in the domain of imaginary numbers? If so, does it have a name? I'd like to find out more about it.
As an aside, the one element and three element cases seem to work also. Mathematica spent a long time trying to find equations for my guess at the four element case $a+b+c+d,a b+a c+a d+b c+b d+c d,a b c+a b d+a c d+b c d,a b c d$, but eventually I aborted it. I suspect if there is a solution, it might generalize to $n$ elements.
Also, I think it works with ordered lists and unordered lists.
Many thanks!
Edit: as DanielWainfleet points out in the comments "if $x^2=4y$ then $g([x,y])={x/2}$ which has only one element. So the above can't go from ordered sets of two elements to unordered sets of two elements.
So I'd like to revise the question to the highly related case of including sets of one OR two elements (unordered sets stay at two elements).
elementary-set-theory
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
$endgroup$
|
show 6 more comments
$begingroup$
Let's write ordered sets as $[a,b,c,...]$ and ${a,b,c,...}$ for unordered sets.
Now let $f(a,b)=[a+b,a b]$. Note $f(a,b)=f(b,a).$
Consider $g([x,y])={a,b}$ where $a+b=x,a b=y$.
Mathematica's Solve function came up with $left{frac{1}{2} left(x-sqrt{x^2-4 y}right),frac{1}{2} left(sqrt{x^2-4 y}+xright)right}$ as the value of ${a,b}$ given $[x,y]$, and in my testing, it worked for everything I threw at it. (On a philosophical note it is interesting that just from the definitions provided there's no implied order for $a$ and $b$ given $[x,y]$.)
Is this a correct one-to-one correspondence from ordered sets of two elements to unordered sets of two elements in the domain of imaginary numbers? If so, does it have a name? I'd like to find out more about it.
As an aside, the one element and three element cases seem to work also. Mathematica spent a long time trying to find equations for my guess at the four element case $a+b+c+d,a b+a c+a d+b c+b d+c d,a b c+a b d+a c d+b c d,a b c d$, but eventually I aborted it. I suspect if there is a solution, it might generalize to $n$ elements.
Also, I think it works with ordered lists and unordered lists.
Many thanks!
Edit: as DanielWainfleet points out in the comments "if $x^2=4y$ then $g([x,y])={x/2}$ which has only one element. So the above can't go from ordered sets of two elements to unordered sets of two elements.
So I'd like to revise the question to the highly related case of including sets of one OR two elements (unordered sets stay at two elements).
elementary-set-theory
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
$endgroup$
$begingroup$
What does $g([x,y])={a,b}$ mean here? The way I would usually interpret it doesn't seem to be what you mean.
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:10
1
$begingroup$
See also en.wikipedia.org/wiki/Elementary_symmetric_polynomial
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:13
$begingroup$
It means that $g$ is a function from an unordered set $[x,y]$ to a set ${a,b}$. Maybe $g:[x,y]mapsto{a,b}$ is more correct. As for symmetric polynomials, they look highly related, but I'm not seeing this particular application.
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 0:54
$begingroup$
(Side note: the above should read "It's supposed to mean...". I didn't mean to imply that was a reasonable way of writing that.)
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 1:04
1
$begingroup$
If $x^2=4y$ then $g([x,y]) ={x/2}$ has only $1$ member.
$endgroup$
– DanielWainfleet
Jun 17 '17 at 2:04
|
show 6 more comments
$begingroup$
Let's write ordered sets as $[a,b,c,...]$ and ${a,b,c,...}$ for unordered sets.
Now let $f(a,b)=[a+b,a b]$. Note $f(a,b)=f(b,a).$
Consider $g([x,y])={a,b}$ where $a+b=x,a b=y$.
Mathematica's Solve function came up with $left{frac{1}{2} left(x-sqrt{x^2-4 y}right),frac{1}{2} left(sqrt{x^2-4 y}+xright)right}$ as the value of ${a,b}$ given $[x,y]$, and in my testing, it worked for everything I threw at it. (On a philosophical note it is interesting that just from the definitions provided there's no implied order for $a$ and $b$ given $[x,y]$.)
Is this a correct one-to-one correspondence from ordered sets of two elements to unordered sets of two elements in the domain of imaginary numbers? If so, does it have a name? I'd like to find out more about it.
As an aside, the one element and three element cases seem to work also. Mathematica spent a long time trying to find equations for my guess at the four element case $a+b+c+d,a b+a c+a d+b c+b d+c d,a b c+a b d+a c d+b c d,a b c d$, but eventually I aborted it. I suspect if there is a solution, it might generalize to $n$ elements.
Also, I think it works with ordered lists and unordered lists.
Many thanks!
Edit: as DanielWainfleet points out in the comments "if $x^2=4y$ then $g([x,y])={x/2}$ which has only one element. So the above can't go from ordered sets of two elements to unordered sets of two elements.
So I'd like to revise the question to the highly related case of including sets of one OR two elements (unordered sets stay at two elements).
elementary-set-theory
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
$endgroup$
Let's write ordered sets as $[a,b,c,...]$ and ${a,b,c,...}$ for unordered sets.
Now let $f(a,b)=[a+b,a b]$. Note $f(a,b)=f(b,a).$
Consider $g([x,y])={a,b}$ where $a+b=x,a b=y$.
Mathematica's Solve function came up with $left{frac{1}{2} left(x-sqrt{x^2-4 y}right),frac{1}{2} left(sqrt{x^2-4 y}+xright)right}$ as the value of ${a,b}$ given $[x,y]$, and in my testing, it worked for everything I threw at it. (On a philosophical note it is interesting that just from the definitions provided there's no implied order for $a$ and $b$ given $[x,y]$.)
Is this a correct one-to-one correspondence from ordered sets of two elements to unordered sets of two elements in the domain of imaginary numbers? If so, does it have a name? I'd like to find out more about it.
As an aside, the one element and three element cases seem to work also. Mathematica spent a long time trying to find equations for my guess at the four element case $a+b+c+d,a b+a c+a d+b c+b d+c d,a b c+a b d+a c d+b c d,a b c d$, but eventually I aborted it. I suspect if there is a solution, it might generalize to $n$ elements.
Also, I think it works with ordered lists and unordered lists.
Many thanks!
Edit: as DanielWainfleet points out in the comments "if $x^2=4y$ then $g([x,y])={x/2}$ which has only one element. So the above can't go from ordered sets of two elements to unordered sets of two elements.
So I'd like to revise the question to the highly related case of including sets of one OR two elements (unordered sets stay at two elements).
elementary-set-theory
elementary-set-theory
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
edited Jun 17 '17 at 19:12
Tyler Spaeth
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
asked Jun 17 '17 at 0:06
Tyler SpaethTyler Spaeth
464
464
$begingroup$
What does $g([x,y])={a,b}$ mean here? The way I would usually interpret it doesn't seem to be what you mean.
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:10
1
$begingroup$
See also en.wikipedia.org/wiki/Elementary_symmetric_polynomial
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:13
$begingroup$
It means that $g$ is a function from an unordered set $[x,y]$ to a set ${a,b}$. Maybe $g:[x,y]mapsto{a,b}$ is more correct. As for symmetric polynomials, they look highly related, but I'm not seeing this particular application.
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 0:54
$begingroup$
(Side note: the above should read "It's supposed to mean...". I didn't mean to imply that was a reasonable way of writing that.)
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 1:04
1
$begingroup$
If $x^2=4y$ then $g([x,y]) ={x/2}$ has only $1$ member.
$endgroup$
– DanielWainfleet
Jun 17 '17 at 2:04
|
show 6 more comments
$begingroup$
What does $g([x,y])={a,b}$ mean here? The way I would usually interpret it doesn't seem to be what you mean.
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:10
1
$begingroup$
See also en.wikipedia.org/wiki/Elementary_symmetric_polynomial
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:13
$begingroup$
It means that $g$ is a function from an unordered set $[x,y]$ to a set ${a,b}$. Maybe $g:[x,y]mapsto{a,b}$ is more correct. As for symmetric polynomials, they look highly related, but I'm not seeing this particular application.
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 0:54
$begingroup$
(Side note: the above should read "It's supposed to mean...". I didn't mean to imply that was a reasonable way of writing that.)
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 1:04
1
$begingroup$
If $x^2=4y$ then $g([x,y]) ={x/2}$ has only $1$ member.
$endgroup$
– DanielWainfleet
Jun 17 '17 at 2:04
$begingroup$
What does $g([x,y])={a,b}$ mean here? The way I would usually interpret it doesn't seem to be what you mean.
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:10
$begingroup$
What does $g([x,y])={a,b}$ mean here? The way I would usually interpret it doesn't seem to be what you mean.
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:10
1
1
$begingroup$
See also en.wikipedia.org/wiki/Elementary_symmetric_polynomial
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:13
$begingroup$
See also en.wikipedia.org/wiki/Elementary_symmetric_polynomial
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:13
$begingroup$
It means that $g$ is a function from an unordered set $[x,y]$ to a set ${a,b}$. Maybe $g:[x,y]mapsto{a,b}$ is more correct. As for symmetric polynomials, they look highly related, but I'm not seeing this particular application.
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 0:54
$begingroup$
It means that $g$ is a function from an unordered set $[x,y]$ to a set ${a,b}$. Maybe $g:[x,y]mapsto{a,b}$ is more correct. As for symmetric polynomials, they look highly related, but I'm not seeing this particular application.
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 0:54
$begingroup$
(Side note: the above should read "It's supposed to mean...". I didn't mean to imply that was a reasonable way of writing that.)
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 1:04
$begingroup$
(Side note: the above should read "It's supposed to mean...". I didn't mean to imply that was a reasonable way of writing that.)
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 1:04
1
1
$begingroup$
If $x^2=4y$ then $g([x,y]) ={x/2}$ has only $1$ member.
$endgroup$
– DanielWainfleet
Jun 17 '17 at 2:04
$begingroup$
If $x^2=4y$ then $g([x,y]) ={x/2}$ has only $1$ member.
$endgroup$
– DanielWainfleet
Jun 17 '17 at 2:04
|
show 6 more comments
1 Answer
1
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oldest
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$begingroup$
Does it have a name? — Yes, it's called the quadratic formula along with Vieta's formulas (often also referred to as Vieta's theorem).
That's because effectively, to find $a$ and $b$, you need to solve the equation $t^2-xt+y=0$. By Vieta's formulas, the coefficients $x$ and $y$ and the roots $a$ and $b$ of this quadratic equation are related precisely in the way you defined $x$ and $y$, i.e. $x=a+b$ and $y=ab$. And the proof of this relation is fairly simple too: the quadratic equation with the roots $a$ and $b$ and the leading coefficient $1$ has to be
$$(t-a)(t-b)=0 iff t^2-at-bt+ab=0 iff t^2-underbrace{(a+b)}_{x}t+underbrace{ab}_{y}=0.$$
What you observed with three and more numbers is the same thing extended to polynomial equations of higher degree: symmetric polynomials in the roots of a polynomial equation are the coefficients of that equation.
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
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$begingroup$
Does it have a name? — Yes, it's called the quadratic formula along with Vieta's formulas (often also referred to as Vieta's theorem).
That's because effectively, to find $a$ and $b$, you need to solve the equation $t^2-xt+y=0$. By Vieta's formulas, the coefficients $x$ and $y$ and the roots $a$ and $b$ of this quadratic equation are related precisely in the way you defined $x$ and $y$, i.e. $x=a+b$ and $y=ab$. And the proof of this relation is fairly simple too: the quadratic equation with the roots $a$ and $b$ and the leading coefficient $1$ has to be
$$(t-a)(t-b)=0 iff t^2-at-bt+ab=0 iff t^2-underbrace{(a+b)}_{x}t+underbrace{ab}_{y}=0.$$
What you observed with three and more numbers is the same thing extended to polynomial equations of higher degree: symmetric polynomials in the roots of a polynomial equation are the coefficients of that equation.
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Does it have a name? — Yes, it's called the quadratic formula along with Vieta's formulas (often also referred to as Vieta's theorem).
That's because effectively, to find $a$ and $b$, you need to solve the equation $t^2-xt+y=0$. By Vieta's formulas, the coefficients $x$ and $y$ and the roots $a$ and $b$ of this quadratic equation are related precisely in the way you defined $x$ and $y$, i.e. $x=a+b$ and $y=ab$. And the proof of this relation is fairly simple too: the quadratic equation with the roots $a$ and $b$ and the leading coefficient $1$ has to be
$$(t-a)(t-b)=0 iff t^2-at-bt+ab=0 iff t^2-underbrace{(a+b)}_{x}t+underbrace{ab}_{y}=0.$$
What you observed with three and more numbers is the same thing extended to polynomial equations of higher degree: symmetric polynomials in the roots of a polynomial equation are the coefficients of that equation.
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Does it have a name? — Yes, it's called the quadratic formula along with Vieta's formulas (often also referred to as Vieta's theorem).
That's because effectively, to find $a$ and $b$, you need to solve the equation $t^2-xt+y=0$. By Vieta's formulas, the coefficients $x$ and $y$ and the roots $a$ and $b$ of this quadratic equation are related precisely in the way you defined $x$ and $y$, i.e. $x=a+b$ and $y=ab$. And the proof of this relation is fairly simple too: the quadratic equation with the roots $a$ and $b$ and the leading coefficient $1$ has to be
$$(t-a)(t-b)=0 iff t^2-at-bt+ab=0 iff t^2-underbrace{(a+b)}_{x}t+underbrace{ab}_{y}=0.$$
What you observed with three and more numbers is the same thing extended to polynomial equations of higher degree: symmetric polynomials in the roots of a polynomial equation are the coefficients of that equation.
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
$endgroup$
Does it have a name? — Yes, it's called the quadratic formula along with Vieta's formulas (often also referred to as Vieta's theorem).
That's because effectively, to find $a$ and $b$, you need to solve the equation $t^2-xt+y=0$. By Vieta's formulas, the coefficients $x$ and $y$ and the roots $a$ and $b$ of this quadratic equation are related precisely in the way you defined $x$ and $y$, i.e. $x=a+b$ and $y=ab$. And the proof of this relation is fairly simple too: the quadratic equation with the roots $a$ and $b$ and the leading coefficient $1$ has to be
$$(t-a)(t-b)=0 iff t^2-at-bt+ab=0 iff t^2-underbrace{(a+b)}_{x}t+underbrace{ab}_{y}=0.$$
What you observed with three and more numbers is the same thing extended to polynomial equations of higher degree: symmetric polynomials in the roots of a polynomial equation are the coefficients of that equation.
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
answered Dec 24 '18 at 17:25
zipirovichzipirovich
11.3k11731
11.3k11731
add a comment |
add a comment |
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$begingroup$
What does $g([x,y])={a,b}$ mean here? The way I would usually interpret it doesn't seem to be what you mean.
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:10
1
$begingroup$
See also en.wikipedia.org/wiki/Elementary_symmetric_polynomial
$endgroup$
– Matt Samuel
Jun 17 '17 at 0:13
$begingroup$
It means that $g$ is a function from an unordered set $[x,y]$ to a set ${a,b}$. Maybe $g:[x,y]mapsto{a,b}$ is more correct. As for symmetric polynomials, they look highly related, but I'm not seeing this particular application.
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 0:54
$begingroup$
(Side note: the above should read "It's supposed to mean...". I didn't mean to imply that was a reasonable way of writing that.)
$endgroup$
– Tyler Spaeth
Jun 17 '17 at 1:04
1
$begingroup$
If $x^2=4y$ then $g([x,y]) ={x/2}$ has only $1$ member.
$endgroup$
– DanielWainfleet
Jun 17 '17 at 2:04