Post History
#2: Post edited
by
Derek Elkins
·
2024-06-04T07:17:31Z (7 months ago)
Make more legible and grammatical.
$(S,\otimes)$ is not entropic, next is an example, that is, a counterexample to the claim of the exercise.$(Z_3,\odot)$, where $\odot$ is defined as $x\odot y=x+2y$, is entropic: $(x\odot y)\odot(z\odot w)=$$(x+2y)+2(z+2w)=$$x+2y+2z+4w=$$(x+2z)+2(y+2w)=$$(x\odot z)\odot(y\odot w)$. Define $\alpha(x):=x^2$ and $\beta(x):=2(x-x^2)$; $(\alpha\odot\beta)(x)=$$\alpha(x)+2\beta(x)=$$x^2+4(x-x^2)=$$x^2+x-x^2=x$. Construe $\otimes$ as $x\otimes y:=\alpha(x)\odot\beta(x)$; given $x,y,z,w\in Z_3$, $(x\otimes y)\otimes(z\otimes w)=$$\alpha(\alpha(x)+2\beta (y))+2\beta(\alpha(z)+2\beta(w))=$$2w^4+2w^3+2w^2z^2+w^2+wz^2+w+$$x^4+x^2y^2+2x^2y+$$y^4+y^3+y^2+2z^4+z^2=$$2w^2+2w^2+2w^2z^2+w^2+wz^2+w+$$x^2+x^2y^2+2x^2y+y^2+$$y^2+y^2+2z^2+z^2=$$2w^2+2w^2z^2+wz^2+w+x^2+x^2y^2+2x^2.$ In conclusion, $(Z_3,\otimes)$ is not entropic, for example, $(1\otimes2)\otimes(0\otimes0)=0eq1=$$(1\otimes0)\otimes(2\otimes0)$.
- $(S,\otimes)$ is not entropic. The following is an example, that is, a counterexample, to the claim of the exercise.
- $(\mathbb Z_3,\odot)$ where $\odot$ is defined as $x\odot y=x+2y$, is entropic: $$\begin{align}
- (x\odot y)\odot(z\odot w)
- & = (x+2y)+2(z+2w) \\\\
- & = x+2y+2z+4w \\\\
- & = (x+2z)+2(y+2w) \\\\
- & = (x\odot z)\odot(y\odot w).
- \end{align}$$
- Define $\alpha(x):=x^2$ and $\beta(x):=2(x-x^2)$. Then
- $$\begin{align}
- (\alpha\odot\beta)(x)
- & = \alpha(x)+2\beta(x) \\\\
- & = x^2+4(x-x^2) \\\\
- & = x^2+x-x^2 \\\\
- & = x.
- \end{align}$$
- Construe $\otimes$ as $x\otimes y:=\alpha(x)\odot\beta(x)$. Given $x,y,z,w\in \mathbb Z_3$, $$\begin{align}
- & (x\otimes y)\otimes(z\otimes w) \\\\
- & = \alpha(\alpha(x)+2\beta (y))+2\beta(\alpha(z)+2\beta(w)) \\\\
- & = 2w^4+2w^3+2w^2z^2+w^2+wz^2+w+x^4+x^2y^2+2x^2y+y^4+y^3+y^2+2z^4+z^2 \\\\
- & = 2w^2+2w^2+2w^2z^2+w^2+wz^2+w+x^2+x^2y^2+2x^2y+y^2+y^2+y^2+2z^2+z^2 \\\\
- & = 2w^2+2w^2z^2+wz^2+w+x^2+x^2y^2+2x^2.
- \end{align}$$
- In conclusion, $(\mathbb Z_3,\otimes)$ is not entropic. For example, $$(1\otimes2)\otimes(0\otimes0)=0
- eq 1=(1\otimes0)\otimes(2\otimes0).$$
#1: Initial revision
by
JohnnyJohn
·
2024-06-02T10:06:22Z (7 months ago)
$(S,\otimes)$ is not entropic, next is an example, that is, a counterexample to the claim of the exercise. $(Z_3,\odot)$, where $\odot$ is defined as $x\odot y=x+2y$, is entropic: $(x\odot y)\odot(z\odot w)=$$(x+2y)+2(z+2w)=$$x+2y+2z+4w=$$(x+2z)+2(y+2w)=$$(x\odot z)\odot(y\odot w)$. Define $\alpha(x):=x^2$ and $\beta(x):=2(x-x^2)$; $(\alpha\odot\beta)(x)=$$\alpha(x)+2\beta(x)=$$x^2+4(x-x^2)=$$x^2+x-x^2=x$. Construe $\otimes$ as $x\otimes y:=\alpha(x)\odot\beta(x)$; given $x,y,z,w\in Z_3$, $(x\otimes y)\otimes(z\otimes w)=$$\alpha(\alpha(x)+2\beta (y))+2\beta(\alpha(z)+2\beta(w))=$$2w^4+2w^3+2w^2z^2+w^2+wz^2+w+$$x^4+x^2y^2+2x^2y+$$y^4+y^3+y^2+2z^4+z^2=$$2w^2+2w^2+2w^2z^2+w^2+wz^2+w+$$x^2+x^2y^2+2x^2y+y^2+$$y^2+y^2+2z^2+z^2=$$2w^2+2w^2z^2+wz^2+w+x^2+x^2y^2+2x^2.$ In conclusion, $(Z_3,\otimes)$ is not entropic, for example, $(1\otimes2)\otimes(0\otimes0)=0\neq1=$$(1\otimes0)\otimes(2\otimes0)$.