Proving $2x^2-3xy+y^2=0$ is transitive and anti-symmetric or symmetric and reflexive.

Question

Let $R$ be the binary relation defined on $\mathbb{R}$ by $xRy$ iff $2x^2-3xy+y^2=0$

For reflexive we get $2x^2=2x^2\implies-x=x$ which means reflexive on $xRx$

$2x^2-3xy+y^2=0$ tried going for $2y^2-yz+z^2=0$ then adding them together but now I'm stuck with a long useless equation any tips of proving this transitive as for anti symmetric i know $2y^2-3yx+x^2=0$ in case $x=y$ so it should be anti symmetric but I don't know how to say it.

Answer 1

Note that\begin{align}x\mathrel Ry&\iff 2x^2-3xy+y^2=0\\&\iff(y-x)(y-2x)=0\\&\iff y=x\vee y=2x.\end{align}So:

• It is not symmetric, since $1\mathrel R2$, but you don't have $2\mathrel R1$.
• It is antisymmetric, since, if $x\neq y$, you cannot have $x\mathrel Ry$ and $y\mathrel Rx$.
• It is not transitive: you have $1\mathrel R2$ and $2\mathrel R4$, but you don't have $1\mathrel R4$.

Answer 2

Hint:

Set $\varphi(x,y)=2x^2-3xy+y^2$. With this notation, $$ x\mathcal R y\stackrel{\text{def}}{\iff}\varphi(x,y)=0. $$ Now the relation $\mathcal R$ is

• reflexive if $\;\varphi(x,x)=0$ for all $x$,
• symmetric if $\;\varphi(x,y)=0\implies\varphi(y,x)=0$ for all $x,y$,
• anti-symmetric if $\;\varphi(x,y)=0$ and $\;\varphi(y,x)=0$ imply $x=y$,
• transitive if, for all $x,y,z$, $\;\varphi(x,y)=0$ and $\;\varphi(y,z)=0$ imply $\;\varphi(x,z)=0$.

Can you prove or find counter-examples for any of these?

Answer 3

$\text{R}$ is a binary relation on $\mathbb{R}$ defined as $x\text{R}y\iff 2x^2-3xy+y^2=0$

Reflexive:

If $x=y$, then $2x^2-3xy+y^2=2x^2-3x^2+x^2= 0$

Hence, we have $x\text{R}x\ \ \forall \ x \in\mathbb{R}$ and the relation is Reflexive

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Now using a technique called Completing the Square we have:

$$\begin{align}x\text{R}y &\iff 2x^2-3xy+y^2=0\\ &\iff x^2-\frac{3}{2}xy+\frac{y^2}{2}=0\\ &\iff x^2-\frac{3}{2}xy+\frac{y^2}{2}+\big(\frac{3y}{4}\big)^2-\big(\frac{3y}{4}\big)^2=0\\ &\iff\big(x-\frac{3y}{4}\big)^2-\big(\frac{y}{4}\big)^2=0\\ &\iff \big(x-\frac{y}{2}\big)\big(x-y\big)=0\\ &\iff x=\frac{y}{2} \ \ \text{Or} \ \ x=y \ \ \ \ \ \ \ \ \ \ \ \ -(1) \end{align}$$

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Antisymmetric:

We will proceed by contradiction. Let us assume $\text{R}$ is not antisymmetric. Then $\exists \ x,y\in\mathbb{R}$ such that $x\neq y$ and both $x\text{R}y$, $y\text{R}x$ are satisfied. Since $x\neq y$, by $(1)$ we have the following $$x\text{R}y\Rightarrow x=\frac{y}{2}$$ $$y\text{R}x\Rightarrow y=\frac{x}{2}$$

Thus,

$$x= y=0$$

Which leads to a contradiction. Hence, $\text{R}$ is antisymmetric.

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Transitive:

The relation is not transitive because $3\text{R}6$ and $6\text{R}12$ but $3\text{R}12$ does not hold.

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Symmetric:

The relation is not symmetric because $6\text{R}12$ but $12\text{R}6$ does not hold.