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.

  • $\begingroup$ What is your question? The title mentions transitive and anti/symmetric (whatever that is) but, in the body of your question, you start by mentioning reflexive. $\endgroup$ May 7, 2019 at 8:48
  • $\begingroup$ edited it.wanted to ask about transitive and anti symmetric then i thought i should see if my reflexive proof is right $\endgroup$
    – oma
    May 7, 2019 at 8:51

3 Answers 3


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 $1\mathrel R4$.


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?

  • $\begingroup$ i know this is not directly related to the question but can i make x=2 y=1 z=1? or all of them should be different numbers. $\endgroup$
    – oma
    May 7, 2019 at 9:20
  • $\begingroup$ For a counter-example, you can specify any values you please. However, the values you mention do not satisfy the relation. $\endgroup$
    – Bernard
    May 7, 2019 at 9:29
  • $\begingroup$ i said not directly because i tried on another equation but if i said x=1 y=2 z=2 i can still get (x,z)=0 in this question but if i went z=4 it wont be transitive anymore $\endgroup$
    – oma
    May 7, 2019 at 9:56
  • $\begingroup$ So you have a counter-example, and transitivity is not satisfied for all $x,y,z$. $\endgroup$
    – Bernard
    May 7, 2019 at 10:01
  • $\begingroup$ all makes sense now so for an equation such as xy-y2-x+y = 0 if we wanted to get (y,z) into zero z must be 1 or 2 which both will satisfy the equation (x,z) making it transitive? $\endgroup$
    – oma
    May 7, 2019 at 10:11

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


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

Now using a technique called Completing the Square we have:

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


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 $$x\text{R}y\Rightarrow x=\frac{y}{2}$$ $$y\text{R}x\Rightarrow y=\frac{x}{2}$$


$$x= y=0$$

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


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


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


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