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

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.

• 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. – José Carlos Santos May 7 at 8:48
• edited it.wanted to ask about transitive and anti symmetric then i thought i should see if my reflexive proof is right – oma May 7 at 8:51

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 $$1\mathrel R4$$.

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?

• 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. – oma May 7 at 9:20
• For a counter-example, you can specify any values you please. However, the values you mention do not satisfy the relation. – Bernard May 7 at 9:29
• 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 – oma May 7 at 9:56
• So you have a counter-example, and transitivity is not satisfied for all $x,y,z$. – Bernard May 7 at 10:01
• 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? – oma May 7 at 10:11

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

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

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}

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.

Transitive:

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

Symmetric:

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