1

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.

balddraz
  • 7,558
oma
  • 81
  • 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 07 '19 at 08:48
  • edited it.wanted to ask about transitive and anti symmetric then i thought i should see if my reflexive proof is right – oma May 07 '19 at 08:51

3 Answers3

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 $1\mathrel R4$.
José Carlos Santos
  • 427,504
0

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?

Bernard
  • 175,478
  • 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 07 '19 at 09:20
  • For a counter-example, you can specify any values you please. However, the values you mention do not satisfy the relation. – Bernard May 07 '19 at 09:29
  • 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 07 '19 at 09:56
  • So you have a counter-example, and transitivity is not satisfied for all $x,y,z$. – Bernard May 07 '19 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 07 '19 at 10:11
  • I'm not sure of what you mean exactly. The relation $x\mathcal R y$ is defined by $xy-y^2-x+y=0$? – Bernard May 07 '19 at 10:17
  • yes my bad. it should be transitive because it does not have a counter example right? – oma May 07 '19 at 10:22
  • Not finding a counter-example doesn't prove an assertion is true. However in the present case, it is indeed transitive. To see it, factorise the polynomial as $(x-y)(y-1)$, so that $x\mathcal R y\iff (x=y)$ or $(y=1)$. – Bernard May 07 '19 at 10:27
  • now this got confusing how can i use $x=y$ or $y=1$ to get my answer for sure (sorry for the many replies just that every time i think i got the answer i get faced with another problem ) – oma May 07 '19 at 10:39
  • You have to examine the different possibilities for $(x,y)$ and for $(y,z)$, combine them in order to deduce each time one of the possibilities for $(x,z)$ and finally organize the proof to have it as short as possible. – Bernard May 07 '19 at 10:51
  • so for one we take $y=1$ we find x and z possibilities and for other we $x=y$ so we have $y=z$ then $x=z$ which is true so we only have to figure out the first one only? – oma May 07 '19 at 11:07
  • Personally, I would start from $z$: if $z=1$ there's nothing to prove. If $z\ne 1$, then $y=z$, so $y\ne 1$ too, &c. – Bernard May 07 '19 at 11:18
  • what do you mean by $z=1$ there is nothing to prove? from what i calculated there is a possibility for $z$ to be 1 so $y$ should also be equal to 1 – oma May 07 '19 at 11:26
  • By the factorisation, if $z=1$, we have automatically $x\mathcal R z$ (and $y\mathcal R z$). – Bernard May 07 '19 at 12:00
0

$\text{R}$ 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.

Community
  • 1
s0ulr3aper07
  • 1,147
  • 7
  • 18