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Consider the relation $R= \{(x, y) \mid x-y = 0 \} \subset \mathbb{R} \times \mathbb{R}$ on the set $\mathbb{R}$. Which of the following is/are true?

  1. $R$ is a transitive relation.

  2. $R$ is a function.

  3. $R$ is not an equivalence relation.

  4. $R$ is a reflexive relation.

  5. $R$ is a symmetric relation.

I've answered this question by selecting 2, 3, and 4. Is my selection of choice correct? Can someone please validate it?

$R$ is a reflexive relation. $\Rightarrow$ because only the matrix's diagonal elements will be 1.

$R$ is not an equivalence relation. $\Rightarrow$ because there is only one relation Reflexive, Transitive, and Symmetric relations are not applicable.

I need help in understanding how $R$ is a function...

Asaf Karagila
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Abbas
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2 Answers2

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$1$. You are correct about the relation being reflexive. But rather than referencing a matrix, you can simply say that: $x-x=0$ for all real numbers and thus $(x, x)\in R$ and $R$ is reflexive.

$2$. As pointed out in the comments, this is an equivalence relation(my bad I didn't really bother checking). This is because $x-y=0$ if and only if $x=y$. Thus $y-x=0$ and the relation is symmetric. This relation is also transitive as if $x-y=0$ and $y-a=0$. Then $a=y=x$ and $x-a=0$ and thus $R$ is transitive. Hence, it is an equivalence relation.

$3$. You should look at the definition of what is a function. A subset $R$ of $S\times T$ is said to be a function

  • If for all $s\in S$, there exists a $t\in T$ such that $(s, t)\in R$.

  • If $(s, t)\in R$, then $(s, x)\notin R$ for $x\ne t$.

In the case of your relation, both conditions are satisfied as $x-y=0$ if and only if $x=y$. Thus for every $x\in \Bbb{R}$, there exists a $y\in \Bbb{R}$ such that $(x, y)\in R$. Also if $(x, y)\in R$, then $(x, a)\notin R$ as that would imply $a=x=y$. Thus your relation is a function.

Seeker
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You have $ ( x , y ) \in R $ if and only if $ y - x = 0 $; this is what the definition of $ R $ says.

  1. To be transitive, you need for real numbers $ x $, $ y $, and $ z $, that if $ ( x , y ) \in R $ and $ ( y , z ) \in R $, then $ ( x , z ) \in R $. In other words, if $ y - x = 0 $ and $ z - y = 0 $, then $ z - x = 0 $. So if you can prove that $ z - x = 0 $ assuming only that $ y - x = 0 $ and $ z - y = 0 $, then $ R $ is transitive; but if you can find even a single example for $ x $, $ y $, and $ z $ such that $ y - x = 0 $ and $ z - y = 0 $ but $ z - x \ne 0 $, then it's not transitive.

  2. To be a function, you need for each $ x $, for some unique $ y $, $ ( x , y ) \in R $. In other words, for each $ x $, there is exactly one $ y $ such that $ y - x = 0 $. It's generally best to express $ y $ using $ x $, if you can (which in this case you can), so do that. Then you prove that $ y - x = 0 $ (using your $ y $), and that if $ y ' - x = 0 $, then $ y = y ' $. If you can do this, then it's a function. On the other hand, if you can find either an $ x $ with no $ y $ such that $ y - x = 0 $, or an $ x $, a $ y $, and a $ y ' $ such that $ y - x = 0 $ and $ y ' - x = 0 $ but $ y \ne y ' $, then it's not a function.

  3. If in (1) you found that $ R $ is not transitive, then you already know that it's not an equivalence relation. But if $ R $ is transitive, then you still need to see if it's reflexive and symmetric. If it is, then it's an equivalence relation; if either of these fails, then it's not. So come back to this one.

  4. To be reflexive, you need that $ ( x , x ) \in R $ always. In other words, that $ x - x = 0 $. Since this is true, you're right, it's reflexive. But if there were an $ x $ such that $ x - x \ne 0 $, then it wouldn't be. (And if it weren't, then you could answer (3) now too.)

  5. To be symmetric, you need that if $ ( x , y ) \in R $, then $ ( y , x ) \in R $ too. In other words, if $ y - x = 0 $, then $ x - y = 0 $. If you can prove that $ x - y = 0 $ assuming only that $ y - x = 0 $, then it's symmetric; but if you can find an $ x $ and a $ y $ such that $ y - x = 0 $ but $ x - y \ne 0 $, then it's not. (And you can answer (3) by now, no matter what.)

Incidentally, this relation often goes by the name $ \Delta $, or $ \Delta _ { \mathbb R } $ if you want to emphasize what set it's on.

Toby Bartels
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