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I am going through past papers for my university exam, and a question in this format appears often:

Define a relation $\sim$ on $\Bbb{R}$ by $x\sim y$ if and only if $x+y \in \Bbb{Q}$. Justify your answers to the following questions. Is this relation (a) reflexive? (b) symmetric? (c) transitive?

I know the definitions to each, but I don’t know where I’d start in proving and therefore explaining the rationale in answering these questions.

Any help would be immensely appreciated.

Blue
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Scottfitzgerald
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    Since you know the definitions for each, try to check whether this relation satisfies each of those definitions. Start with "The relation ~ is reflexive because if $x$ ~$y$ then ..." – Ethan Bolker May 05 '22 at 14:58
  • To start with, note that $\sqrt{2}$ IS in R but $\sqrt{2}+ \sqrt{2}$ is NOT in Q so $\sqrt{2}~\sqrt{2}$ is not true and ~ is NOT reflexive! – George Ivey May 05 '22 at 16:02

2 Answers2

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Try checking each of the attributes one by one.

As an example (for reflexivity)

$$\sim \text{is reflective} \iff \forall x \in \mathbb{R}: x \sim x$$

Let $x = \sqrt{2} \in \mathbb{R}$. Now the definition states, that $$ x \sim x \iff x + x \in \mathbb{Q}$$ However: $$ \sqrt{2} + \sqrt{2} = 2 \cdot \sqrt{2} \notin \mathbb{Q}$$ As one can see the relation does not seem to be reflexive (note that in this case I wanted to disprove the definition and as such I used an example. If one wanted to prove it one would have to let $x \in \mathbb{R}$ 'arbitrary' and prove the statement for the arbitrary x.

Try this for symmetry and transitivity.

(The definitions are as follows: $$ \sim \text{ is symmetrical} \iff \forall x \sim y \in \mathbb{R}: y \sim x $$ $$ \sim \text{ is transitive} \iff \forall x \sim y, y \sim z \in \mathbb{R}: y \sim z $$ (As I said trying chosing x,y (and z) as arbitrary in $\mathbb{R}$ and then show the statement)

Raoul Luqué
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A relation is reflexive if for every element x in the set you have $x\sim x$. In this case, this is false because adding any irrational number to itself makes it irrational. $\sqrt7 + \sqrt7$.

It should be symmetric because it doesn't matter which order the x and y are put in (commutative property), $x\sim y = y\sim x$.

To prove that the relation isn't transitve we have to show a case such that if $x\sim y$ and $y\sim z$ are valid, but not $x\sim z$. If you add a irrational number to a rational number you get an irrational. I think it's transitive because I can't think of a single case where the sum of z and x is irrational without violating the previous relationships.

Graham Kemp
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Ethan _kom
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  • To see it isn't transitive, pick your favorite irrational number $r$ and consider $x=-y=z=r$ – Mor A. May 06 '22 at 08:56