Following relation is defined on $\mathbb{N}^{+}$:

$$xRy :\Leftrightarrow x \cdot y \text{ is a square number }$$

This relation is an equivalence relation. True or false?

I first need know how you get a square number when you have two factors.

Wikipedia say: "[...] is the product of some integer with itself."

But this is confused because example $9$ is square number. How you get it when you have two factors? You get it with $3 \cdot 3$ like Wikipedia say correct but you can also get it with $1 \cdot 9$ or $9 \cdot 1$.

For the reason the relation isn't equivalence relation because example:

$1 R 9$ and $9 R 1$, we have relation $\left\{(1,9),(9,1)\right\}$ this isn't equivalence relation because it's no reflexive, also no transitive.

I do all good or all wrong?

  • 1
    $\begingroup$ You are right that $1R9$ and $9R1$, so the relation $R$ contains $(1, 9)$ and $(9,1)$. However, $R$ contains more than that. We also have $1R1$ and $9R9$, so the relation also contains $(1, 1)$ and $(9,9)$. Thus you still haven't found a counterwitness to transitivity or reflexifity. $\endgroup$ – Arthur Nov 13 '17 at 11:48

For all $k, l \in \mathbb {N}^+$:

Reflexive: We can easily prove $x Rx $.

Symmetric: If $xRy $ is square, $xy = k^2$ $\implies yx = k^2$ $\implies yRx $. (Due to commutativity of multiplication)

Transitive: If $xRy $, that is, $xy = k^2$ and $yRz$, that is, $yz = l^2$, can you show that $xRz $ also holds??


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.