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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?

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    $\begingroup$ 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
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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??

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