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Show that the set of natural numbers may be chosen as a set of representatives for the equivalence classes of the kernel relations $\textbf{ker}sq$, where $sq=\mathbb{Z} \to \mathbb{Z};n \mapsto n^2$. I understand that the set of natural numbers is a set of representatives, i'm just unsure how to prove it. I am mainly just looking for a hint to get me started, any help is appreciated thanks!

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Two elements $a,b$ are related under the relation $\sim_{sq}$ if and only if $a^2 = b^2 \implies a=\pm b$. Hence, any equivalence class $[a]$ contains only the elements $\{a,-a\}$. Now, consider the map $a \in \mathbb N, a \to \{ a,-a\}$. This is a map from $\mathbb N$ to the set of equivalence classes under $\sim_{sq}$.

It is injective : If $\{ -a, a\} = \{ b,-b\}$, then $a= \pm b$, but since both are natural numbers, either $a=b$ or $a=b=0$, in which case they are equal anyway.

Checking surjectivity is even easier, I leave you to see this.

Hence, $\mathbb N$ can be identified with the equivalence classes under $\sim_{sq}$, so every equivalence class can be represented with a natural number.

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    Could I similarly assume $sq(n_1)=sq(n_2)$ then use the following argument. $n_1=\sqrt{sq(n_1)}=\sqrt{sq(n_2)}=n_2$ – 2017-02-14
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    Yes, except that the last equality is wrong : $\sqrt{sq(n_2)} = \pm n_2$. – 2017-02-14
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    Ah yes true, then could I rule out the negative as the set of representatives is the natural numbers? – 2017-02-14
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    @JustinStevenson I get what you say, and yes, that is right. – 2017-02-14
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    Okay great, thanks for your help sir! – 2017-02-14
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    I, er, do not call me Sir. It is harsh on you, you must be better than me. Look at my profile for more details! – 2017-02-14