I don't know how to go about proving symmetry. I have proven that the relation is reflexive. But I have no idea how to start with proving the symmetry of a given relation.
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$$(a^2-b^2)(a^2b^2-1)=0\iff (b^2-a^2)(b^2a^2-1)=0.$$
Walace
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Yes, I know that. But can ⟨a²-b²) be the same as ⟨b²-a²)? That's where I got stuck. – za stacka2345 Apr 25 '20 at 11:56
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@zastacka2345: No, they are not the same, nevertheless, $(a^2-b^2)(a^2b^2-1)=(b^2-a^2)(b^2a^2-1)$ since both are zero since $a\sim b$. – Walace Apr 25 '20 at 12:00
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Considering $$y=\left(a^2-b^2\right)\left(a^2b^2-1\right)$$ let $b=k a$ to get $$y=a^2 \left(1-k^2\right) \left(a^4 k^2-1\right)$$ Expand it as a polynomial in $(k-1)$ to get $$y=\left(2 a^2-2 a^6\right) (k-1)+\left(a^2-5 a^6\right) (k-1)^2-4 a^6 (k-1)^3-a^6 (k-1)^4$$ Now, if $b \sim a$, $k\sim 1$.
Claude Leibovici
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