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24 votes
4 answers
2k views

A reinterpretation of the $abc$ - conjecture in terms of metric spaces?

I hope it is appropriate to ask this question here: One formulation of the abc-conjecture is $$ c < \text{rad}(abc)^2$$ where $\gcd(a,b)=1$ and $c=a+b$. This is equivalent to ($a,b$ being ...
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0 votes
1 answer
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Prove that $(v^Tx)^2-(u^Tx)^2 < 1-(u^Tv)^2$ for any unit vectors $u$, $v$, $x$

Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$. That is $\lVert uu^T-vv^T\rVert^2_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,...
Dan Feldman's user avatar