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$\textbf{Lemma:}$ if $a, b\in\mathbb{R}$ then $|a-b| \ge |a| - |b|$ $\textbf{Proof:}$ by the triangle inequality $$ |a| = |(a-b) + b| \le |a-b| + |b|\quad\Longrightarrow\quad |a-b| \ge |a| - |b|$$ $\textbf{Claim:}$ if $a, b\in\mathbb{R}$ and $|a-b|\le b^2$ then $|\frac ab|\le |b|+1$. $\textbf{Proof:}$ Dividing the condition in the claim by $|b|$ we get $$\left|\frac ab-1\right|\le |b|$$ According to the lemma $$\left|\frac ab\right| - 1 \le \left|\frac ab - 1\right|$$ Putting the two together we get what we need.
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Sun, 28 Oct 2018 13:06 GMT