deletions | additions       diff --git a/Assume_we_are_comparing_two__.tex b/Assume_we_are_comparing_two__.tex  index 0d7e8bb..de84251 100644  --- a/Assume_we_are_comparing_two__.tex  +++ b/Assume_we_are_comparing_two__.tex 
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Our function is continuous, so by the intermediate value theorem there has to be some value in the range of the function where $f(x)=0$. This implies that at some angle of the scanline $x$, $\|a_x\| = \|b_x\|$. In polar coordinates we've uniquely defined a point in the plane that both intervals occupy. This contradicts the non-intersection of our intervals.  \end{proof}  This lemma indicates that it is sufficient to determine the ordering of compare  intervalsalong a scan line  at one point. angle of the scanline.