# bachotex2008-david-kastrup-pearl2.tex

bachotex2008-david-kastrup-pearl2.tex — TeX document, 1 KB (1268 bytes)

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%%% Is 2in=1in+1in? \emergencystretch 3em \def\stripit#1>{} \def\showx#1{\def\x{#1}{\tt \expandafter\stripit\meaning\x}} \TeX's arithmetic forgets to round when dealing with units. For example, 1in is defined to be exactly 72.27pt. We can check that by writing \showx{\dimen0=100in \the\dimen0} and getting \x. % compile to see what is \x So far, so good. But when we write \showx{\dimen0=1in \the\dimen0}, we get \x. Oops. This becomes worse since a number called 72.27pt actually exists, as witnessed by \showx{\dimen0=72.27pt \the\dimen0} which gives us \x. So we have the unfortunate situation that {\tt 1in} (which should be exactly 72.27pt) gives a number different from the number actually called {\tt 72.27pt}. This is because \TeX\ truncates when working with the fractions representing exact units, but rounds when working with decimal fractions. \TeX\ calculates {\tt 1in} as $$\left\lfloor 1\times{7227\over100}\times 2^{16}\right\rfloor2^{-16}$$ namely truncating the fractional calculation rather than rounding it. Other artifacts of \TeX's fractional representation of units mean that $\tt 2in \neq 1in+1in$: Indeed \showx{\dimen0=1in \advance\dimen0 1in \the\dimen0 \dimen0=2in =\the\dimen0} leaves us with \x. \end

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