\newcount\x \newcount\nx
\def\mysqrt#1{\x=#1 \nx=\x \isqrt{#1}
\big\lfloor\sqrt{#1}\big\rfloor=\number\x}
\def\isqrt#1{\nx=#1
\divide\nx by\x \advance\nx by\x \divide\nx by2
\ifnum\x=\nx \else\x=\nx \isqrt{#1}\fi}
$\mysqrt{361}$
KTUGFaq
KTUG FAQ
You have an ability to sense and know higher truth.
LittleTree/ReadingTeXbook/2006-04 › Karnes/2007-04 › NSIS › TeXmacs › Karnes/2012-02 › ÀÛÀº³ª¹«/2008-03