\(1 > 2\) !

Level pending

\(\frac {1}{10} > \frac {1}{100}\)
\(ln \frac {1}{10} > ln \frac {1}{100}\)
\(ln \frac {1}{10} > ln (\frac {1}{10})^2\)
\(ln \frac {1}{10} > 2 \cdot ln (\frac {1}{10})\)

Cancelling \(ln(\frac {1}{10})\), we obtain \(1 > 2\)