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'Adding in Rows' printed from https://nrich.maths.org/
Congratulations to Soh Yong Sheng from Raffles Institution,
Singapore for this excellent solution.
We have $0 < a < b$ which means $1/a > 1/b$ and so $3 +
1/a > 3 + 1/b$. Flipping over the fraction, we will get 1/(3 +
1/a) < 1/(3 + 1/b) and the inequality remains the same way round
when 2 is added. Flipping over again for the last time we get \[
\frac{1}{2+\frac{1}{3+\frac{1}{a}}} \] is greater than \[
\frac{1}{2+\frac{1}{3+\frac{1}{b}}} \]
The second part is a further expansion of the first, and in the
process of repeating the above we know that it involves just one
more flipping over of the fraction, thus \[
\frac{1}{2+\frac{1}{3+\frac{1}{4 + \frac{1}{a}}}} \] is less than
the same thing with $b$ in place of $a$ as the inequality would be
reversed again.
Lastly the continued fractions are expanded all the way down to
$100 + 1/a$ and $100 + 1/b$. Observe the above process, we can tell
that if the last or biggest number is odd then the continued
fraction with a in it is bigger. If the last or biggest number is
even then the continued fraction with $b$ in it is bigger. Each
successive continued fraction involves one more 'flipping over' and
reverses the inequality one more time. The following continued
fraction is smaller than the same thing with $b$ in place of $a$:
$${1\over\displaystyle 2 + { 1 \over \displaystyle 3+ { 1\over
\displaystyle 4 + \dots + {1\over\displaystyle 99+ {1\over
\displaystyle {100 + {1 \over \displaystyle a}} }}}}}$$