Exercise 4.1.30 ($(2a)!(2b)! /(a!b!(a+b)!)$ is an integer)

Question

Show that $(2a)!(2b)! /(a!b!(a+b)!)$ is an integer.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

\begin{proof} 
Using the de Polignac's formula, we obtain for every prime $p$
\begin{align}

u_p((2a)!(2b)!) &= \sum_{i=1}^\infty \left( \left \lfloor \frac{2a}{p^i} ight floor +  \left \lfloor \frac{2b}{p^i} ight floor ight),\

u_p(a!b!(a+b)!) & =  \sum_{i=1}^\infty \left(   \left \lfloor \frac{a}{p^i} ight floor +  \left \lfloor \frac{b}{p^i}  ight floor +  \left \lfloor \frac{a+b}{p^i}ight floor ight).
\end{align}
By Problem 16, for every $\alpha \in \R$ and $\beta \in \R$,
$$\lfloor 2\alpha floor + \lfloor 2\beta floor \geq \lfloor \alpha floor+ \lfloor \beta floor + \lfloor \alpha + \beta floor.$$
By applying this  inequality to $\alpha= \frac{a}{p^i},\ \beta = \frac{b}{p^i}$, we obtain for every positive integer $i$
$$ \left \lfloor \frac{2a}{p^i} ight floor +  \left \lfloor \frac{2b}{p^i} ight floor \geq \left \lfloor \frac{a}{p^i} ight floor +  \left \lfloor \frac{b}{p^i}  ight floor +  \left \lfloor \frac{a+b}{p^i}ight floor.$$
By equations (1) and (2), for every prime $p$,
$$
u_p((2a)!(2b)!)  \geq 
u_p(a!b!(a+b)!) ,$$
This shows that $a!b!(a+b)! \mid (2a)!(2b)! $, so $(2a)!(2b)! /(a!b!(a+b)!)$ is an integer.
\end{proof}

Exercise 4.1.30 ($(2a)!(2b)! /(a!b!(a+b)!)$ is an integer)

Answers

Comments

Exercise 4.1.30 ($(2a)!(2b)! /(a!b!(a+b)!)$ is an integer)

Answers

Comments

Add answer