Exercise 4.1.17* ($n!(n-1)! \mid (2n-2)!$)

Proof. By Problem 1.4.25, we know that

We show that the left member is even. Since $m^2\equiv m(\mathit{mod}2)$ for every integer $m$, we obtain modulo $2$

Therefore, for every integer $n\ge 0$,

so

If we replace $n$ by $n-1$, we obtain that for every positive integer $n$,

Therefore $n{((n-1)!)}^2\mid (2n-2)!$, that is

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Proof. For every prime number $p$,

We prove now that for every positive integer $i$,

(We note that this is false if $i=0$.)

Put $\alpha =\frac{1}{p^i}$. Then $0<\alpha \le \frac{1}{2}$. We want to prove

We write $\mathit{n\alpha }=k+\nu$, where $k\in \mathbb{Z}$ and $0\le \nu <1$, so that $k=\lfloor \mathit{n\alpha }\rfloor$. Then

Since $0<\alpha \le 1∕2$ and $0\le \nu <1$, then $-1∕2\le \nu -\alpha <1$. We examine three cases (the uncomfortable case $-1\le \nu -\alpha <-1∕2$ cannot occur here).

• If $-1∕2\le \nu -\alpha <0$, then $-1\le 2(\nu -\alpha )<0$, thus

  $$\lfloor \mathit{n\alpha }\rfloor +\lfloor (n-1)\alpha \rfloor =2k-1=\lfloor 2(n-1)\alpha \rfloor .$$

• If $0\le \nu -\alpha <1∕2$, then $0\le 2(\nu -\alpha )<1$, thus

  $$\lfloor \mathit{n\alpha }\rfloor +\lfloor (n-1)\alpha \rfloor =2k=\lfloor 2(n-1)\alpha \rfloor .$$

• If $1∕2\le \nu -\alpha <1$, then $1\le 2(\nu -\alpha )<2$, thus

  $$\lfloor \mathit{n\alpha }\rfloor +\lfloor (n-1)\alpha \rfloor =2k<2k+1=\lfloor 2(n-1)\alpha \rfloor .$$

In all cases, $\lfloor \mathit{n\alpha }\rfloor +\lfloor (n-1)\alpha \rfloor \le \lfloor 2(n-1)\alpha \rfloor$, so the inequalities (4) and (3) are proven. Then equations (1) and (2) shows that

Since this is true for every prime number $p$,

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