4

I have to prove the following theorem:

Prove that the product of $r$ consecutive positive integers in divisible by $r!$

I am having a hard time getting a generalization down for the full set of real numbers, if I start from 1 and work up to r, I have the following:

$$r!k=\prod_{i=1}^{r}n_i$$

Can easily prove the base case of this, (n=1), and then go in to prove:

$$(r+1)!k=\prod_{i=1}^{r+1}n_i$$

Expand that out and get:

$$(r+1)r!k=n(n+1)(n+2).....(n+r)(n+r+1)$$

Can say that the product of the first $r$ elements in equal to $r!k$ by our base case. Leaving using with:

$$(r+1)k=(n+r+1)$$

Not sure where I can go from here, n is the integer that we start at, so how can I get it to work out to be equal to our induction hypothesis?

Martin Sleziak
  • 53,687
Richard P
  • 751
  • 4
  • 18

2 Answers2

5

You can do this by simultaneous induction on $r$ and $n$. Note that

$$\begin{align} (n+1)\cdots(n+r)&=(n+1)\cdots(n+r-1)n\quad+\quad(n+1)\cdots(n+r-1)r\\ &=((n-1)+1)\cdots((n-1)+r)\quad+\quad(n+1)\cdots(n+(r-1))r \end{align}$$

(I inserted a little extra space around the central plus signs to make the key pieces easier to see.) By induction on $n$, $r!$ divides $((n-1)+1)\cdots((n-1)+r)$, and by induction on $r$, $(r-1)!$ divides $(n+1)\cdots(n+(r-1))$, hence $r!$ divides $(n+1)\cdots(n+(r-1))r$.

(Please note, I'm glossing over all the fine points of getting the inductions started.)

Barry Cipra
  • 79,832
  • Sometimes induction doesn't need to be started. If $\forall n\in N (; [ \forall m\in N ;( m<n \to ( \psi (m);) ] ; \to \psi (n) ;)$ then $\forall n\in N; (\psi (n).$ – DanielWainfleet Jan 20 '16 at 17:29
1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ n\pars{n + 1}\ldots\pars{n + r - 1}={\pars{n + r -1}! \over \pars{n - 1}!} ={\pars{n + r -1}! \over \pars{n - 1}!r!}\,r! = {n + r - 1 \choose r}r! $$

$$ {n\pars{n + 1}\ldots\pars{n + r - 1} \over r!} = {n + r - 1 \choose r} \quad\mbox{which is an integer !!!} $$

Felix Marin
  • 89,464