Proof using induction: $r!$ divides the product of $r$ consecutive integers

Question

Prove by induction that the product $$n(n+1)...(n+r-1)$$ of any $r$ consecutive numbers is divisible by $r!$.

In the inductive case i've$$n(n+1)...(n+k-1)(n+k)$$ I was not able to find a way to prove this is divisible by (k+1)!. I try to use distributive property in (n+k).

Isn't it easier to avoid induction and just notice that $\binom{n+r-1}{r}=\frac{n(n+1)\cdot\ldots\cdot(n+r-1)}{r!}$ is obviously an integer number? — Jack D'Aurizio, Sep 09 '16 at 11:04
@Jack D' Aurizio ,since it is integer we won't need induction am I right? — Sathasivam K, Sep 10 '16 at 17:07

Sarvesh Ravichandran Iyer · Answer 1 · 2016-09-09T11:17:28.453

1

The product of $r$ consecutive numbers is divisible by $r$.

Proof: Okay, I've just remodeled my answer. For $r=1$ it is obvious.

Let it be true for $r=k$. For $r=k+1$, we will induct by the integer $m$, while showing that $(k+1)! | m(m+1)...(m+k)$. So suppose that $(k+1)! | m(m+1)...(m+k)$.We want to show that $(k+1)! | (m+1)(m+2)...(m+k+1)$.

How we do this is as follows: define $m_k = (m)(m+1)...(m+k)$. Then, $$ \begin{equation} \begin{split} & (m+1)_k = (m+1)_k -m_k + m_k\\ & =(m+1)...(m+k)(m+k+1-m) + m_k\\ & =(k+1)(m+1)_{k-1} + m_{k} \end{split} \end{equation} $$

Now, note that $k! | (m+1)_{k-1}$ by induction on $k$, hence $(k+1)! |(k+1)(m+1)_{k-1}$. On the other hand, by induction on $m$, $(k+1)! | m_{k}$. Hence, $(k+1)!$ divides their sum, which is $(m+1)_k$. Now, we are done by induction on $m$, and then by induction on $k$.

edited Sep 09 '16 at 11:17

answered Sep 09 '16 at 10:41

Sarvesh Ravichandran Iyer

76,105

1

OP wants it divisible by $r$-factorial, not just by $r$. – Gerry Myerson Sep 09 '16 at 10:57
To show that $(k+1)!$ divides $A$, it does not suffice to show that both $k+1$ and $k!$ divide $A$. E.g. if $k=3$, $6 \mid A$ and $4 \mid A$ together do not imply that $24 \mid A$. – user133281 Sep 09 '16 at 10:59
@user133281 I understand. I wrote this much of the answer, I''m still editing it. – Sarvesh Ravichandran Iyer Sep 09 '16 at 11:01
You prove that $k! | n(n+1)\ldots(n+k)$ and then that $k+1 | n(n+1)\ldots(n+k)$, but how do you conclude $k!(k+1) | n(n+1)\ldots(n+k)$? It seems you can only conclude $\operatorname{lcm} (k!, k+1) | n(n+1)\ldots(n+k)$ from that. – Joel Cohen Sep 09 '16 at 11:20
@JoelCohen Corrected. I reformed my entire answer using a double induction argument. – Sarvesh Ravichandran Iyer Sep 09 '16 at 11:21

Caleb Stanford · Answer 2 · 2016-09-10T19:15:39.737

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Without induction: The expression $$ n(n+1)\cdots(n+r-1) $$ counts the number of words of length $r$ with all distinct letters, from an alphabet of $n+r-1$ letters. But this is also equal to the number of sets of $r$ letters, times $r!$ since there are $r!$ ways of arranging the set into a word. Thus, the above expression is divisible by $r!$.

edited Sep 10 '16 at 19:15

answered Sep 09 '16 at 11:07

Caleb Stanford

45,895

score 1 · Answer 3 · answered Sep 09 '16 at 11:45

If $r=1$ the theorem obviously holds true.
Let $r$ be the least natural number for which the theorem does not hold true and $a$ the least natural such that $R_a=(a+2)(a+3)\cdots (a+r+1)$ is not divisble by $r!$.
Obviously, $R_a>r!$ and from now we suppose that any product of $r-1$ numbers is a multiple of $(r-1)!$

We can see that $R_a-R_{a-1}=$
$(a+2)(a+3)\cdots (a+r)[a+r+1-(a+1)]=(a+2)(a+3)\cdots (a+r)\cdot r$.

It is obvious that $(a+2)(a+3)\cdots (a+r)$ is a product of $r-1$ numbers which means it is divisible by $(r-1)!$ (from induction hypothesis)so, multiplying by $r$ , it becomes a multiple of $r!$.
Hence $r!\mid R_a-R_{a-1}$ and $r!\mid R_{a-1}$ so, $r!\mid R_a$ which contradicts the definition of $a$.

Proof using induction: $r!$ divides the product of $r$ consecutive integers

3 Answers3