# Condition for Increasing Binomial Coefficients

## Theorem

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

Let $\dbinom n k$ denote a binomial coefficient for $k \in \N$.

Then:

$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$

## Proof 1

 $\ds \dbinom n k$ $<$ $\ds \dbinom n {k + 1}$ $\ds \leadstoandfrom \ \$ $\ds \frac {n!} {\paren {n - k}! k!}$ $<$ $\ds \frac {n!} {\paren {n - k - 1}! \paren {k + 1}!}$ Definition of Binomial Coefficient $\ds \leadstoandfrom \ \$ $\ds k + 1$ $<$ $\ds n - k$ Multiplying both sides by $\dfrac {\paren {n - k}! \paren {k + 1}!} {n!}$ $\ds \leadstoandfrom \ \$ $\ds 2 k$ $<$ $\ds n - 1$ $\ds \leadstoandfrom \ \$ $\ds k$ $<$ $\ds \frac {n - 1} 2$

$\blacksquare$

## Proof 2

The proof proceeds by induction on $n$.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$

First we investigate the edge case.

Let $n = 1$.

Then we have:

 $\ds \dbinom 1 0$ $=$ $\ds 1$ Binomial Coefficient with Zero $\ds \dbinom 1 1$ $=$ $\ds 1$ Binomial Coefficient with Self

Thus we see:

there are no $k$ such that $0 \le k < \dfrac {1 - 1} 2 = 0$

and:

there are no $k$ such that $\dbinom 1 k < \dbinom 1 {k + 1}$

Thus $\map P 1$ is seen to hold.

### Basis for the Induction

Let $n = 2$.

Then we have:

 $\ds \dbinom 2 0$ $=$ $\ds 1$ Binomial Coefficient with Zero $\ds \dbinom 2 1$ $=$ $\ds 2$ Binomial Coefficient with One $\ds \dbinom 2 2$ $=$ $\ds 1$ Binomial Coefficient with Self

Thus we see:

there is one $k$ such that $0 \le k < \dfrac {2 - 1} 2 = \dfrac 1 2$, and that is $k = 0$

and:

$\dbinom 2 k < \dbinom 2 {k + 1}$ holds for exactly $k = 0$.

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:

$\dbinom r k < \dbinom r {k + 1} \iff 0 \le k < \dfrac {r - 1} 2$

from which it is to be shown that:

$\dbinom {r + 1} k < \dbinom {r + 1} {k + 1} \iff 0 \le k < \dfrac r 2$

### Induction Step

This is the induction step:

 $\ds \dbinom {r + 1} k$ $=$ $\ds \dbinom r k + \dbinom r {k - 1}$ Pascal's Rule $\ds$ $<$ $\ds \dbinom r k + \dbinom r k$ $\ds \iff 0 \le {k - 1} < \dfrac {r - 1} 2$ Induction Hypothesis $\ds$ $<$ $\ds \dbinom r {k + 1} + \dbinom r k$ $\ds \iff 0 \le {k - 1} < \dfrac {r - 1} 2 \text { and } 0 \le k < \dfrac {r - 1} 2$ Induction Hypothesis $\ds$ $<$ $\ds \dbinom r {k + 1} + \dbinom r k$ $\ds \iff 0 \le k < \dfrac {r - 1} 2$ $\ds$ $=$ $\ds \dbinom {r + 1} {k + 1}$ $\ds \iff 0 \le k < \dfrac {r - 1} 2$ Pascal's Rule

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{> 0}: \dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$