# 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 |

This needs considerable tedious hard slog to complete it.In particular: Can't get my head round the inequalities on $k$, needs more thoughtTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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

Therefore:

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

## Sources

- 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $1 \ \text{(a)}$