Condition for Increasing Binomial Coefficients

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Theorem

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

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


Then:

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


Proof

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:

\(\displaystyle \dbinom 1 0\) \(=\) \(\displaystyle 1\) Binomial Coefficient with Zero
\(\displaystyle \dbinom 1 1\) \(=\) \(\displaystyle 1\) 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:

\(\displaystyle \dbinom 2 0\) \(=\) \(\displaystyle 1\) Binomial Coefficient with Zero
\(\displaystyle \dbinom 2 1\) \(=\) \(\displaystyle 2\) Binomial Coefficient with One
\(\displaystyle \dbinom 2 2\) \(=\) \(\displaystyle 1\) 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:

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



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


Therefore:

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


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