Bertrand-Chebyshev Theorem/Lemma 2

Lemma for Bertrand-Chebyshev Theorem
For all $m \in \N$:
 * $\ds \prod_{p \mathop \le m} p \le 2^{2 m}$

where the product is taken over all prime numbers $p \le m$.

Proof
The proof proceeds by strong induction.

For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
 * $\ds \prod_{p \mathop \le m} p \le 2^{2 m}$

$\map P 0$ and $\map P 1$ are the vacuous cases:

Basis for the Induction
$\map P 2$ is the case:

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 $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.

This is the induction hypothesis:
 * $\ds \prod_{p \mathop \le k} p \le 2^{2 k}$

from which it is to be shown that:
 * $\ds \prod_{p \mathop \le k + 1} p \le 2^{2 \paren {k + 1} }$

Induction Step
This is the induction step:

Let $m > 2$ be even.

Then:

and $\map P k$ seen to hold.

Let $m = 2 k + 1$ be odd.

By Sum of Binomial Coefficients over Lower Index:


 * $\ds \sum_{r \mathop = 0}^{2 k + 1} \binom {2 k + 1} r = 2^{2 k + 1}$

Therefore:

Let $p$ be a prime number such that:
 * $k + 2 \le p \le 2 k + 1$

Consider:
 * $q = \dbinom {2 k + 1} k = \dfrac {\paren {2 k + 1}!} {k! \paren {k + 1}!}$

by definition of binomial coefficient.

We note that $p$ is greater than all of the divisors of the denominator of $q$.

Hence $p$ is a divisor of the numerator.

Hence:
 * $p \divides q$

Thus:

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

Therefore:
 * $\forall m \in \N: \ds \prod_{p \mathop \le m} p \le 2^{2 m}$