Editing Proof of Bertrand's postulate (section)

===Lemma 3=== 
If <math>p</math> is an [[parity (mathematics)|odd]] prime and <math>\frac{2n}{3} < p \leq n</math>, then <math>R(n,p) = 0.</math>

'''Proof:''' There are exactly two factors of <math>p</math> in the numerator of the expression <math>\tbinom{2n}{n}=(2n)!/(n!)^2</math>, coming from the two terms <math>p</math> and <math>2p</math> in <math>(2n)!</math>, and also two factors of <math>p</math> in the denominator from one copy of the term <math>p</math> in each of the two factors of <math>n!</math>. These factors all cancel, leaving no factors of <math>p</math> in <math>\tbinom{2n}{n}</math>. (The bound on <math>p</math> in the preconditions of the lemma ensures that <math>3p</math> is too large to be a term of the numerator, and the assumption that <math>p</math> is odd is needed to ensure that <math>2p</math> contributes only one factor of <math>p</math> to the numerator.)