Understanding Prime Power Proofs

[mlsbbe]

Hi, I am having trouble understanding this proof.


Statement
If pⁿ is the nth prime number, then p_n $$\leq$$ 2^2n-1


Proof:

Let us proceed by induction on n, the asserted inequality being clearly true when n=1. As the hypothesis of the induction, we assume n>1 and the result holds for all integers up to n.

Then

p_n+1 $$\leq$$ p₁p₂...p_n + 1

p_n+1 $$\leq$$ 2.2²...2^2n-1 + 1 = 2^{1 + 2 + 22+ ...+ 2n-1}

Recalling the indentity 1 + 2 +2²+ ...+2^n-1=2ⁿ-1

Hence

p_n+1 $$\leq$$ 2^2n-1+1

But 1 $$\leq$$ 2^2n-1 for all n; whence

p_n+1 $$\leq$$2^2n-1+2^2n-1
=2.2^2n-1
=2²ⁿ
completing the induction step, and the argument.



What I don't understand is why the proof uses p₁, p₂, etc as powers of two. What is the nature of the p_n? Are they prime or what? Why use powers?

 

[srijithju]

In the proof , it is not that p1 , p2 , p3 have been replaced by powers of 2 . All that it is saying is that :

p1.p2.p3.p4...pn < 2 . (2^2) . (2^4) ... ( 2 ^ (2^n - 1 ) ).

This is because it is assuming the theorem to be true for p1 , p2 .. upto pn .