Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$1+ord_{p}n \le 2^{ord_{p}n} \le p^{(ord_{p}n)/x}$ for $p> 2^{x},n\ge 2 \in \mathbb{N}$; $x>0 \in \mathbb{R}$ It was written here that this is clearly the case; I don't see it. Does somebody see it and cares to explain?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The first inequality is a consequence of the fact that if $k$ is a positive integer, then $1+k \le 2^k$. Just take $k=\text{ord}_p(n)$.

If you wanted to, you could prove that $1+k\le 2^k$ by induction on $k$. But there is no need, the result is clear. At $k=1$ we have equality, but then $2^k$ grows much faster than $1+k$.

For the second inequality, take the logarithm of both sides. It is cleanest to use logarithms to the base $2$. So we want to prove that $$\text{ord}_p(n) \le \frac{\text{ord}_p(n)}{x}\log_2(p).$$ Since $p>2^x$, we have $\log_2(p)>x$, and the result follows.

share|improve this answer
    
Thanks, André Nicolas! –  VVV Mar 8 '12 at 23:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.