Question

Show that for any positive integers n and any base b , there must some power of b lying in the range $\left[b,bn\right]$.

Short answer

To show that some power of b falls in the range of $\left[n,bn\right]$

Step-by-step solution

Introduction

With each integer values n and so any base b , basic process of designing algorithms is used to argue that there must be some power of b lying in the range $\left[n,bn\right]$. Each algorithm's execution duration is determined by the randomness of our inputs.

In the above question any positive integers either it is n and any base as b , it must be power of b lying with the range of n and bn . To check or understand see base information theory.

Condition of origin:

Here,n is a positive integer and b would be any base, as shown by numerically solving.

Assume n=1 and b as the basis.

As a result, there is a power of b that falls inside the range$\left[1,b\right]$ .

Proving

Hypothesis via Induction:

Look at the fact that for each given base number b and n=k, there must be a factor of b inside the range $\left[k,bk\right]$.

Within interval $\left[k,bk\right]$, there persists a strength of b , say, bp , for some base b and positive integer k .

The proof:

Its proven assertion holds with $n=k+1$, i.e., for each set of base number b, a power of b within range $\left[\left(k+1\right),b\left(k+1\right)\right]$ exists at all times.

To get all the bottom limits' area, do the following:

By inductive inference, if $bp>k$, then k is a positive integer and bp should be in the range $\left[k,bk\right]$.

As a result, the following integer as in k value is k+1 .

As a result, the sentence $bp>k$may be written $bp\ge \left(k+1\right)$.

To have the higher boundaries' area, do the following:

Assume the logical premise bp=k .

Multiply either system of equations by b.

This is undeniably true.

$$\begin{gathered} b\left(b^p\right)=bk \\ bk<b\left(k+1\right) \\ \to b{b}^p<b\left(k+1\right) \\ \to b^{p+1}<b\left(k+1\right) \\ \to \left(k+1\right)\le b^p\le b^{p+1}<\left(k+1\right) \\ \to \left(k+1\right)\le b^p<b\left(k+1\right) \end{gathered}$$

As a result, it is shown that for each positive integer $\left(k+1\right)$ and b , there exists some power of b in the region $\left(k+1\right),b\left(k+1\right)$.

As a result, there exists some power of b in the range $\left[n,bn\right]$ for each positive integer n and also any base b.