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

Short Answer

Expert verified

To show that some power of b falls in the range of n,bn

Step by step solution

01

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 n,bn. 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 range1,b .

02

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 k,bk.

Within interval k,bk, 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 k+1,bk+1 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 k,bk.

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

As a result, the sentence bp>kmay be written bpk+1.

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.

bbp=bkbk<bk+1bbp<bk+1bp+1<bk+1k+1bpbp+1<k+1k+1bp<bk+1

As a result, it is shown that for each positive integer k+1 and b , there exists some power of b in the region k+1,bk+1.

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

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