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Let {ak} be the sequence a1=3,a2=3.1,a3=3.14,a4=3.141,... That is, each term ak contains the first k decimal digits of ฯ€.

(a) Explain why ak is a rational number for each positive integer k.

(b) Explain why the sequence {ak} is increasing.

(c) Provide an upper bound for the sequence {ak}.

(d) What is the least upper bound of the sequence {ak} ?

(e) Use this sequence to explain why the Least Upper Bound Axiom does not apply to the set of rational numbers.

Short Answer

Expert verified

Part a. For each k, the number ak has a finite decimal representation. Hence, ak is a rational number for each positive integer k.

Part b. The sequence is increasing because each ak+1 is obtained from ak by adding one more significant place after the decimal.

Part c. An upper bound for the sequence {ak} is 3.2

Part d. The least upper bound of the sequence {ak} is ฯ€.

Part e. The Least Upper Bound Axiom does not apply to the set of rational numbers because the sequence {a1,a2,a3,a4,...} is set of rational numbers bounded above, but its least upper bound is ฯ€ which is not rational.

Step by step solution

01

Part (a) Step 1. Given Information

We are given a sequence {ak} such as a1=3,a2=3.1,a3=3.14,a4=3.141,... where each term ak contains the first k decimal digits of ฯ€.

The objective is to know why ak is a rational number for each positive integer k.

02

Part (a) Step 2. Explanation

For each k, the number ak has a finite decimal representation. Any number having finite decimal representation of the form above is given by

x10kโˆ’1

where x is an integer. But then, x10kโˆ’1 is the ratio of two integers, hence, a rational number.

03

Part (b) Step 2. Given Information

We are given a sequence {ak} such as a1=3,a2=3.1,a3=3.14,a4=3.141,... where each term ak contains the first k decimal digits of ฯ€.

The objective is to know why the sequence is increasing.

04

Part (b) Step 2. Explanation

The sequence is increasing because each ak+1 is obtained from ak by adding one more significant place after the decimal, so that if ak+1โ‰ ak then

ak+1โˆ’akโ‰ฅ10โˆ’k.

05

Part (c) Step 1. Provide an upper bound

Every number ak is of the form 3.141... therefore it is 3+ fraction; this fraction part is โ‰ค0.2.

Therefore, an upper bound for the sequence is 3.2.

06

Part (d) Step 1. Find the least upper bound

For the given sequence the terms are given as: a1=3,a2=3.1,a3=3.14,a4=3.141,... and so on. It can be seen that each term ak contains the first k decimal digits of ฯ€.

So as the term continues the infinite will tends to the value ฯ€.

So the least upper bound is ฯ€.

07

Part (e) Step 1. Explanation

The given sequence {a1,a2,a3,a4,...} is set of rational numbers bounded above, but its least upper bound is ฯ€ which is not rational.

This is happening because even though the rational numbers are increasing but it does not converge in the set of rational numbers.

Hence, the Least Upper Bound Axiom does not apply to the set of rational numbers.

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