Question

Let \(\left \{ a_k \right \}\) be the sequence \(a_1=3, a_2=3.1, a_3=3.14, a_4=3.141, ...\) That is, each term \(a_k\) contains the first \(k\) decimal digits of \(\pi\).

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

(b) Explain why the sequence \(\left \{ a_k \right \}\) is increasing.

(c) Provide an upper bound for the sequence \(\left \{ a_k \right \}\).

(d) What is the least upper bound of the sequence \(\left \{ a_k \right \}\) ?

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

Short answer

Part a. For each \(k\), the number \(a_{k}\) has a finite decimal representation. Hence, \(a_k\) is a rational number for each positive integer \(k\).

Part b. The sequence is increasing because each \(a_{k+1}\) is obtained from \(a_k\) by adding one more significant place after the decimal.

Part c. An upper bound for the sequence \(\left \{ a_k \right \}\) is \(3.2\)

Part d. The least upper bound of the sequence \(\left \{ a_k \right \}\) is \(\pi\).

Part e. The Least Upper Bound Axiom does not apply to the set of rational numbers because the sequence \(\left \{ a_1,a_2,a_3,a_4,... \right \}\) is set of rational numbers bounded above, but its least upper bound is \(\pi\) which is not rational.

Step-by-step solution

Part (a) Step 1. Given Information

We are given a sequence \(\left \{ a_k \right \}\) such as \(a_1=3, a_2=3.1, a_3=3.14, a_4=3.141, ...\) where each term \(a_k\) contains the first \(k\) decimal digits of \(\pi\).

The objective is to know why \(a_k\) is a rational number for each positive integer \(k\).

Part (a) Step 2. Explanation

For each \(k\), the number \(a_{k}\) has a finite decimal representation. Any number having finite decimal representation of the form above is given by

\(\frac{x}{10^{k-1}}\)

where \(x\) is an integer. But then, \(\frac{x}{10^{k-1}}\) is the ratio of two integers, hence, a rational number.

Part (b) Step 2. Given Information

We are given a sequence \(\left \{ a_k \right \}\) such as \(a_1=3, a_2=3.1, a_3=3.14, a_4=3.141, ...\) where each term \(a_k\) contains the first \(k\) decimal digits of \(\pi\).

The objective is to know why the sequence is increasing.

Part (b) Step 2. Explanation

The sequence is increasing because each \(a_{k+1}\) is obtained from \(a_k\) by adding one more significant place after the decimal, so that if \(a_{k+1}\neq a_k\) then

\(a_{k+1}-a_k\geq 10^{-k}\).

Part (c) Step 1. Provide an upper bound

Every number \(a_k\) is of the form \(3.141...\) therefore it is \(3+\) fraction; this fraction part is \(\leq 0.2\).

Therefore, an upper bound for the sequence is \(3.2\).

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

For the given sequence the terms are given as: \(a_1=3, a_2=3.1, a_3=3.14, a_4=3.141, ...\) and so on. It can be seen that each term \(a_k\) contains the first \(k\) decimal digits of \(\pi\).

So as the term continues the infinite will tends to the value \(\pi\).

So the least upper bound is \(\pi\).

Part (e) Step 1. Explanation

The given sequence \(\left \{ a_1,a_2,a_3,a_4,... \right \}\) is set of rational numbers bounded above, but its least upper bound is \(\pi\) 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.