Answer to Question #105341 in Real Analysis for Nimesh

Question

How can I prove that the sequence of irrational numbers is unbounded

Answer 1

In general, the question is not true.

We have many counter examples.

Before giving counter example, recall some definition.

SEQUENCE: A sequence of real numbers is a function defined on the set $N=$ {1,2,3,.......} of natural numbers whose range is contained in the set $R$ of real number.

BOUNDED SEQUENCE: A sequence $X=(x_n)$ of real number is said to be bounded if there exist a real number $M$ $>0$ such that $|x_n|\leqslant M$ for all $n\in N$ .

Example 1. Let $(x_n)=\sqrt{5}$ is a sequence of irrational number bounded by $\sqrt{5}$

Example 2. Let $(x_n)=\sqrt{2}+\frac{1}{n}$ is a sequence of irrational number bounded by $\sqrt{2}+1$