Question #105341
How can I prove that the sequence of irrational numbers is unbounded
1
Expert's answer
2020-03-13T10:35:35-0400

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=N= {1,2,3,.......} of natural numbers whose range is contained in the set RR of real number.

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

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

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



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