Answer to Question #105341 in Real Analysis for Nimesh

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=" {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"



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