# Real Analysis of Irrational Numbers

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Proof:

Given any two real numbers a<b ,there exists an irrational number t satisfying a<t<b

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##### Solution Summary

The expert provides a proof for given any two real numbers a<b ,there exists an irrational number t satisfying a<t<b

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Let decimal expansions of two real numbers a and b (a<b) first differ in the nth digit.

First, if the nth digit is not the last digit of b, consider the number bn obtained from b by cutting all the digits of b after the nth, then: a<bn<b.

there is two cases here:

(1)b has non zero digits after the nth, Append to bn one zero digit, give the result number to t;

(2)b has zero digits right after the nth. Let bm be ...

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