<PRE><B>Question 189409</B>
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I&#39;ll do you one better.  Let&#39;s prove that the sum of &lt;i&gt;&lt;b&gt;any&lt;/b&gt;&lt;/i&gt; rational number and &lt;i&gt;&lt;b&gt;any&lt;/b&gt;&lt;/i&gt; irrational number is irrational.


Assume that there exists an irrational number &lt;i&gt;x&lt;/i&gt; and a rational number *[tex \Large \frac{a}{b}], *[tex \Large a, b\ \in\ \I] such that the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x + \frac{a}{b}] is rational.


Then there exists integers &lt;i&gt;c&lt;/i&gt; and &lt;i&gt;d&lt;/i&gt; such that *[tex \Large \frac{c}{d}] is a rational number and,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x + \frac{a}{b} = \frac{c}{d}]


But then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x = \frac{c}{d} - \frac{a}{b} = \frac{bc - ad}{bd}]


and since &lt;i&gt;a&lt;/i&gt;, &lt;i&gt;b&lt;/i&gt;, &lt;i&gt;c&lt;/i&gt;, and &lt;i&gt;d&lt;/i&gt; are integers, &lt;i&gt;bd&lt;/i&gt;, &lt;i&gt;bc&lt;/i&gt;, and &lt;i&gt;ad&lt;/i&gt; are integers, and further &lt;i&gt;bc&lt;/i&gt; - &lt;i&gt;ad&lt;/i&gt; is an integer, hence *[tex \Large \frac{bc - ad}{bd}] is rational by definition.


But that leads to a contradiction because &lt;i&gt;x&lt;/i&gt; was originally assumed to be irrational.


Therefore, reductio ad absurdum, 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x + \frac{a}{b}] is irrational.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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