Question #c9e1c

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Question #c9e1c

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Answer 1 · 2015-05-22T13:44:08

OK, because the problem statement is ambiguous, I'd rather try to prove a more general and--presumably--more useful result.

I shall prove that, if $\frac{a}{b}$ $a/b$ is rational and $x$ $x$ is irrational, then their sum $\frac{a}{b} + x$ $a/b + x$ is irrational. ( $a$ $a$ and $b$ $b$ are integers and $b \neq 0$ $b!=0$ )

I'll use a proof by contradiction: let's assume that the sum of the two numbers is rational. This means that the sum can be written as a ratio of two integers $c$ $c$ and $d$ $d$ , $d \neq 0$ $d!=0$ :

$\frac{a}{b} + x = \frac{c}{d} \Leftrightarrow x = \frac{c}{d} - \frac{a}{b} \Leftrightarrow x = \frac{b c - a d}{b d}$ $a/b + x = c/d <=> x = c/d - a/b <=> x = (bc - ad)/(bd)$

Note that, because $a, b, c$ $a, b, c$ and $d$ $d$ are integers, then $b d$ $bd$ is an integer and $b c - a d$ $bc - ad$ is also an integer. Therefore, $x$ $x$ is the ratio of two integers, i.e. $x$ $x$ is rational, which contradicts the hypothesis that $x$ $x$ is irrational.

Thus, the assumption that $\frac{a}{b} + x$ $a/b + x$ is a rational number led us to a contradiction. Hence, its opposite must be true, that is $\frac{a}{b} + x$ $a/b + x$ is an irrational number.

This kind of reasoning can be applied to any particular cases.

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