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Well yeah, how do you formally prove if the set of Irrational numbers is a subset of the Real numbers?

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closed as not a real question by William, Bruno Joyal, Austin Mohr, Andres Caicedo, BenjaLim Sep 22 '12 at 8:43

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That depends entirely on how you’ve formally defined the two sets. –  Brian M. Scott Sep 22 '12 at 4:43
@BrianM.Scott: I'm not very sure about that - the exercise in the book doesn't actually tell me anything else. –  Zol Tun Kul Sep 22 '12 at 4:45
If the exercise doesn't tell you, maybe you should look earlier for the definition of the irrationals, or maybe you are expected to use the definition Brian M. Scott gives. –  Ross Millikan Sep 22 '12 at 4:50
@Omega What book and exercise? –  Bill Dubuque Sep 22 '12 at 5:13

2 Answers 2

That depends entirely on how you’ve formally defined the two sets. In many developments there’s nothing to prove: starting from the set $\Bbb Q$ of rational numbers, you construct the set $\Bbb R$ of real numbers, and define the set of irrational numbers to be $\Bbb R\setminus\Bbb Q$.

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Are the irrational numbers not defined to be those real numbers which cannot be written as a ratio of integers? Are they not real by definition?

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Yes, of course, but how do you formally prove it? –  Zol Tun Kul Sep 22 '12 at 4:46
There does not appear to be much to prove. Let $\mathbb{I}$ be the set of irrational numbers. $\mathbb{I}$ is defined to be {$r \in \mathbb{R}$ : $r$ cannot be written as $\frac{a}{b}$ where $a,b \in \mathbb{Z}$ and $b \neq 0$}. Then for any $r \in \mathbb{I}$, $r \in \mathbb{R}$. Therefore $\mathbb{I} \subseteq \mathbb{R}$. –  Shankman Sep 22 '12 at 4:50

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