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I want to do a proof by contradiction. You guys let me know if I goofing up.

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Suppose $f$ is non-constant. Since $f$ is continuous, it satisfies the Intermediate value theorem [in the most general sense, $f$ satisfies the theorem as long as $f$ is a mapping from any connected space $M$ to $\mathbb R$].

Pick any two irrational numbers $a in the image of $f$. Since $f$ satisfies the Intermediate value theorem, then $f$ attains all the intermediate values from $[a,b]$. We know that between any two irrationals, lies a rational--a contradiction. $f$ has a value that is not irrational.

This means the assumption of $f$ being non-constant is false. QED

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    Your reasoning is correct. Just a linguistic quibble - the word "consecutive" would mean that there are no irrational numbers between $a$ and $b$, whereas I suspect you just meant to say that $a.2012-11-05
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    First, it makes no sense to talk about *consecutive* irrational numbers: between any two irrational numbers there is another irrational number, so they can’t be consecutive. Apart from that, the basic idea is sound, provided that $M$ is given to be a subset of $\Bbb R$; if not, you can’t use the intermediate value theorem.2012-11-05
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    @BrianM.Scott The intermediate value theorem applies for any connected topological space $M$: the image of $M$ under a continuous map $M\to\mathbb{R}$ is connected, and every connected subspace of $\mathbb{R}$ is an interval.2012-11-05
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    @Brad: No, the theorem that continuous functions preserve connectedness applies. The intermediate value theorem, so named, is specifically a theorem about real-valued functions on $\Bbb R$.2012-11-05
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    I don't think it's uncommon to refer to the more general result as "the intermediate value theorem", e.g. in James Munkres's book Topology it is referred to as such.2012-11-05
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    @Brad: thanks for the pointing that out! Yes, you are right, I should have mentioned $a instead of two consecutive. Let me edit that. If we do that, then are we fully home? I have seen much detailed proofs, but this line of reasoning seemed intuitive to me. I mean how can you go from one irrational to the other without skipping values?2012-11-05
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    @Brad: I consider that a minor solecism on his part; it’s not the only one, though the book is certainly one of the best available. But then I also think it silly to bother, as he does, with the special case of continuous functions to linearly ordered spaces, since it’s a trivial consequence of preservation of connectedness by continuous maps.2012-11-05
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    @BrianM.Scott: I think I agree with Brad here Brian. I have found a few text books where they generalize the theorem to real valued function from any connected space. I see you are always more thorough, what recommendations would you have for this proof?2012-11-05
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    My only objection, apart from the bit about ‘adjacent’ irrationals, was to the terminology; the argument is fine, so long as you meant the more general result when you wrote ‘intermediate value theorem’, as apparently you did.2012-11-05
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    @BrianM.Scott: Apart from stating that $a, is there a more "appropriate" term you would use?2012-11-05
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    No, it’s fine the way you’ve worded it now, as long as your readers know which intermediate value theorem you’re talking about.2012-11-05