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Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?

consider R as a vector space over Q then what is the dimension and how do we prove it

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marked as duplicate by Akhil Mathew Mar 21 '11 at 14:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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See math.stackexchange.com/questions/6244/… –  JavaMan Mar 21 '11 at 8:12
    
If you're asking specifically what which infinite cardinal number, i.e., the dimension of a basis, then the answers at the duplicate don't directly answer your question. However, you can extend the ideas in Arturo's answer there to see that the dimension is $2^{\aleph_0}$, as Pete L. Clark commented on Gadi A.'s answer below. You could also see this by witnessing a linearly independent set of cardinality $2^{\aleph_0}$, as I commented with link on Arturo's answer to the other question. –  Jonas Meyer Mar 21 '11 at 22:55

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The dimension is infinite; it suffices to show that the powers of, say, $\pi$ are linearly independent (a nontrivial finite linear combinations of powers of $\pi$ that equals 0 would show $\pi$ to be an algebraic number, which it is not).

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The dimension of a vector space over a field is a cardinal number. "Infinite" or "infinity" is not a cardinal number. The answer here is that the dimension is $c = 2^{\aleph_0}$. –  Pete L. Clark Mar 21 '11 at 13:14

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