Definition: A set X is closed under addition if, for p and q both in X (with p and q not necessarily distinct!), p+q is always in X.

Easy question: Is there a non-empty subset of the irrationals that is closed under addition?

Tricky question? Is there an uncountably infinite non-empty subset of the irrationals that is closed under addition?

For example, the set {sqrt(2)} fails because it is not closed; taking p=sqrt(2), q=sqrt(2), we find that p+q=2*sqrt(2) is not in the set. The set of reals fails because it is not a subset of the irrationals.

I came up with this problem today, solved it, and I hope it makes you think. :)

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## Comments

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TopNewestThe set of reals \(\mathbb{R}\) is uncountably infinitely-dimensional as a vector space over \(\mathbb{Q}\). Let \(X\) be a Hamel basis for \(\mathbb{R}\) which contains \(1\), and let \(Y = X \backslash \{1\}\) be everything in the Hamel basis except \(1\). Certainly \(Y\) is uncountable.

Let \(Z\) be the set of finite sums of elements of \(Y\), so that \(Z\) is the set of real numbers of the form \[ z \; = \; n_1y_1 + n_2y_2 + \cdots + n_my_m \] where \(m \in \mathbb{N}\) and \(n_1,n_2,\ldots,n_m \in \mathbb{N}\), while \(y_1,y_2,\ldots,y_m \in Y\). It is clear that:

\(Z\) is uncountable, since it has \(Y\) as a subset,

\(Z\) is closed under addition,

No element of \(Z\) is rational. If \(Z \cap \mathbb{Q} \neq \varnothing\), we could find \(m,n_1,n_2,\ldots,n_m \in \mathbb{N}\) and distinct \(y_1,y_2,\ldots,y_m \in Y\) and \(q \in \mathbb{Q}\) such that \[ n_1y_1 + n_2y_2 + \cdots + n_my_m - q\times1 \; = \; 0 \] But this tells us that there is a nontrivial rational linear dependence relation between the elements \(y_1,y_2,\ldots,y_m,1\) of \(X\), which is not possible.

Thus \(Z\) does the job. I would not want to attempt to write down a concrete example of a set \(Z\)!

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Nice! ^__^ And much better explained that I ever would have. :P

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The set with elements multiples of root 2 (only positive multiples)

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Yes, this satisfies the first one. :D It obviously does not satisfy the second one, though; I suggest trying that one too. :)

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I tried the 2nd one bt i could not, plz tell me the answer...

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2nd 1 is really difficult...

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