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$\{10^z|z\in \mathbb Z\}$ looks like a basis of $\mathbb R$ over some field. However, it is definitely not a Hamel basis over $\mathbb Q$ due to

1.It only has countable elements compared with any Hamel basis actually has uncountable many.

2.Most real number needs to be represented as an infinite sum.

3.The most lethal one, itself is not independent over $\mathbb Q$.

So my question is is it another type of basis?

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    @AsafKaragila Because Legendre polynomials looks like an orthogonal basis on a vector space, and it has countable members. $\{10^z|z\in \mathbb Z\}$ also has countable members which are all polynomials too.2012-11-16

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No this is nothing like a basis. As you noted the elements are dependent over $\Bbb Q$, and since this is the prime field in characteristic $0$, any candidate field will contain $\Bbb Q$, and the elements given will remain linearly dependent over any such field.

If you are thinking of decimal representation of real numbers, it is not really like expression of vectors in a basis in any formal sense. Besides it has some surprises of the kind $0.99999999\ldots=1.00000000\ldots$, which never happen for vector space bases.

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    Okay, I remember. Legendre polynomials is an orthogonalization of $\{x^n|n \in \mathbb N\}$, but $x^n$ is a bit defferent from $10^n$.2012-11-17