Why is $ (a, b) = 0 $ for distinct $ a, b \in A $, the Hamel basis of vector space $ \mathbb{R} $ over $ \mathbb{Q} $?

Question

Let us consider the set of real numbers $ \mathbb{R} $ as a vector space over the set of rationals $ \mathbb{Q} $.

We know that this vector space has a basis known as the Hamel basis. Let the Hamel basis be $ A $.

Now see this post at https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/ > sections "R as a vector space over Q" and "Periodicity of (x,a)". In the section "Periodicity of (x,a)", the article says,

Moreover $ (a,b)=0 $ for distinct $ a,b \in A $.

How is this true? From what I have understood so far is that the notation $ (x, a) $ is just a notation for a rational number as an ordered pair, i.e., $ (x, a) = \frac{x}{a} $ where $ a \ne 0 $. If I am right about this, then the above quoted statement implies that $ (a, b) = (b, a) = 0 $, i.e., $ \frac{a}{b} = \frac{b}{a} = 0 $. Now that cannot be true.

What am I missing? Have I understood the $ (a, b) $ notation incorrectly? Or am I making some other mistake?

Please help me understand how the quoted statement is true.

Answer 1

No, in their notation $(x,a)$ does not mean that, i.e. it does not stand for $\frac{x}{a}$.

Recall that $A$ is a basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$. By the very definition of a basis of a vector space, every element of the vector space can be written uniquely as a finite linear combination of the basis elements. In topological vector spaces, this is usually called a Hamel basis rather than just a basis, to distinguish them from topological bases that allow convergent infinite linear combination.

Anyways, the notation $x=\sum\limits_{a\in A}(x,a)a$ here is simply a statement representing such a linear combination, i.e. the unique expansion of $x$ with respect to the basis $A$. Each $(x,a)$ is a coefficient of this expansion, i.e. each $(x,a)$ is the rational number that appears in the expansion of $x$ as the coefficient of $a$. Note that by definition, for a given $x$, only finitely many of $(x,a)$ are nonzero, because all linear combinations have to be finite.

There's no standard notation for coefficients in expansions with respect to a basis. Very often this would be written as $x=\sum\limits_{a\in A}x_aa$, or in some other way. But they chose to write these coefficients this way — well, why not.

Note that the notation $(x,a)$ does not suggest any practical way for finding the values of these coefficients.

And finally, $(a,b)=0$ simply because $a$ and $b$ are two different elements of a basis of a vector space. For any $a\in A$, the unique expansion of $a$ with respect to the basis $A$ is $a=1a+\sum\limits_{b\in A,b\ne a}0b$, i.e. $(a,a)=1$ and $(a,b)=0$ for all $b\in A,b\ne a$.