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

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?

• It is true that $0/1=0/2$, so at least $(a,b)=0$ for different $b\in A$. Jan 20, 2019 at 18:18
• @Dog_69 I am not sure what your point is. What exactly do you mean by different $b \in A$. Different from what? Also, how is it related to the question? Jan 21, 2019 at 1:59
• Different to each other. I was assuming your notation. Jan 21, 2019 at 7:33

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 $$b\in A,b\ne a$$.