# A step with respect to showing equivalence of displacements

I am currently studying Hardy's A Course of Pure Mathematics and am on Chapter 3, Section 35, Equivalence of displacements. Multiplication of displacements by numbers.

In this chapter, displacement's notation is given by $[x,y]$ where $x$ is the displacement along the x-axis and likewise for $y$.

In this section, Hardy had shown that $-[x,y]=[-x,-y]$ Then he says, "it is natural to agree further that $\alpha[x,y]=[\alpha x,\alpha y]$ where $\alpha$ is any real number, positive or negative"

What I don't understand is how that is a natural extension from what he had previously shown. Essentially what he showed is that $-1 \times[x,y]=[-1 \times x,-1 \times y]$ which is a single case, not a general form which he extended it to.

The question I would like to ask is: am I missing something either logically or intuitively?

-

We seem to be looking at different editions of the book. In the online version at archive.org, it explicitly states that $-[x,y]=[-x,-y]$ is a definition, which makes a lot more sense, since it's unclear on what basis this could be "shown".
Thus, $\alpha[x,y]=[\alpha x,\alpha y]$ is a definition that naturally fits with the definition $-[x,y]=[-x,-y]$.