Geometric transformation $ (x_1,x_2) \rightarrow\left(\frac{x_1}{x_2}, x_2\right) $

Question

I am trying to understand the geometric interpretation of the transformation obtained by the inequalities

$$0 < x_1 < x_2 < 1$$

when there is a transformation

$$y_1 = \frac{x_1}{x_2} \quad \text{and} \quad y_2 = x_2$$

I understand how to get $0 < y_1 < 1$ but I don't know what the lower bound of $y_2$ is.

I know that it has to be larger than 0 but I don't know if there is any relationship between $y_1$ and $y_2$.

Geometrically I tried to transform the vertices

$$A(0,0), \quad B(1,1) \quad \text{and} \quad C(0,1)$$

and try to obtain the new shape but point $A$ is giving me trouble with the division by $0$.

May I have some assistance?

Answer 1

Define subsets $A,B$ of $\mathbb{R}^2$ by

• $A=(x_1,x_2)\in\mathbb{R}^2$ such that $0 < x_1 < x_2 < 1$.$\\[4pt]$
• $B=f(A)$, where $f:A\to\mathbb{R}^2$ is given by $f(x_1,x_2)=\left({\large{\frac{x_1}{x_2}}},x_2\right)$.

Geometrically, $A$ is the open triangular region with vertices $(0,0),(0,1),(1,1)$.

It's clear that $f$ maps $A$ bijectively onto $B$.

Claim:$\;B=S$, where $S$ is the open unit square region with vertices $(0,0),(1,0),(1,1),(0,1)$.

Proof:

Let $y=(y_1,y_2)\in B$.

Then for some $(x_1,x_2)\in A$, we have $y_1=\left({\large{\frac{x_1}{x_2}}}\right)$, and $y_2=x_2$.

Since $0 < x_1 < x_2$, we get $0 < y_1 < 1$ and $0 < y_2 < 1$, hence $y\in S$.

Thus $B\subseteq S$.

For the reverse inclusion, let $s=(s_1,s_2)\in S$.

Since $s\in S$, we have $0 < s_1 < 1$ and $0 < s_2 < 1$, hence $0 < s_1s_2 < s_2 < 1$.

Then letting $x_1=s_1s_2$, and $x_2=s_2$, we have $x=(x_1,x_2)\in A$, and $$f(x)=f(x_1,x_2)=\left(\frac{x_1}{x_2},x_2\right)=\left(\frac{s_1s_2}{s_2},s_2\right)=(s_1,s_2)=s$$ hence $s\in B$.

Thus $S\subseteq B$.

Therefore $B=S$, as claimed.

Interpreted geometrically, $f$ transforms the triangular region $A$ bijectively onto the square region $B$.