Is this an acceptable proof that $x^2<y^2$ when $x$ and $y$ are positive reals with $x<y$?

Question

The question is

Prove the following statement:

Suppose $x, y \in \mathbb{R}$ and that $x$ and $y$ are positive. If $x < y$ then $x^{2} < y^{2}$.

Proof:

Suppose there is a value $r$ between $x$ and $y$ so that $$x < r < y \tag1$$ where $r \in \mathbb{R}$ and $ r > 0$. Therefore, $$x<r \Rightarrow x^{2}<xr \tag2$$ and $$r<y \Rightarrow yr < y^{2} \tag3$$ Moreover, if we multiply equation $(1)$ by $r$, we get $$xr < r^{2} < yr \tag4$$ Combining equations $(2)$, $(3)$, and $(4)$, we get $$x^{2} < xr < r^{2} < xy < y^{2} \tag5$$ Therefore, $x^{2} < y^{2}$. $\blacksquare$

If necessary, we can even define $r=$ $x+y\over{2}$

Is there anything wrong with this approach? I'm aware of the normal method of direct proof.

Answer 1

No need to introduce $r$; to wit:

$x = x; \tag 1$

$x < y; \tag 2$

thus

$x^2 < xy; \tag 3$

also,

$y = y; \tag 4$

from (2) and (4),

$xy < y^2; \tag 5$

combining (3) and (5):

$x^2 < xy < y^2. \tag 5$

$OE\Delta$.

Answer 2

Note that this is like proving $1 < 3$ by showing $1 < 2$ and $2 < 3$. You basically end up proving the result along the way except you wrote down a bunch of extra steps.

As soon as you have $x^{2} < xr$, you have the result immediately from the hypothesis: $x^{2} < xr < yr < y^{2}$ as none of them are zero.

Just in case you didn't prove this concisely using a direct proof, you need to prove that $(y+x)(y-x)=y^{2}-x^{2} > 0$ where $y,x > 0$ and $y-x > 0$ which is clear since $y+x > 0$.