Q:

Simple Direct Proof

In Hammack's 'Book of Proof' I had a simple direct proof to write:
"If x and y are positive reals and x < y then, x^2 < y^2
I approached it by saying:
x<y    # from proof defintion
x-y<0  # x-y is negative

x^2 - y^2 < 0
(x+y)(x-y) < 0  #by factoring

(x+y) is positive # x & y are positive reals
(x-y) is negative # shown above 

In all cases a postive * negative is a negative
therefore (x+y)(x-y) < 0
therefore x^2 - y^2 < 0
therefore x^2 < y^2

Not sure if this the right approach

A:

A direct proof would begin by assuming $x,y$ are positive reals, and that $x<y$.  It would END with $x^2<y^2$.  Your proof did not do this; you have $x^2<y^2$ in the middle.  Consequently your narrative structure is all messed up.
Here is a correct version:

$x<y$ and $0<x$, so $x^2<xy$ (multiplying both sides by positive $x$ preserves $<$).
$x<y$ and $0<y$, so $xy<y^2$ (multiplying both sides by positive $y$ preserves $<$).
$x^2<xy<y^2$, combining the previous two results.
Hence $x^2<y^2$.

