Proof Exercise II: Real Numbers and Squares

[reenmachine]

Homework Statement

Suppose $x$ and $y$ are positive real numbers.If $x < y$ , then $x^2 < y^2$.

Homework Equations

Assume $x<y$ with $x,y \in ℝ^+$.This implies that $\exists z \in ℝ^+$ such that $x+z=y$.We have $y^2 = (x+z)^2 = x^2 + 2xz + z^2$ , and this proves that if $x < y$ with $x,y \in ℝ^+$ , then $x^2 < y^2$.

any thoughts on that one? Thank you!

 

[tiny-tim]

hi reenmachine!

yes, that looks fine

(alternatively, you could factor y² - x²)

 

[hilbert2]

You only need to know that if $x>0$, $y>0$ and $x<y$ then $x+y>0$ and $x-y<0$.

Also, if $a>0$ and $b<0$ then $ab<0$.

 

[pasmith]

You need to explain why 2xz + z^2 is strictly positive. It is obvious, but then so is the result you're trying to prove.

An alternative proof is that if 0 &lt; x &lt; y then, since multiplying by a positive number preserves inequalities, we have x^2 &lt; xy on multiplying by x and xy &lt; y^2 on multiplying by y. Putting these together we have x^2 &lt; xy &lt; y^2.

 

[reenmachine]

hi reenmachine!

yes, that looks fine

(alternatively, you could factor y² - x²)

Hi! :)

thank you for taking the time to respond!

 

[reenmachine]

You need to explain why 2xz + z^2 is strictly positive. It is obvious, but then so is the result you're trying to prove.

An alternative proof is that if 0 &lt; x &lt; y then, since multiplying by a positive number preserves inequalities, we have x^2 &lt; xy on multiplying by x and xy &lt; y^2 on multiplying by y. Putting these together we have x^2 &lt; xy &lt; y^2.

I thought the fact that 2xz + z^2 is positive was obvious from the fact x and z are in $R^+$...

I like your way of doing it , pretty good road to the proof.

thank you!