I have an exercise in my last assignment of calculus:
Show using ordering axioms that $$x^2 < y^2$$ for $x, y \in
> \mathbb{Q}$, with $0 < x < y$
This is my solution:
We have that $x < y$.
Since $x > 0$, we can multiply both sides of the previous inequality by $x$ (Order axiom compatability), and we obtain:
- $$x \cdot x < y \cdot x$$
Now, we know also that $y > 0$, so also in this case we can multiply both sides of the inequality $x < y$ by $y$:
- $$x \cdot y < y \cdot y$$
Since all field axioms apply also to the set $\mathbb{Q}$, we can use them; in particular the field axiom 2 (Field axiom commutativity). We have that:
$$x \cdot y = y \cdot x$$
If we replace $y \cdot x$ by $x \cdot y$ in inequality $1.$, we obtain:
$$x \cdot x < x \cdot y$$
Since we have both $x \cdot x < x \cdot y$ and $x \cdot y < y \cdot y$, by transitivity we have:
$$x \cdot x < y \cdot y$$
Since $x \cdot x = x^2$ and $y \cdot y = y^2$, we have:
$$x^2 < y^2$$
The thing I am not convinced about is my suppositions that $a \cdot a = a^2$. Should I also show it?
Best Answer
Yes, your proof looks good.
Alternative Approach
If you have defined $u<v$ as "$v-u$ is a positive number," you might have also proved or assumed that the positive rationals are closed under addition and multiplication.
Then you can show that $y^2-x^2 = (y-x)(y+x)$, which is true by the distributive law and commutativity of multiplication. So since $y-x$ is positive (since $x<y$) and $y+x$ is positive (since the sum of positive numbers is positive) we have that $y^2-x^2$ is positive (since positives are closed under multiplication.)
So $x^2<y^2$.
This proof is, essentially, the same as your proof, but more basic.