1. **State the problem:** We want to prove that for positive numbers $x$ and $y$, if $x^2 < y^2$, then $x < y$. 2. **Recall the properties:** Since $x$ and $y$ are positive, both $x > 0$ and $y > 0$. 3. **Start with the given inequality:** $$x^2 < y^2$$ 4. **Use the fact that both sides are positive:** Since $x > 0$ and $y > 0$, we can take the positive square root of both sides without reversing the inequality: $$\sqrt{x^2} < \sqrt{y^2}$$ 5. **Simplify the square roots:** $$x < y$$ 6. **Conclusion:** We have shown that if $x^2 < y^2$ and both $x$ and $y$ are positive, then $x < y$. This completes the proof.