We are tasked with proving that if $0 < x < y$, then $x^2 < y^2$.

Proof:

1. Given condition: We are given that $0 < x < y$, which means both $x$ and $y$ are positive, and $x$ is less than $y$.

2. Inequality properties: Since $0 < x < y$, subtracting $x$ from both sides of the inequality $x < y$ gives: $y - x > 0.$

3. Multiply both sides by $x + y$: Now, observe that both $x + y > 0$ because $x > 0$ and $y > 0$. So, we can multiply both sides of $y - x > 0$ by $x + y$ without changing the direction of the inequality: $(y - x)(x + y) > 0.$

4. Expand the expression: Expanding the left-hand side: $(y - x)(x + y) = y^2 - x^2.$ Thus, we have: $y^2 - x^2 > 0.$

5. Conclusion: From $y^2 - x^2 > 0$, we conclude that: $y^2 > x^2.$ Therefore, $x^2 < y^2$ as required.

Thus, the proof is complete.

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Would you like to see more details or ask any questions? Here are some related questions to deepen your understanding:

1. What happens to the inequality if $x < 0$ and $y > 0$?
2. How would the proof change if we allowed $x = 0$?
3. Can you prove this inequality using another method, such as the concept of functions?
4. How does this proof extend to other powers, say $x^3$ and $y^3$?
5. What role does positivity play in preserving the inequality when squaring numbers?

Tip: When dealing with inequalities, always be cautious when multiplying or dividing by variables—ensure their sign does not flip the inequality!