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!

Understand the proof that if 0 < x < y, then x^2 < y^2 using algebraic properties such as inequalities and the difference of squares formula. This step-by-step guide is designed for high school students studying algebra and inequalities.

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