To prove that \(\sqrt{2}\) is irrational, we can use a proof by contradiction.

1. **Assumption**: Assume that \(\sqrt{2}\) is rational. This means it can be expressed as a fraction of two integers \(a\) and \(b\) in simplest form, such that:
   \[
   \sqrt{2} = \frac{a}{b}
   \]
   where \(a\) and \(b\) have no common factors (other than 1) and \(b \neq 0\).

2. **Square Both Sides**: Squaring both sides gives:
   \[
   2 = \frac{a^2}{b^2}
   \]
   Multiplying both sides by \(b^2\) yields:
   \[
   a^2 = 2b^2
   \]

3. **Analyze \(a^2\)**: This equation shows that \(a^2\) is even because it is equal to \(2b^2\) (which is clearly even).

4. **Conclusion About \(a\)**: Since \(a^2\) is even, \(a\) must also be even (the square of an odd number is odd). Thus, we can express \(a\) as:
   \[
   a = 2k
   \]
   for some integer \(k\).

5. **Substitute \(a\)**: Substituting back into the equation \(a^2 = 2b^2\):
   \[
   (2k)^2 = 2b^2 \quad \Rightarrow \quad 4k^2 = 2b^2 \quad \Rightarrow \quad 2k^2 = b^2
   \]

6. **Conclusion About \(b\)**: This shows that \(b^2\) is also even, which means \(b\) is even.

7. **Contradiction**: If both \(a\) and \(b\) are even, they share at least the common factor of 2. This contradicts our initial assumption that \(a\) and \(b\) have no common factors.

8. **Conclusion**: Therefore, our assumption that \(\sqrt{2}\) is rational must be false. Thus, \(\sqrt{2}\) is irrational.

If you have any questions or need further clarification, feel free to ask! Here are some related questions you might consider:

1. What are some other examples of irrational numbers?
2. How can we prove the irrationality of other square roots, like \(\sqrt{3}\)?
3. What is the significance of rational vs. irrational numbers in mathematics?
4. Can you provide a visual representation of this proof?
5. What are some applications of irrational numbers in real life?

**Tip**: Familiarize yourself with properties of even and odd numbers, as they are often used in proofs involving integers.

Discover the proof that \(\sqrt{2}\) is irrational through a step-by-step contradiction approach. This proof covers key concepts in number theory and properties of even and odd integers, making it suitable for high school mathematics students.

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