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To prove that \( \sqrt{2} \) is irrational by contradiction, we follow these steps:

### Proof by Contradiction:

1. **Assumption (for contradiction)**: 
   Assume \( \sqrt{2} \) is **rational**. By definition, a rational number can be written as a fraction of two integers. So, we can write:
   \[
   \sqrt{2} = \frac{p}{q}
   \]
   where \( p \) and \( q \) are integers with no common factors (i.e., the fraction is in its simplest form), and \( q \neq 0 \).

2. **Squaring both sides**: 
   Squaring both sides of the equation \( \sqrt{2} = \frac{p}{q} \), we get:
   \[
   2 = \frac{p^2}{q^2}
   \]
   Multiply both sides by \( q^2 \) to eliminate the denominator:
   \[
   2q^2 = p^2
   \]

3. **Implication for \( p^2 \)**: 
   From the equation \( 2q^2 = p^2 \), it follows that \( p^2 \) is even, since it is equal to \( 2 \times q^2 \) (which is clearly even). 

4. **Conclusion about \( p \)**: 
   If \( p^2 \) is even, then \( p \) must also be even (since the square of an odd number is odd). So, let \( p = 2k \) for some integer \( k \).

5. **Substitute \( p = 2k \) into the equation**: 
   Substituting \( p = 2k \) into the equation \( 2q^2 = p^2 \), we get:
   \[
   2q^2 = (2k)^2 = 4k^2
   \]
   Simplifying this, we get:
   \[
   q^2 = 2k^2
   \]

6. **Implication for \( q^2 \)**: 
   This shows that \( q^2 \) is even, which means \( q \) must also be even (since the square of an odd number is odd).

7. **Contradiction**: 
   If both \( p \) and \( q \) are even, they have a common factor of 2. However, this contradicts our original assumption that \( \frac{p}{q} \) is in its simplest form, where \( p \) and \( q \) have no common factors.

### Conclusion:
Since assuming that \( \sqrt{2} \) is rational leads to a contradiction, we conclude that \( \sqrt{2} \) must be **irrational**.

Would you like further clarification or more details on any part of the proof?

Here are 5 related questions:
1. Can you apply this contradiction method to prove other square roots are irrational?
2. What is the significance of proving a number is irrational?
3. How does this proof relate to the properties of prime numbers?
4. Why is it important that \( p \) and \( q \) are coprime in this proof?
5. What would change if we considered \( \sqrt{3} \) instead of \( \sqrt{2} \)?

**Tip:** When dealing with proofs, always make sure your assumptions align with the conditions of the problem.

Prove that √2 is irrational by contradiction.

This page provides a detailed step-by-step solution to prove that √2 is irrational using a proof by contradiction method. It covers key mathematical concepts such as rational and irrational numbers and involves important algebraic manipulation. Suitable for high school students in Grades 9-12.

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