Proof by Contradiction: $\sqrt{2}$ is Irrational

We will prove that $\sqrt{2}$ is irrational using a proof by contradiction. This method assumes the opposite of what we want to prove and then shows that this assumption leads to a contradiction.

Step 1: Assume $\sqrt{2}$ is Rational

Suppose $\sqrt{2}$ is rational. This means we can express $\sqrt{2}$ as a ratio of two integers in simplest form:

$\sqrt{2} = \frac{p}{q}$

where $p$ and $q$ are integers, and the fraction $\frac{p}{q}$ is in its simplest form (i.e., $p$ and $q$ have no common factors other than 1, or they are coprime).

Step 2: Square Both Sides

Next, we square both sides of the equation to eliminate the square root:

$2 = \frac{p^2}{q^2}$

Multiplying both sides by $q^2$ to clear the denominator:

$2q^2 = p^2$

Step 3: Analyze the Equation $2q^2 = p^2$

From this equation, we can conclude that $p^2$ is even (since it is equal to $2q^2$, which is clearly even). Since $p^2$ is even, $p$ itself must be even (because the square of an odd number is odd).

Step 4: Express $p$ as an Even Integer

Since $p$ is even, we can write $p = 2k$ for some integer $k$. Substituting $p = 2k$ into the equation $2q^2 = p^2$:

$2q^2 = (2k)^2$

$2q^2 = 4k^2$

Dividing both sides by 2:

$q^2 = 2k^2$

Step 5: Analyze the Equation $q^2 = 2k^2$

This equation tells us that $q^2$ is also even, which implies that $q$ is even (for the same reason as before: the square of an odd number is odd).

Step 6: Contradiction

We have now shown that both $p$ and $q$ are even. But this contradicts our assumption that $\frac{p}{q}$ is in its simplest form, because if both $p$ and $q$ are even, they share a common factor of 2 and thus the fraction could be simplified further.

Step 7: Conclusion

Since our assumption that $\sqrt{2}$ is rational led to a contradiction, we conclude that $\sqrt{2}$ cannot be rational. Therefore, $\sqrt{2}$ is irrational.

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Would you like more details, or do you have any specific questions about this proof?

Here are 5 follow-up questions:

1. Can you generalize this proof to show that $\sqrt{n}$ is irrational for any non-square $n$?
2. How can the concept of irrational numbers be applied to real-life measurements?
3. Why is it important that $p$ and $q$ are in simplest form for this proof to work?
4. What other irrational numbers can be proved using a similar method of contradiction?
5. How does this proof relate to the ancient Greeks' discovery of irrational numbers?

Tip: When working with proofs, always carefully examine your assumptions—they are often key to identifying contradictions.