To prove that $\sqrt{2}$ is irrational by contradiction, let's proceed step by step:

Proof by Contradiction:

1. Assume the opposite: Suppose $\sqrt{2}$ is rational. By definition of a rational number, $\sqrt{2}$ can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers, and the fraction is in its simplest form (i.e., $\gcd(a, b) = 1$).

   $\sqrt{2} = \frac{a}{b}$

2. Square both sides: Squaring both sides of $\sqrt{2} = \frac{a}{b}$ gives:

   $2 = \frac{a^2}{b^2}$

   Multiplying both sides by $b^2$ to eliminate the denominator:

   $a^2 = 2b^2$

3. Analyze the equation: The equation $a^2 = 2b^2$ implies that $a^2$ is even, because it is equal to $2$ times another integer. If $a^2$ is even, then $a$ must also be even (since the square of an odd number is odd).

   Therefore, we can express $a$ as $a = 2k$ for some integer $k$.

4. Substitute $a = 2k$ into the equation: Substituting $a = 2k$ into the equation $a^2 = 2b^2$, we get:

   $(2k)^2 = 2b^2$

   Simplifying:

   $4k^2 = 2b^2$

   Dividing both sides by 2:

   $2k^2 = b^2$

   This implies that $b^2$ is also even, and thus $b$ must be even as well.

5. Contradiction: If both $a$ and $b$ are even, then $\frac{a}{b}$ is not in its simplest form, which contradicts our original assumption that $\frac{a}{b}$ is in lowest terms.

6. Conclusion: Therefore, the assumption that $\sqrt{2}$ is rational leads to a contradiction. Hence, $\sqrt{2}$ must be irrational.

Do you want more details, or have any questions about the steps?

Relative Questions:

1. Why do we assume $\sqrt{2}$ is rational to start the contradiction proof?
2. What does it mean for a fraction to be in its simplest form?
3. How do we know that $a$ and $b$ must both be even in the proof?
4. Can this method of proof by contradiction be applied to other square roots?
5. What is the significance of proving that certain numbers are irrational?

Tip:

Always verify that your assumption aligns with known properties of numbers when doing a proof by contradiction.