To prove that \(\sqrt{2}\) is not a rational number, we use a proof by contradiction. Here's the step-by-step process:

### Step 1: Assume the Opposite
Assume that \(\sqrt{2}\) is rational. If \(\sqrt{2}\) is rational, then it can be expressed as a fraction of two integers \(a\) and \(b\) in simplest form (i.e., the fraction is fully reduced with no common factors other than 1 between \(a\) and \(b\)):

\[
\sqrt{2} = \frac{a}{b}
\]

where \(a\) and \(b\) are integers, \(b \neq 0\), and \(\gcd(a, b) = 1\).

### Step 2: Square Both Sides
Square both sides of the equation to eliminate the square root:

\[
2 = \frac{a^2}{b^2}
\]

### Step 3: Multiply by \(b^2\)
To remove the fraction, multiply both sides by \(b^2\):

\[
2b^2 = a^2
\]

This equation implies that \(a^2\) is an even number because it is equal to \(2b^2\), which is clearly even (since it's two times an integer).

### Step 4: Analyze the Parity of \(a\)
If \(a^2\) is even, then \(a\) itself must also be even (since the square of an odd number is odd). Therefore, we can write \(a\) as \(a = 2k\) for some integer \(k\).

### Step 5: Substitute \(a = 2k\) into the Equation
Substitute \(a = 2k\) into the equation \(2b^2 = a^2\):

\[
2b^2 = (2k)^2
\]

\[
2b^2 = 4k^2
\]

### Step 6: Simplify the Equation
Divide both sides of the equation by 2:

\[
b^2 = 2k^2
\]

This equation implies that \(b^2\) is also even, so \(b\) must also be even.

### Step 7: Reach a Contradiction
Now we have found that both \(a\) and \(b\) are even. This means that \(a\) and \(b\) have at least 2 as a common factor. But this contradicts our original assumption that \(\frac{a}{b}\) is in the simplest form (i.e., \(\gcd(a, b) = 1\)).

### Conclusion
The contradiction arises from our initial assumption that \(\sqrt{2}\) is rational. Therefore, we must conclude that \(\sqrt{2}\) is **not** a rational number.

Would you like further details or have any questions?

### Related Questions:
1. Can you prove that \(\sqrt{3}\) is not rational?
2. What is the difference between rational and irrational numbers?
3. How do we prove that \(\pi\) is irrational?
4. Can the square root of any non-perfect square be a rational number?
5. How does the proof for \(\sqrt{2}\) being irrational differ from proving \(\sqrt{4}\) is rational?

### Tip:
When proving by contradiction, always ensure that your assumption contradicts a known fact or leads to an impossible situation. This method is powerful for proving irrationality and other mathematical properties.

Learn how to prove that √2 is not a rational number using a proof by contradiction method. This detailed step-by-step solution explores number theory concepts, including irrational numbers and proof techniques. Suitable for Grades 10-12.

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