Verify by the method of contradiction that `sqrt(7)` is irrational.

Answer

To verify by the method of contradiction that \(\sqrt{7}\) is irrational, we will follow these steps:

### Step 1: Assume the opposite
Assume that \(\sqrt{7}\) is a rational number. By definition of rational numbers, this means that we can express \(\sqrt{7}\) as a fraction of two integers \(a\) and \(b\) (where \(b \neq 0\)):
\[ \sqrt{7} = \frac{a}{b} \]
where \(a\) and \(b\) have no common factors other than 1 (i.e., they are coprime).

### Step 2: Square both sides
Squaring both sides of the equation gives:
\[ 7 = \frac{a^2}{b^2} \]
Multiplying both sides by \(b^2\) results in:
\[ 7b^2 = a^2 \]

### Step 3: Analyze the equation
From the equation \(7b^2 = a^2\), we can deduce that \(a^2\) is divisible by 7. This implies that \(a\) must also be divisible by 7 (since if a prime number divides the square of a number, it must also divide the number itself).

### Step 4: Express \(a\) in terms of 7
Let \(a = 7k\) for some integer \(k\). Substituting this back into the equation gives:
\[ 7b^2 = (7k)^2 \]
This simplifies to:
\[ 7b^2 = 49k^2 \]
Dividing both sides by 7 results in:
\[ b^2 = 7k^2 \]

### Step 5: Analyze \(b^2\)
From the equation \(b^2 = 7k^2\), we can see that \(b^2\) is also divisible by 7. Therefore, \(b\) must also be divisible by 7.

### Step 6: Reach a contradiction
Since both \(a\) and \(b\) are divisible by 7, this means that they have a common factor of 7. However, this contradicts our initial assumption that \(a\) and \(b\) are coprime (having no common factors other than 1).

### Step 7: Conclusion
Since our assumption that \(\sqrt{7}\) is rational leads to a contradiction, we conclude that \(\sqrt{7}\) must be irrational.

Thus, we have verified by contradiction that \(\sqrt{7}\) is irrational.

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