To prove that \(\sqrt{7}\) is irrational, we can use a proof by contradiction. Here are the steps:

### Step 1: Assume \(\sqrt{7}\) is rational
- Suppose that \(\sqrt{7}\) is a rational number.
- If \(\sqrt{7}\) is rational, then we can express it as a fraction of two integers:
  \[
  \sqrt{7} = \frac{a}{b}
  \]
  where \(a\) and \(b\) are integers, \(b \neq 0\), and the fraction \(\frac{a}{b}\) is in its simplest form (i.e., \(a\) and \(b\) have no common divisors other than 1).

### Step 2: Square both sides
- Squaring both sides of the equation gives:
  \[
  7 = \frac{a^2}{b^2}
  \]
- Multiply both sides by \(b^2\) to eliminate the denominator:
  \[
  7b^2 = a^2
  \]
- This equation implies that \(a^2\) is divisible by 7.

### Step 3: Analyze the divisibility of \(a\)
- If \(a^2\) is divisible by 7, then \(a\) must also be divisible by 7. This is because if a prime number divides a square, it must divide the base of that square.
- Therefore, we can write \(a = 7k\) for some integer \(k\).

### Step 4: Substitute \(a = 7k\) into the equation
- Substitute \(a = 7k\) into the equation \(7b^2 = a^2\):
  \[
  7b^2 = (7k)^2
  \]
  \[
  7b^2 = 49k^2
  \]
- Divide both sides by 7:
  \[
  b^2 = 7k^2
  \]
- This equation implies that \(b^2\) is divisible by 7, and thus \(b\) must also be divisible by 7.

### Step 5: Contradiction
- We now have that both \(a\) and \(b\) are divisible by 7. However, this contradicts our original assumption that \(\frac{a}{b}\) is in its simplest form, as \(a\) and \(b\) have a common divisor (7).

### Conclusion
- Since assuming \(\sqrt{7}\) is rational leads to a contradiction, the original assumption must be false. Therefore, \(\sqrt{7}\) is **irrational**.

Would you like more details or have any questions?

---

Here are 5 related questions to expand your understanding:

1. How can we generalize this proof to other square roots of non-perfect squares?
2. What is the difference between rational and irrational numbers?
3. How does the Fundamental Theorem of Arithmetic support this proof?
4. Can \(\sqrt{49}\) be considered irrational, and why?
5. How does this proof relate to the proof of the irrationality of \(\sqrt{2}\)?

**Tip**: For any non-perfect square \(n\), the square root \(\sqrt{n}\) is irrational.

Discover a detailed proof that the square root of 7 is irrational using proof by contradiction. This solution covers essential number theory concepts, the Fundamental Theorem of Arithmetic, and properties of rational and irrational numbers. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation