22.11 Theorem The polynomial-2 has no zeros in the rational numbers. Thus sqrt(2) is not a rational number.

The statement you're referring to connects the irrationality of \(\sqrt{2}\) with the properties of a specific polynomial.

### Explanation:

**Theorem 22.11** (as stated):  
"The polynomial \(x^2 - 2\) has no zeros in the rational numbers. Thus, \(\sqrt{2}\) is not a rational number."

This is a classic proof of the irrationality of \(\sqrt{2}\) based on the fact that if a rational number is a root of a polynomial with integer coefficients, it must satisfy certain conditions.

#### Proof Outline:

1. **The Polynomial**:  
   Consider the polynomial \(f(x) = x^2 - 2\). This is a quadratic equation, and its roots are \(x = \pm \sqrt{2}\).

2. **Assume a Rational Root**:  
   To prove by contradiction, assume \(\sqrt{2}\) is a rational number. A rational number can be written as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers with no common factors (i.e., in lowest terms).

   Then:
   \[
   \left(\frac{p}{q}\right)^2 = 2
   \]
   Multiplying both sides by \(q^2\):
   \[
   p^2 = 2q^2
   \]
   This implies that \(p^2\) is an even number (since it is equal to \(2q^2\), which is clearly even).

3. **\(p\) Must Be Even**:  
   If \(p^2\) is even, then \(p\) must also be even (since the square of an odd number is odd). Therefore, we can write \(p = 2k\) for some integer \(k\).

4. **Substitute \(p = 2k\)**:  
   Substituting \(p = 2k\) into the equation \(p^2 = 2q^2\):
   \[
   (2k)^2 = 2q^2
   \]
   \[
   4k^2 = 2q^2
   \]
   Dividing both sides by 2:
   \[
   2k^2 = q^2
   \]
   This implies that \(q^2\) is also even, and thus \(q\) must also be even.

5. **Contradiction**:  
   If both \(p\) and \(q\) are even, then \(\frac{p}{q}\) is not in its lowest terms, contradicting our assumption that \(\frac{p}{q}\) was a fraction in its simplest form.

Thus, our original assumption that \(\sqrt{2}\) is rational must be false. Therefore, \(\sqrt{2}\) is **irrational**.

### Conclusion:
The fact that the polynomial \(x^2 - 2\) has no rational zeros is a direct way to conclude that \(\sqrt{2}\) is not a rational number. This is because the Rational Root Theorem tells us that any rational root of a polynomial must be a fraction where the numerator divides the constant term, and the denominator divides the leading coefficient. In this case, there are no such rational numbers that satisfy the equation \(x^2 - 2 = 0\).

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Would you like more details or have any questions on this? Here are some follow-up questions to help you explore further:

1. What is the Rational Root Theorem, and how does it apply here?
2. How can we generalize this proof to other roots, such as \(\sqrt{3}\)?
3. What is the history behind the discovery of the irrationality of \(\sqrt{2}\)?
4. Can you explain how the Pythagorean Theorem relates to the irrationality of \(\sqrt{2}\)?
5. How do we define a number to be irrational, and what are other famous examples?

**Tip**: The proof of irrationality for numbers like \(\sqrt{2}\) is one of the earliest known examples of mathematical contradiction, a very powerful proof technique.

22.11 Theorem: The polynomial x^2 - 2 has no zeros in the rational numbers. Thus sqrt(2) is not a rational number.

Explore the classic proof of the irrationality of sqrt(2) through the polynomial x^2 - 2. This detailed explanation applies the Rational Root Theorem and covers key concepts in algebra and number theory. Suitable for students in Grades 9-12.

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