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.