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## Section 4.4 Example

Welcome to PreTeXt. Let’s type up some math. According to Pythagoras, \(a^2 + b^2 = c^2\text{.}\) But he also thought \(\sqrt{2}\) was rational, so maybe we shouldn’t take his word for it. Let’s see a proof by picture.
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### Theorem 4.4.1. Pythagorean Theorem.

Suppose a triangle has side lengths \(a, b\) and hypotenuse length \(c\text{.}\) Then
\begin{gather*} a^2 + b^2 = c^2. \end{gather*}
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### Proof.

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Right triangle with legs labeled \(a\) and \(b\) and hypotenuse labeled \(c\text{.}\) Squares are appended to each side of the triangle.
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### Definition 4.4.2.

A real number \(x \in \R\) is called rational if there are integers \(a, b\) so that \(x = \frac{a}{b}\text{.}\) Otherwise, \(x\) is called irrational, no matter what Pythagoras says.
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### Example 4.4.3.

\(\frac{1}{2}\) is rational.
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### Example 4.4.4.

Is \(\sqrt{2}\) rational?
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Hint 1.
Why does \(\pi = \frac{\sqrt{2}}{1}\) not show that \(\sqrt{2}\) is rational?
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Hint 2.
What would happen if you assumed \(\sqrt{2}\) is rational?
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Answer.
No!
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Solution.
Suppose that \(\sqrt{2} = \frac{a}{b}\) for relatively prime integers \(a, b\text{.}\) Then:
\begin{align*} 2 \amp = \frac{a^2}{b^2}\\ 2b^2 \amp = a^2 \end{align*}
But, after taking the prime factorization of both sides, the left side has an odd number of prime factors and the right side has an even number. This is a contradiction, so \(\sqrt{2}\) is irrational.
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### Exercise 4.4.5.

Prove that \(\sqrt{3}\) is irrational.
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### Exercises Exercises

#### 1.

An exercise.
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#### 2.

An exercise.
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#### 3.

An exercise.
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#### 4.

An exercise.
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