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*}
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.
Proof.
Right triangle with legs labeled \(a\) and \(b\) and hypotenuse labeled \(c\text{.}\) Squares are appended to each side of the triangle.
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.
Example 4.4.3.
\(\frac{1}{2}\) is rational.
Example 4.4.4.
Is \(\sqrt{2}\) rational?
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.
Exercises Exercises
1.
An exercise.
2.
An exercise.
3.
An exercise.
4.
An exercise.
