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An irrational number is a number that cannot be expressed as a ratio of two integers.
This means it cannot be written in the form \(\frac{a}{b}\) where both \(a\) and \(b\) are integers and \(b \eq 0\).
Some common examples of irrational numbers include \( \pi \) (pi) and \(\text{e}\) (Euler's number).
The decimal representation of an irrational number is non-repeating and non-terminating, which means it goes on forever without forming a repeating pattern.
For instance, \( \pi \) is approximately 3.14159, but its digits continue indefinitely without repetition.
Another famous irrational number is \( \sqrt{2}\) because it cannot be expressed as a ratio of two integers.
Understanding this distinction is essential when proving properties of irrational numbers, such as proving that \( \sqrt{x} \) is also irrational if \(x\) itself is irrational.

In contrast to irrational numbers, a rational number can be expressed as a ratio of two integers.
This means it can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \eq 0\).
For example, \( \frac{1}{2}\), 3, and \( -\frac{4}{5}\) are all rational numbers.
The decimal representation of rational numbers either terminates or repeats.
For instance, \( \frac{1}{2}\) can be written as 0.5 (terminating decimal), and \( \frac{1}{3}\) can be written as approximately 0.3333... (repeating decimal).
When we assume a number like \( \sqrt{x}\) is rational, we assume it can be written in the form of \( \frac{a}{b}\), leading us to test whether this assumption leads to a logical conclusion or a contradiction.

The square root of a number \(y\), represented as \( \sqrt{y}\), is a value that, when multiplied by itself, gives \(y\).
For example, \( \sqrt{9} = 3\) because \(3 \times 3 = 9\).
Square roots can be rational or irrational.
If \(x\) is a perfect square (like 4, 9, 16), then \( \sqrt{x}\) is rational.
However, if \(x\) is not a perfect square (like 2, 3, or 5), then \( \sqrt{x}\) is irrational.
To determine if \( \sqrt{x}\) is rational or irrational, we can use proof by contradiction, like in the given exercise.
By assuming \( \sqrt{x}\) is rational and showing this leads to a contradiction, we effectively prove that \( \sqrt{x}\) must be irrational if \(x\) is irrational.