Question

classify each number below as a rational number or an irrational number.

	rational	irrational
27.45	○	○
-16π	○	○
79.69	○	○
√36	○	○
√26	○	○

Explanation:

Step1: Recall definitions

A rational number is a number that can be expressed as $\frac{p}{q}$ where $p,q$ are integers and $q
eq0$. It includes integers, fractions, terminating decimals, repeating decimals, and roots of perfect squares. An irrational number is a non - repeating, non - terminating decimal, and roots of non - perfect squares.

Step2: Classify \(27.\overline{45}\)

\(27.\overline{45}\) is a repeating decimal. Repeating decimals can be expressed as a fraction. Let \(x = 27.\overline{45}=27.454545\cdots\). Then \(100x=2745.4545\cdots\), and \(100x - x=2745.4545\cdots - 27.4545\cdots\), \(99x = 2718\), \(x=\frac{2718}{99}\). So \(27.\overline{45}\) is rational.

Step3: Classify \(- 16\pi\)

\(\pi\) is an irrational number (\(\pi = 3.1415926\cdots\), non - repeating, non - terminating). When we multiply an irrational number by a non - zero integer (\(- 16\) here), the result is still irrational. So \(-16\pi\) is irrational.

Step4: Classify \(79.69\)

\(79.69\) is a terminating decimal. Terminating decimals can be written as a fraction (e.g., \(79.69=\frac{7969}{100}\)). So \(79.69\) is rational.

Step5: Classify \(\sqrt{36}\)

\(\sqrt{36}=6\), and \(6\) is an integer. Integers are rational numbers (since \(6=\frac{6}{1}\)). So \(\sqrt{36}\) is rational.

Step6: Classify \(\sqrt{26}\)

\(26\) is not a perfect square (since \(5^2 = 25\) and \(6^2=36\)), so \(\sqrt{26}\) is a non - repeating, non - terminating decimal. Thus, \(\sqrt{26}\) is irrational.

Answer:

• \(27.\overline{45}\): rational
• \(-16\pi\): irrational
• \(79.69\): rational
• \(\sqrt{36}\): rational
• \(\sqrt{26}\): irrational