Classifying Real Numbers 7B00A8

1. **Problem:** Classify each number in the first set as (a) natural number, (b) integer, (c) rational number, or (d) irrational number. 2. **Recall definitions:** - Natural numbers: positive whole numbers starting from 1 (1, 2, 3, ...). - Integers: whole numbers including negatives, zero, and positives (..., -2, -1, 0, 1, 2, ...). - Rational numbers: numbers that can be expressed as a fraction \(\frac{p}{q}\) where \(p,q\) are integers and \(q \neq 0\). - Irrational numbers: numbers that cannot be expressed as a simple fraction, their decimal expansions are non-repeating and non-terminating. 3. **Classify each number in set 1:** - \(\frac{3}{5}\): fraction, so rational. - \(-4\): integer (negative), so integer and rational. - \(0\): integer and rational. - \(\sqrt{2}\): known irrational number. - \(52\): positive whole number, natural number, integer, rational. - \(-\frac{1}{8}\): rational (fraction), integer? No, because not whole. - \(\sqrt{9}\): \(\sqrt{9} = 3\), which is natural, integer, rational. 4. **Summary for set 1:** - Natural numbers: \(52, 3\) - Integers: \(-4, 0, 52, 3\) - Rational numbers: \(\frac{3}{5}, -4, 0, 52, -\frac{1}{8}, 3\) - Irrational numbers: \(\sqrt{2}\) 5. **Classify each number in set 2:** - \(98\): natural, integer, rational. - \(-141\): integer, rational. - \(-\frac{7}{8}\): rational. - \(3.99\): decimal, but can be written as \(\frac{399}{100}\), rational. - \(-\sqrt{12}\): \(\sqrt{12} = 2\sqrt{3}\), \(\sqrt{3}\) is irrational, so irrational. - \(-\frac{54}{11}\): rational. 6. **Summary for set 2:** - Natural numbers: \(98\) - Integers: \(98, -141\) - Rational numbers: \(98, -141, -\frac{7}{8}, 3.99, -\frac{54}{11}\) - Irrational numbers: \(-\sqrt{12}\)