Real Numbers

Real numbers include both rational and irrational numbers. In exams remember the whole number system on the real number line.

( \sqrt{5} ) is irrational and every irrational number is real. It can be shown on the real number line.

Natural numbers are a part of real numbers. Smaller sets can be included in a larger number system.

( \frac{-3}{8} ) is a ratio of two integers with non-zero denominator. So it is both rational and real.

The governing concept is decimal classification. The decimal 0.625 terminates after three digits, so it can be converted into a fraction: 0.625 = 625/1000. Dividing numerator and denominator by 125 gives 5/8. Since 5 and 8 are integers and the denominator is non-zero, 0.625 is rational. It is also real because every rational number belongs to the real-number system. Therefore option B is correct. It is not irrational, since irrational decimals are non-terminating and non-repeating. It is not non-real or merely imaginary; the given finite decimal has an ordinary position on the real number line.

The bar over 6 means that 6 repeats indefinitely: 0.666... A repeating decimal is rational because it can be expressed as a fraction. Let x = 0.666.... Then 10x = 6.666.... Subtracting the first equation from the second gives 9x = 6, so x = 6/9 = 2/3. Thus the number is rational, and every rational number is real. Option C is correct. It is not irrational because irrational decimals do not repeat in a fixed pattern. It is not a natural number, since 2/3 is not a counting number, and it is certainly not undefined.

The governing rule is that rational numbers have decimal expansions that either terminate or repeat a fixed pattern. A decimal that continues forever without any recurring block cannot be written as p/q for integers p and q with q non-zero. Such a number is irrational. It is nevertheless real, because real numbers include both rational and irrational numbers. Therefore option B is correct. Option A describes terminating or repeating decimals, not this type. A whole number has no fractional part and cannot generally have this decimal form, while zero is a particular rational number equal to 0/1. The absence of repetition is the decisive clue.

When the denominator of a fraction is zero, the division is undefined. There is no real number that gives 5 when multiplied by 0, so \(\frac{5}{0}\) is not a real number. Exam tip: division by zero is always undefined.

( -12 ) is an integer and every integer is a real number. An integer can be written as ( \frac{p}{1} ).

The governing concept is the hierarchy of number systems. Natural numbers, whether the convention begins with 0 or 1, are counting numbers and are all located on the real number line. Therefore the natural-number set is a subset of the real-number set: N ⊂ R. Real numbers include rational numbers and irrational numbers; natural numbers are included among the integers, then rationals, and finally reals. Hence option B is correct. Natural numbers are not only irrational, because every natural number is rational, for example 3 = 3/1. They are neither undefined nor restricted to negative values, since natural numbers are non-negative or positive by convention.

\(\sqrt{25}\) denotes the principal, or non-negative, square root of 25. Since \(5^2=25\), \(\sqrt{25}=5\). Although \((-5)^2\) is also 25, the principal square root is not negative. Exam tip: remember that \(\sqrt{x^2}=|x|\), not always \(x\).

( -2.5 ) is negative and lies between ( -3 ) and ( -2 ). Be careful with the left direction for negative numbers.

Since (1^2<2<2^2), ( \sqrt{2} ) lies between (1) and (2). Compare squares to locate square roots.

(3^2=9) and (4^2=16), so ( \sqrt{10} ) lies between (3) and (4). First check nearby perfect squares.

Real numbers are closed under addition. The sum of two real numbers is always a real number.

Real numbers are closed under multiplication, so the product of any two real numbers is also a real number. The product may be an integer, rational number, or irrational number; for example, \(\sqrt{2}\times 1=\sqrt{2}\), which is irrational but still real. Therefore, option C is incorrect. Exam tip: real numbers are closed under addition, subtraction, and multiplication.

Like terms are added, so ( \sqrt{3}+\sqrt{3}=2\sqrt{3} ). Do not add numbers inside square roots directly.

The governing concept is evaluation of principal square roots. The principal square root of a positive number is its non-negative root. Since 3 × 3 = 9, √9 = 3, and since 2 × 2 = 4, √4 = 2. Therefore, √9 + √4 = 3 + 2 = 5, so option A is correct. Option B, √13, comes from the incorrect assumption that √a + √b equals √(a + b); square roots generally cannot be combined across addition. Option C does not follow from the two evaluated roots, and option D is also inconsistent with the calculation. The exact value is obtained by evaluating each square root separately and then adding the results.

To simplify a square root, look for a factor inside the radical that is a perfect square. Here, write 18 as the product of 9 and 2: \\(18=9\\times2\\). Since the square root of 9 is 3, the square root of the product can be separated as \\(\\sqrt{18}=\\sqrt{9\\times2}=\\sqrt{9}\\sqrt{2}=3\\sqrt{2}\\). The remaining factor 2 has no square factor greater than 1, so this is the simplest radical form.

Option A is correct because it gives \\(3\\sqrt{2}\\). The expression \\(2\\sqrt{3}\\) would square to 12, not 18, so it is not equivalent. Also, \\(9\\sqrt{2}\\) is too large, and \\(\\sqrt{9}=3\\) is not equal to \\(\\sqrt{18}\\). The essential step is separating the perfect-square factor 9 from the number under the square root.

Since \(8=4\times2\), we get \(\sqrt{8}=\sqrt{4\times2}=\sqrt{4}\sqrt{2}=2\sqrt{2}\). Therefore, option B is correct. In options A and D, the coefficient is incorrectly made larger, while option C omits the factor \(\sqrt{4}=2\). Exam tip: factor the number inside the square root and take the greatest perfect-square factor outside the radical.