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.

Direct answer: Option C, between 3 and 4. To locate a square root, compare the number under the root with nearby perfect squares. We have \(3^2=9\) and \(4^2=16\). Since \(9<10<16\), taking square roots gives \(3<\sqrt{10}<4\), because square root is increasing for non-negative numbers. Therefore \(\sqrt{10}\) lies between 3 and 4. Option A, between 1 and 2, is wrong because the squares of 1 and 2 are 1 and 4, and 10 is not between them. Option B, between 2 and 3, is wrong because 10 is greater than \(3^2=9\), so its square root is greater than 3. Option C is correct because 10 lies between \(3^2\) and \(4^2\). Option D, between 4 and 5, is wrong because \(10<16=4^2\), so the root must be less than 4. Remember: bracket a square root by finding the two consecutive perfect squares around the radicand.

Direct answer: Option A, a real number. Real numbers include rational numbers such as integers, fractions, and terminating or recurring decimals, as well as irrational numbers such as \(\sqrt2\) and \(\pi\). The real number system is closed under addition. This means that whenever two real numbers are added, the result is again a real number. If the numbers are \(x\) and \(y\), then \(x,y\in\mathbb R\) implies \(x+y\in\mathbb R\). Option A is correct because it states this closure property. Option B, only natural number, is wrong: for example, \(\frac12+\frac13=\frac56\), which is real but not a natural number. Option C, only irrational number, is wrong: \(2+3=5\), and 5 is rational. Option D, undefined number, is wrong because addition of real numbers is defined. Remember: real numbers are closed under addition, subtraction, and multiplication, although division needs a non-zero divisor.

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.

( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 ). In multiplication, numbers inside square roots can be multiplied.

Using the quotient rule for square roots, \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}=\sqrt{9}=3\). Therefore, option A is correct. Option C results from forgetting to take the square root of 9, while option D is only the simplified form of \(\sqrt{27}\) and does not account for division by \(\sqrt{3}\). Exam tip: when the radicands are positive, combine the quotient under one square root before simplifying.

Since \(2^2=4\) and \(3^2=9\), and \(4<6<9\), it follows that \(2<\sqrt{6}<3\). Therefore, \(\sqrt{6}\) is correct. Exam tip: for positive numbers, compare the numbers under the square roots with the squares of the given bounds.

( \sqrt{49}=7 ), and (7) is a rational real number. The square root of a perfect square can be an integer.

( \pi ) is an irrational number and lies on the real number line. Do not treat it as exactly equal to ( \frac{22}{7} ).