Real Numbers

Each point on the real number line represents a real number. Understanding the number line helps in comparison.

( \frac{6}{0} ) is undefined because the denominator is zero. A real number must be defined.

Since \(81=9^2\), \(\sqrt{81}=9\). The principal square root is the positive number whose square is 81. The squares of 8 and 18 are 64 and 324, respectively, so they are incorrect; 81 itself is the radicand, not its square root. Exam tip: Memorising common perfect squares helps solve such questions quickly.

An irrational number has a decimal expansion that is non-terminating and non-repeating, so option B is correct. Option C describes rational numbers, which can be written as the ratio of two integers. Option A applies to rational numbers with terminating decimal expansions. Exam tip: terminating or non-terminating repeating decimals are rational, whereas non-terminating non-repeating decimals are irrational.

(0.125) is a terminating decimal, so it is rational. A terminating decimal can be changed into a fraction.

( \frac{-11}{6} ) is a ratio of two integers with non-zero denominator. So it is a rational real number.

Since \(32=16\times2\) and \(16\) is a perfect square, \(\sqrt{32}=\sqrt{16\times2}=\sqrt{16}\times\sqrt{2}=4\sqrt{2}\). Therefore, option C is correct. Exam tip: To simplify a surd, factor the number using its largest perfect-square factor.

\(45=9\times5=3^2\times5\). Therefore, \(\sqrt{45}=\sqrt{3^2\times5}=3\sqrt{5}\), so option A is correct. Option B results from incorrectly treating 45 as \(5^2\times3\), but \(5^2\times3=75\). Exam tip: factor the number into the largest perfect square multiplied by the remaining factor, then take the square root of the perfect square outside the radical.

Since \(75=25\times3\) and \(25\) is a perfect square, \(\sqrt{75}=\sqrt{25\times3}=\sqrt{25}\times\sqrt{3}=5\sqrt{3}\). Option C, \(3\sqrt{5}\), results from separating the factors incorrectly. Exam tip: take the largest perfect-square factor outside the square root.

The governing concept is the product property of square roots for non-negative numbers: √a × √b = √(ab). Applying it gives √2 × √18 = √(2 × 18) = √36. The principal square root of 36 is 6 because 6² = 36 and the principal root is non-negative. Therefore option A is correct. Option B, √20, results from an incorrect operation and is not equal to the product. Option C simply rewrites part of the expression without using the correct product property, while option D is incorrect because 9² = 81, not 36. This example also illustrates that multiplying two irrational square roots can produce a rational integer.

The governing concept is the quotient property of square roots: for non-negative radicands and a non-zero denominator, √a ÷ √b = √(a/b). Applying it gives √48 ÷ √3 = √(48/3) = √16 = 4, so option B is correct. A second method confirms the result. Since 48 = 16 × 3, √48 = 4√3; therefore (4√3) ÷ √3 = 4 because √3 is non-zero. Options 2, 8, and 16 do not equal the quotient. They may arise from treating the radicands or the radicals incorrectly, such as dividing 48 by an unsuitable number or forgetting that √16 equals 4. Both valid simplifications independently produce 4, establishing B as the unique answer.

Option B is correct. In \(0.272727\ldots\), the block 27 repeats indefinitely, so it is a recurring decimal and can be expressed as a ratio of two integers; therefore, it is rational. The student's mistake is assuming that every non-terminating decimal is irrational. Non-terminating non-recurring decimals are irrational, whereas recurring decimals are rational. Exam tip: a terminating or recurring decimal represents a rational number.

Using the Pythagorean theorem for the right triangle formed by two sides of the square, \(d^2=6^2+6^2=72\). Hence, \(d=\sqrt{72}=6\sqrt{2}\) metres, so option C is correct. Option A is only the side length, while option B is the sum of two sides, not the diagonal. Exam tip: the diagonal of a square is side \(\times\sqrt{2}\); therefore, this diagonal is also an irrational real number.

Since \(27=9\times3\) and \(12=4\times3\), we get \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Therefore, \(\sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3}\), so option A is correct. Option C simplifies only \(\sqrt{27}\) and does not include the second term. In an exam, first factor out perfect-square factors from each radical, then combine like surds.

The student's statement is incorrect. In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Therefore, its decimal expansion is non-terminating and non-recurring, which makes the number irrational. Option A is wrong because only terminating or recurring decimals are rational. Exam tip: For an infinite decimal, always check whether a fixed digit pattern repeats.

Since 5 and 6 are positive numbers, compare their squares: 5² = 25 and 6² = 36. Because 25 < 30 < 36, we get 5 < \(\sqrt{30}\) < 6. Therefore, option B is correct. Exam tip: to locate a square root, compare the number under the radical with the squares of the surrounding integers. \(\sqrt{24}\) is less than 5, while \(\sqrt{37}\) and \(\sqrt{50}\) are greater than 6.

\(\sqrt{9}=3\) because the principal square root is taken as non-negative. The negative sign outside the radical then gives \( -\sqrt{9}=-3 \). Option B ignores the outside negative sign. Exam tip: evaluate the square root first, then apply any sign written outside the radical.

Absolute value represents a number’s distance from 0 on the number line, so it is positive for every non-zero number. Therefore, \( |-12|=12 \). Option A incorrectly retains the negative sign. Exam tip: when finding an absolute value, remove the sign of a negative number.

First subtract inside the absolute-value bars: \(4-9=-5\). Absolute value is a number’s distance from zero, so \(|-5|=5\). Option A is the value obtained before applying absolute value. Exam tip: The absolute value of a real number is never negative.

The decimal 1.25 has two digits after the decimal point, so write it over 100: \\(1.25=\frac{125}{100}\\). Now simplify by dividing numerator and denominator by their common factor 25: \\(\frac{125}{100}=\frac{5}{4}\\). Therefore option A is correct. The denominator 100 is used because there are two decimal places; it is then reduced to the simplest form.

A useful check is that \\(\frac{5}{4}=1.25\\), since 5 divided by 4 equals 1.25. Option \\(\frac{25}{10}\\) also has numerical value 2.5, not 1.25, while \\(\frac{1}{25}\\) and \\(\frac45\\) are different values. Thus the correct fraction is \\(\frac54\\), and the supplied answer A is mathematically consistent. Writing a terminating decimal over a power of 10 is the standard conversion method.