Real Numbers

\(\sqrt{81}=9\) because \(9^2=81\), and \(\sqrt{1}=1\) because \(1^2=1\). Therefore, \(\sqrt{81}+\sqrt{1}=9+1=10\). Option 9 is only the value of \(\sqrt{81}\); it misses the addition of \(\sqrt{1}\). Exam tip: the square root of a perfect square is the positive number whose square equals the given number.

Since \(20=4\times5\), \(\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}\). Therefore, \(\sqrt{20}+\sqrt{5}=2\sqrt{5}+\sqrt{5}=3\sqrt{5}\), so option A is correct. \(2\sqrt{5}\) is only the simplified form of \(\sqrt{20}\), not of the complete sum. Exam tip: simplify surds first, then combine like surds.

\(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\) and \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). Therefore, \(\sqrt{50}-\sqrt{8}=5\sqrt{2}-2\sqrt{2}=3\sqrt{2}\), so option C is correct. \(\sqrt{42}\) is incorrect because the difference of square roots is not generally equal to \(\sqrt{50-8}\). Exam tip: simplify each surd by factoring out its greatest perfect-square factor first.

\(0.\overline{27}=0.272727\ldots\) is a non-terminating recurring decimal, and it equals \(\frac{27}{99}=\frac{3}{11}\). Hence, it is rational even though its decimal expansion does not end. \(\sqrt{2}\), \(\pi\), and \(0.101001000100001\ldots\) have non-terminating, non-recurring decimal expansions, so they are irrational. Exam tip: A non-terminating recurring decimal is rational; a non-terminating non-recurring decimal is irrational.

Here, \(x\times x=(\sqrt{6})\times(\sqrt{6})=(\sqrt{6})^2=6\). Squaring the principal square root of a positive number gives the number itself. \(\sqrt{12}\) is incorrect because \(\sqrt{6}\times\sqrt{6}=\sqrt{36}=6\). Exam tip: recognise directly that \(\sqrt{a}\times\sqrt{a}=a\).

\(0.27272727\ldots\) is a non-terminating recurring decimal because the block 27 repeats indefinitely. Every recurring decimal is rational; in fact, \(0.272727\ldots=\frac{3}{11}\). Hence, it disproves the student's claim. In option B, the number of zeros keeps increasing, so there is no fixed repeating block and it is irrational. Exam tip: For a non-terminating decimal, check whether a repeating pattern exists before deciding if it is rational or irrational.

Given \(a=\sqrt{3}\), we get \(2a+\sqrt{3}=2\sqrt{3}+\sqrt{3}=3\sqrt{3}\). Hence, \(3\sqrt{3}\) is correct. \(2\sqrt{3}\) is only the value of \(2a\); the remaining \(\sqrt{3}\) must still be added. Exam tip: for like surds, add their coefficients.

Irrational numbers are not rational numbers, so statement C is incorrect. A rational number can be written in the form \(p/q\), where \(p\) and \(q\) are integers and \(q\ne0\). Integers, natural numbers, and whole numbers are all included in the set of real numbers. Exam tip: Remember the inclusion order: natural ⊂ whole ⊂ integers ⊂ rational ⊂ real.

Since \(24=4\times6\) and \(54=9\times6\), \(\sqrt{24}=2\sqrt{6}\) and \(\sqrt{54}=3\sqrt{6}\). Therefore, \(\sqrt{24}+\sqrt{54}=2\sqrt{6}+3\sqrt{6}=5\sqrt{6}\). A result such as \(7\sqrt{6}\) can arise from adding the coefficients incorrectly. Exam tip: first take out perfect-square factors from each surd, then add the coefficients of like surds.

\(\frac{9}{2}=4.5\), which is not an integer because integers do not have a fractional or decimal part. In contrast, \(-8\), \(0\), and \(\sqrt{121}=11\) are all integers. Exam tip: simplify a square-root option first; if its value is a whole number, it is an integer.

A rational number has a decimal expansion that either terminates or repeats after some point. Here, the number of zeros between the 1s keeps increasing, so no fixed block of digits repeats periodically. Hence \(0.101001000100001\ldots\) is non-terminating and non-repeating, making it irrational. Using only the digits 0 and 1 does not make a number rational. Exam tip: When a decimal has a pattern, check whether a fixed block actually repeats after a certain point.

Since \(72=36\times2\) and \(36\) is a perfect square, \(\sqrt{72}=\sqrt{36\times2}=\sqrt{36}\sqrt{2}=6\sqrt{2}\). \(4\sqrt{2}\) is incorrect because its square is \(32\), not \(72\). Exam tip: To simplify a surd, first identify the greatest perfect-square factor of the number.

Since \(125=25\times5\), \(\sqrt{125}=\sqrt{25\times5}=5\sqrt{5}\). Therefore, \(\sqrt{125}-\sqrt{5}=5\sqrt{5}-\sqrt{5}=4\sqrt{5}\), so option B is correct. \(\sqrt{120}\) is not equal to \(\sqrt{125}-\sqrt{5}\), because subtraction of square roots cannot be combined into one square root. Exam tip: simplify each surd by identifying its largest perfect-square factor first.

Given \(x=\sqrt{14}\), we get \(x^2=(\sqrt{14})^2=14\). Therefore, \(x^2-10=14-10=4\), so option D is correct. The value 14 is only \(x^2\); 10 still has to be subtracted. Exam tip: squaring a square root gives the original non-negative number.

Since \(18=9\times2\) and \(8=4\times2\), \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\). Therefore, \(\sqrt{18}+\sqrt{8}=3\sqrt{2}+2\sqrt{2}=5\sqrt{2}\). \(3\sqrt{2}\) is only the simplified form of \(\sqrt{18}\), not of the sum. Exam tip: simplify each surd by taking out perfect-square factors before adding like surds.

\(\sqrt{81}=9\) and \(\sqrt{9}=3\). Therefore, \(\frac{\sqrt{81}}{\sqrt{9}}=\frac{9}{3}=3\). Hence, option B is correct. The value 9 is only the square root of the numerator; division by the denominator is still required. Exam tip: evaluate each square root first, then simplify the fraction.

The set of real numbers contains all rational numbers and all irrational numbers. A rational number can be written in the form \(p/q\), where \(q\ne0\), whereas an irrational number such as \(\sqrt{2}\) cannot be written in this form. Integers and natural numbers are only subsets of rational numbers, so they do not give the complete classification of real numbers. Exam tip: remember that real numbers = rational numbers + irrational numbers.

(\sqrt{20}=2\sqrt{5}) and (\sqrt{45}=3\sqrt{5}), so the sum is (5\sqrt{5}). Simplify surds before adding.

In simplest form the denominator contains 3 along with 2, so the decimal will not terminate. Check prime factors of the denominator.

Since \(27=9\times3\) and \(12=4\times3\), \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Therefore, \(x=3\sqrt{3}-2\sqrt{3}=\sqrt{3}\). \(2\sqrt{3}\) is only the simplified value of \(\sqrt{12}\), not the value of the difference. Exam tip: simplify both surds to like radicals before subtracting.