Real Numbers

Adding irrational ( \sqrt{6} ) to rational (4) gives an irrational number. It is also a real number.

Subtracting an irrational number from a rational number gives an irrational result. Thus (7-\sqrt{3}) is an irrational real number.

Multiplying the same square root by itself gives the number inside. Therefore ( \sqrt{13}\times\sqrt{13}=13 ).

The principal square root √15 is the non-negative number whose square is 15. Therefore the inverse relationship between squaring and taking the principal square root gives (√15)² = 15. More generally, for every non-negative real number a, (√a)² = a. Hence option A is correct. It is important not to square 15 again: 225 would be 15², not the square of √15. Option B doubles the number, and option D applies an unrelated product rule. Because 15 is positive, there is no issue involving the sign or the distinction between √(x²) and (√x)²; the latter directly returns x here.

The absolute value represents a number’s distance from zero, so it is always non-negative. The distance of \(-18\) from zero is 18; therefore, \(\lvert -18\rvert=18\). Option A is incorrect because it is the original number, not its absolute value. Exam tip: To find the absolute value of a negative number, remove its negative sign.

First calculate \(6-14=-8\). The absolute value represents a number’s distance from zero, so \(|-8|=8\). Therefore, option C is correct. Remember that an absolute value is never negative, so
-8
is not the answer.

Since (3^2<10<4^2), ( \sqrt{10} ) lies between (3) and (4). Since (10) is not a perfect square, it is irrational.

Since (4^2<22<5^2), ( \sqrt{22} ) lies between (4) and (5). Since (22) is not a perfect square, it is irrational.

There are infinitely many irrational numbers between two distinct real numbers. This is a density property of the real number line.

The real number line is continuous, so infinitely many real numbers lie between two distinct points. Remember this basic property.

\(\sqrt{0}=0\) and \(\sqrt{100}=10\), because \(10\times10=100\). Hence, \(\sqrt{0}+\sqrt{100}=0+10=10\), so option B is correct. Exam tip: the principal square root of a positive perfect square is taken as positive.

Since \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\), we get \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\). Because \(\sqrt{2}\) is irrational, multiplying it by the non-zero rational number 3 still gives an irrational number. Therefore, the student’s statement is incorrect. Option A reflects the misconception that a square-root sign automatically makes a number rational. Exam tip: simplify the surds first and then combine like surds.

Since \(121=11^2\) and \(169=13^2\), we have \(\sqrt{121}=11\) and \(\sqrt{169}=13\). Therefore, \(11+13=24\), so option B is correct. Exam tip: evaluate each square root first and then perform the addition.

Since \(225=15^2\) and \(9=3^2\), we have \(\sqrt{225}=15\) and \(\sqrt{9}=3\). Therefore, \(15\div3=5\), so option C is correct. Options B and D result from an incorrect division calculation. In an exam, evaluate both square roots first and then divide.

Multiply the coefficients and the surd parts separately: \(2\times 3\times \sqrt{3}\times\sqrt{3}=6\times 3=18\), since \(\sqrt{3}\times\sqrt{3}=3\). Therefore, \(18\) is correct. Exam tip: \(\sqrt{a}\times\sqrt{a}=a\) for \(a\geq 0\).

A terminating decimal can always be written in the form c(p/qc) by taking a denominator such as 10, 100, or 1000, so it is rational. Option B is incorrect because some non-terminating decimals are recurring and rational, such as c(0.333...=1/3c). Exam tip: rational numbers have terminating or non-terminating recurring decimal expansions, whereas irrational numbers have non-terminating, non-recurring expansions.

A number is rational only when its decimal expansion is either terminating or repeating in a fixed pattern. In 0.101001000100001…, the number of zeros between successive 1s keeps increasing, so no fixed block repeats and the decimal expansion does not terminate. Therefore, the number is irrational. Merely using the digits 0 and 1 does not make a number rational. Exam tip: terminating or recurring decimals are rational; non-terminating, non-recurring decimals are irrational.

( \sqrt{18}=3\sqrt{2} ), so ( \sqrt{2}+3\sqrt{2}=4\sqrt{2} ). First simplify the surd and then add like terms.

Since \(125=25\times5\) and \(25\) is a perfect square, \(\sqrt{125}=\sqrt{25\times5}=5\sqrt{5}\). Therefore, option A is correct. Option B has an extra factor of \(5\), while option C is not the simplified form and equals \(25\). Exam tip: factor the radicand using its largest perfect-square factor before simplifying the square root.

To simplify a square root, factor the radicand into the largest possible perfect-square factor and the remaining factor. Here 147 = 49 × 3 = 7² × 3. Therefore √147 = √(49 × 3) = √49 × √3 = 7√3. Thus option B is correct. Option C corresponds to √63 rather than √147, while option A incorrectly uses 21 as the coefficient and option D treats 49 itself as the coefficient after taking a square root. The simplified form should have no square factor remaining under the radical; since 3 has no perfect-square factor greater than 1, 7√3 is fully simplified.