Irrational numbers and real numbers

Subtracting an irrational number from a rational number gives an irrational result. (\sqrt{6}) is irrational because (6) is not a perfect square.

(\sqrt{3}\times\sqrt{27}=\sqrt{81}=9). The product of two irrational numbers can be rational.

\(\sqrt{32}=\sqrt{16\times2}=\sqrt{16}\cdot\sqrt{2}=4\sqrt{2}\). Hence the simplified form is \(4\sqrt{2}\). Option B (\(8\sqrt{2}\)) is much larger and incorrect; option C (\(4\)) equals \(\sqrt{16}\), not \(\sqrt{32}\); option D is the unsimplified radical. Exam tip: always factor out the largest perfect square from under the root.

Since \(75=25\times3\), and \(25\) is a perfect square, \(\sqrt{75}=\sqrt{25}\times\sqrt{3}=5\sqrt{3}\). \(3\sqrt{5}\) is incorrect because \((3\sqrt{5})^2=45\), not 75. Exam tip: To simplify a surd, identify and factor out the greatest perfect-square factor.

\(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\) and \(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\). Adding like terms gives \(4\sqrt{3}+2\sqrt{3}=6\sqrt{3}\). The distractor \(2\sqrt{15}\) comes from wrongly treating the sum as \(\sqrt{48+12}=\sqrt{60}=2\sqrt{15}\); you cannot combine square roots across addition that way. Exam tip: simplify each radical into simplest surd form first, then add or subtract only like surds.

\sqrt{45}=\sqrt{9\cdot5}=3\sqrt{5} and \sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}. Subtracting gives 3\sqrt{5}-2\sqrt{5}=\sqrt{5}. The closest distractor D (2\sqrt{5}) is incorrect because it equals \sqrt{20}, not the difference. Exam tip: factor common square factors (like \sqrt{5}) to simplify radicals before adding or subtracting.

\(\sqrt{19}\) is correct because \(4^2=16\) and \(5^2=25\), and since \(16<19<25\), we have \(4<\sqrt{19}<5\). Also 19 is not a perfect square, so \(\sqrt{19}\) is irrational. The closest distractor \(\frac{9}{2}=4.5\) lies between 4 and 5 but is rational. \(\sqrt{16}=4\) and \(\sqrt{25}=5\) are the end points, not inside the open interval (4,5). Exam tip: to check whether a square root is between two integers, compare the radicand with the squares of those integers and check if it's a perfect square.

The decimal never terminates and has no fixed repeating block: the number of zeros between successive 4s increases (1, 2, 3, ...), so there is no periodic pattern. Therefore it is non-terminating and non-repeating (an irrational-type decimal). Option C is incorrect because a repeating decimal must have a fixed repeating cycle, which this expansion lacks. Exam tip: a decimal is rational only if it terminates or eventually repeats; check for a constant periodic block to decide.

The decimal is repeating ('18' repeats), so it must be rational because repeating decimals can be written as fractions. For example, let \(x=2.\overline{18}\). Then \(100x=218.\overline{18}\), subtracting gives \(99x=216\), so \(x=\dfrac{216}{99}=\dfrac{24}{11}\). Thus the number is rational. Option B (irrational) is wrong because irrational numbers have non‑repeating, non‑terminating decimals; here the decimal repeats. Options C and D are also incorrect: the number is real (not non‑real) and it is not an integer. Exam tip: convert repeating decimals to fractions by multiplying by an appropriate power of 10 equal to the repeating block length and subtracting.

Since \((\sqrt{13})^2=13\), squaring the square root returns the original nonnegative number. Option B is just the original \(x\), not \(x^2\). Option C (169) results from incorrectly squaring 13 itself. Option D is wrong because \(x^2\) cannot be negative here. Exam tip: whenever \(x=\sqrt{a}\) for \(a\ge0\), you can directly replace \(x^2\) with \(a\).

(\frac{3}{\sqrt{3}}=\sqrt{3}) because (3=\sqrt{3}\times\sqrt{3}). Practice simplifying denominators with roots.

Multiplying any real number — whether rational or irrational — by zero always yields zero (0). Zero is rational because it can be written as \(0=0/1\). Hence the correct choice is (0), which is rational. The closest distractor is B (Always irrational), which is incorrect since zero is not irrational. Exam tip: Whenever you see "×0" in a question, the product is immediately 0, regardless of the other factor's type.

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but their sum \(\sqrt{5}+(-\sqrt{5})=0\) is rational, so A is correct. Closest distractor B, \(\sqrt{2}+\sqrt{7}\), remains irrational (sums of distinct square‑free roots are generally irrational). C gives \(\sqrt{3}+\sqrt{3}=2\sqrt{3}\), still irrational. D is a sum of a rational and an irrational, which is irrational. Exam tip: look for additive inverses — if both terms cancel each other, the sum is rational (often 0).

(\sqrt{2}\times\sqrt{5}=\sqrt{10}), which is irrational. Multiplying equal roots can often give a rational number.

Since \(0.09=\tfrac{9}{100}\), we have \(\sqrt{0.09}=\sqrt{\tfrac{9}{100}}=\tfrac{3}{10}=0.3\). The square root symbol denotes the principal (non‑negative) root, so \(-0.3\) is not the value of \(\sqrt{0.09}\). Exam tip: convert decimals to fractions (e.g. \(0.09=9/100\)) to make square roots easier to compute.

\(\sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\frac{4}{5}\). The principal square root is non-negative, so the value is \(\frac{4}{5}\). Option B is the negative of the principal root and hence incorrect; option C is the original fraction, not its square root; option D is a wrong computation. Exam tip: unless ± is specified, take the principal (non-negative) square root.

\(\sqrt[3]{8}=2\). Since 8 is a perfect cube, its cube root is an integer (2), and every integer is rational (for example \(2=\frac{2}{1}\)). Thus option A is correct. Option B is incorrect because irrational numbers are non‑terminating, non‑repeating decimals, which does not apply to 2. Option C is wrong because 2 is a real number (not a non‑real/complex‑only number). Option D is incorrect because 2 is a terminating decimal (2.0), not a non‑terminating decimal. Exam tip: first check if the radicand is a perfect power; a perfect cube gives an integer cube root which is rational.

Assume \(\sqrt[3]{4}\) is rational: \(\sqrt[3]{4}=p/q\) in lowest terms (gcd\((p,q)=1\)). Cubing both sides gives \(4q^{3}=p^{3}\). Hence \(p^{3}\) is even so \(p\) is even; write \(p=2k\). Substituting yields \(4q^{3}=8k^{3}\) ⇒ \(q^{3}=2k^{3}\), so \(q\) is even too, contradicting gcd\((p,q)=1\). Therefore \(\sqrt[3]{4}\) is irrational. Option B (2) is incorrect because \(2^{3}=8\), not 4; C and D are also obviously wrong. Exam tip: for cube roots of integers, check whether the radicand is a perfect cube — if not, the cube root is irrational.

0 is a rational number and adding 0 does not change the value (a+0=a). Rational numbers are closed under addition, so a+0 remains rational. Option B is wrong because adding 0 cannot turn a rational into an irrational; C is wrong because rationals are real numbers, not non‑real; D is wrong because a rational need not be a natural number (it can be a fraction or negative). Exam tip: remember the additive identity (0) and the closure property of rational numbers under addition.

Adding 0 does not change a number — 0 is the additive identity, so \(b+0=b\). Since \(b\) is given as irrational, it remains irrational after adding 0. Option B is wrong because it assumes \(b\) is rational, contrary to the premise; C and D are incorrect because integers are a subset of rationals and "always zero" would require \(b=0\), but 0 is rational, not irrational. Exam tip: remember the additive identity (0) — adding 0 leaves any number unchanged.