Irrational numbers and real numbers

\(\sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\frac{4}{5}\). 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.

(\sqrt{23}) is real but irrational because (23) is not a perfect square. So it is not rational.

\(\frac{3}{8}\) is rational because it can be expressed as a ratio of integers \(\frac{p}{q}\) with \(q\neq0\), and it is not a whole number (whole numbers are 0,1,2,...). Options B (6) and C (0) are whole numbers, so they are incorrect. Option D (\(\sqrt{2}\)) is irrational, so it is not a rational number. Exam tip: first check whether the number is an integer/whole number; if not, check if it can be written as \(p/q\) with \(q\neq0\).

The number systems in the first option form a nested chain. Natural numbers are the counting numbers, such as 1, 2, 3, and so on; depending on convention, 0 may also be included. Whole numbers contain the natural numbers together with 0. Integers contain all whole numbers and also their negative counterparts, such as -1 and -2. Thus every natural number is a whole number, and every whole number is an integer.

Symbolically, the inclusion is \(\mathbb{N}\subseteq\mathbb{W}\subseteq\mathbb{Z}\). The reverse statement is false because negative integers are not natural numbers. Irrational numbers are not integers; examples include \(\sqrt{2}\) and \(\pi\). Real numbers include both rational and irrational numbers, so they are much more than only natural numbers. Therefore option A gives the correct order.

\(0.125\) is a terminating decimal, so it is a rational number. Converting to a fraction gives \(0.125 = \dfrac{125}{1000} = \dfrac{1}{8}\), which shows it equals \(p/q\). Option B is wrong because irrational numbers cannot be written as \(p/q\). Option C is wrong because \(0.125\) is a real number. Option D is wrong because integers have no fractional part, whereas \(0.125\) does. Exam tip: Convert a terminating decimal to a simplified fraction to confirm it is rational.

\(0.\overline{12}\) is a non-terminating recurring decimal; every recurring decimal is rational because it can be expressed as a fraction. For example \(0.\overline{12}=\frac{12}{99}=\frac{4}{33}\), so it is rational and non-terminating. Option B (\(0.875\)) is rational but terminating (\(0.875=\frac{875}{1000}=\frac{7}{8}\)), so it does not meet the "non-terminating" condition. Options C (\(\sqrt{12}=2\sqrt{3}\)) and D (\(\pi\)) are irrational. Exam tip: convert recurring decimals to fractions using 9, 99, 999... under the recurring block to quickly test rationality.

Rational numbers are closed under addition. If \(m=\frac{a}{b}\) and \(n=\frac{c}{d}\) with integers \(a,b,c,d\) and \(b,d\neq0\), then \(m+n=\frac{ad+bc}{bd}\), which is a ratio of integers and thus rational. Option B is incorrect because the sum of two rationals is not generally irrational. Options C and D hold only in special cases, not in general. Exam tip: express numbers as fractions and add to verify rationality quickly.

Write rationals as fractions: let \(m=\frac{a}{b}\) and \(n=\frac{c}{d}\) with integers \(a,b,c,d\) and \(b,d\neq0\). Then \(m\times n=\frac{ac}{bd}\), a ratio of integers, so it is rational. Option C is wrong because the product need not be an integer (e.g. \(1/2\times1/3=1/6\)); option D is wrong because the product need not be negative. Exam tip: test closure by expressing numbers as fractions and multiplying numerator with numerator, denominator with denominator.

\(\sqrt{28}=\sqrt{4\times7}=\sqrt{4}\cdot\sqrt{7}=2\sqrt{7}\). The perfect square 4 is taken outside the radical, giving the principal (positive) square root. Option D is incorrect because the principal square root is non-negative, so \(-2\sqrt{7}\) is not the simplified value. Option C (\(\sqrt{14}\)) comes from a mistaken grouping and does not equal \(2\sqrt{7}\). Option B (\(4\sqrt{7}\)) would imply \(28=16\times7\), which is false. Exam tip: factor out the largest perfect-square factor to simplify radicals efficiently.

\(\sqrt{147}=\sqrt{49\times3}=\sqrt{49}\times\sqrt{3}=7\sqrt{3}\). Thus the simplified form is \(7\sqrt{3}\). Option B is incorrect because \(3\sqrt{7}\) would arise from \(\sqrt{9\times7}\), which is not the factorization here. Option C is just the unsimplified radical and D is three times the correct value. Exam tip: always factor the radicand to find the largest perfect square and take its square root outside the radical (here 49 → 7).

(-\frac{1}{2}) lies between (-1) and (0) and is rational. Place negative fractions carefully on the number line.

(\frac{\sqrt{2}}{2}) is irrational and its value lies between (0) and (1). The root part makes it irrational.

From \(\sqrt{k}=12\), square both sides to get \(k=12^2=144\). Hence \(k\) is a perfect square (and an integer). Why other options are wrong: B and C give values (24 or 6) that do not satisfy the equation; D incorrectly states \(k=12\) and calls it irrational. Exam tip: when given \(\sqrt{\,\cdot\,}\), square both sides to remove the root—remember the principal square root is nonnegative.

The square root of a positive integer is rational only when the number is a perfect square. This is a direct MCQ rule.

(\sqrt{29}) is irrational because (29) is not a perfect square. All others can be written in rational form.

\(\sqrt{80}=\sqrt{16\times5}=\sqrt{16}\,\sqrt{5}=4\sqrt{5}\), so option A is correct. Option C (\(2\sqrt{5}\)) is only half of the correct value and option B (\(8\sqrt{5}\)) is twice as large, so both are incorrect. Option D (\(10\sqrt{2}\)) is numerically different (about 14.14) from \(4\sqrt{5}\) (about 8.94). Exam tip: always pull out the largest perfect square factor from under the root (here 16) and use \(\sqrt{a\times b}=\sqrt a\,\sqrt b\).

(\sqrt{2}+(-\sqrt{2})=0). Remember that the sum of two irrational numbers can sometimes be rational.

(\sqrt{22}) is irrational because (22) is not a perfect square. In exams first check perfect squares.

A rational number can be written as \\(\frac{p}{q}\\) with integers p, q and q ≠ 0. \\(\sqrt{34}\\) is irrational because 34 is not a perfect square; its decimal expansion is non-terminating and non-repeating, so it cannot be expressed as \\(\frac{p}{q}\\). The other choices are rational: \\(-\\frac{11}{4}\\) is already a fraction, \\(6.2 = \\frac{31}{5}\\), and \\(0.\overline{3} = \\frac{1}{3}\\). Exam tip: check for perfect squares under square roots and convert repeating/terminating decimals to fractions to test rationality.

A number is rational if it can be written as p/q, where p and q are integers and q is not zero. Since 0.875 is a terminating decimal, 0.875 = 875/1000 = 7/8, which has this form. Therefore it is a rational real number. It is not irrational, non-real, or an integer, so option A is correct.

Why: \(\sqrt{144}=12\), and 12 can be written as \(12/1\), a ratio of integers, so it is rational. The closest distractor, "irrational number," is incorrect because irrational numbers cannot be expressed as a ratio \(p/q\) of integers, which does not apply here. "Non real number" is wrong because 12 is a real number, and "Negative number" is wrong because 12 is positive. Exam tip: First check if the radicand is a perfect square — the square root of a perfect square is an integer and therefore rational.