Irrational numbers and real numbers

14 is a natural number because it is a positive integer, and it can be written as \(\frac{14}{1}\), so it is rational as well (a rational number can be expressed as \(\frac{p}{q}\), \(q\neq0\)). \(\sqrt{14}\) is irrational, -14 is not a natural number (it's negative), and 0.14 is rational but not a natural number. Exam tip: natural numbers are positive integers (1,2,3,...); any number expressible as \(\frac{p}{q}\) with integer p,q (q≠0) is rational.

The square root of a positive integer is irrational when it is not a perfect square. Remember this simple rule.

Since \(0.4\times0.4=0.16\), \(\sqrt{0.16}=0.4\). Note that the square-root symbol denotes the principal (non-negative) root, so although \((-0.4)^2=0.16\), the principal square root is positive 0.4. The other options are incorrect because \(0.04^2=0.0016\) and \(4^2=16\). Exam tip: when asked for \(\sqrt{\,\cdot\,}\) without sign, take the non-negative root.

\(\sqrt{41}\) is real and since 41 is not a perfect square, \(\sqrt{41}\) is irrational. Check the other options: \(\frac{41}{1}=41\) is an integer (rational), 4.1 = \(\frac{41}{10}\) is rational, and 0.\overline{41} is a repeating decimal equal to \(\frac{41}{99}\), hence rational. Exam tip: a decimal that terminates or repeats is rational; the square root of a non‑perfect square is typically irrational.

In option A, \(\frac{3}{5}\) is rational and \(\sqrt{10}\) is irrational, so A is explicitly a sum of a rational and an irrational number. Such a sum is generally irrational, so A is correct. In option B both terms (\(\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\)) are irrational — this is not a rational+irrational case. Options C and D are sums of rational numbers only, so they are not examples of rational plus irrational. Exam tip: identify each term's type first — rational+irrational is usually irrational; sum of two rationals is always rational, while sum of two irrationals can be either but is not the case asked here.

Correct because \(\sqrt{150}=\sqrt{25\times6}=5\sqrt{6}\); 25 is the largest perfect square factor and can be taken outside the root. Why the closest distractor is wrong: \(3\sqrt{50}=3\times5\sqrt{2}=15\sqrt{2}\), not equal to \(5\sqrt{6}\). Exam tip: always factor the radicand into the largest perfect square times the remainder to simplify square roots quickly.

(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}). Taking out the greatest perfect square is a good method.

Since \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\), we have \(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\). Option B, \(\sqrt{20}=2\sqrt{5}\), is not equal to \(4\sqrt{2}\). Option C equals only \(\sqrt{18}\) (i.e. \(3\sqrt{2}\)), and option D is too small. Exam tip: always simplify each radical by extracting perfect squares first, then combine like (same-radical) terms.

Since \(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\), we have \(\sqrt{3}+\sqrt{75}=\sqrt{3}+5\sqrt{3}=6\sqrt{3}\). The common wrong idea \(\sqrt{78}\) assumes \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\), which is not true in general. Exam tip: factor out perfect squares first and then add like radical terms.

Zero can be expressed as \\(0=\frac{0}{1}\\), so it is the ratio of two integers and therefore rational. Zero also lies on the number line, so it is a real number. Option B is incorrect because irrational numbers cannot be written as a ratio of integers, but zero can. Option C is incorrect because zero is included among real numbers. Option D is misleading: many conventions define natural numbers as 1,2,3,... so zero is not necessarily a natural number (but check the definition used). Exam tip: always confirm the convention for 'natural numbers' in the question context or syllabus.

The decimal is formed by concatenating the natural numbers: 0.1 2 3 4 5 6 7 8 9 10 11 12 ... . It never terminates and does not settle into any fixed periodic block, so it is non‑terminating and non‑repeating. A non‑terminating, non‑repeating decimal represents an irrational number. The closest distractor (C) is wrong because a repeating decimal would show a constant repeating pattern after some point, which this sequence does not. Options B and D are also incorrect: it does not terminate and is not an integer. Exam tip: to test rationality, look for a repeating block — its presence implies rationality, its absence (for infinite decimals) implies irrationality.

All numbers in the first set can be written in (\frac{p}{q}) form. The other sets contain an irrational number.

(\sqrt{3}), (\sqrt{5}), and (\sqrt{12}) are all irrational. Remove options with perfect squares and rational numbers.

A number of the form \(\frac{p}{q}\) with integers \(p,q\) and \(q\neq0\) is, by definition, a rational number. Rational numbers have decimal expansions that either terminate or repeat. Option C (real number) is true in a broader sense because every rational is real, but it is not the specific name asked for. Option B (irrational) is incorrect since irrationals cannot be written as \(\frac{p}{q}\). Option D (natural number) is incorrect because natural numbers are positive integers, not general fractions. Exam tip: remember the defining condition — expressible as \(\frac{p}{q}\) with integer numerator and nonzero integer denominator; look for terminating/repeating decimals as a quick check.

A real number that is not rational is called irrational. It can also be identified through its decimal form.

Since 11 is not a perfect square, \(\sqrt{11}\) is irrational. The difference of a rational number (3) and an irrational number is irrational: if \(3-\sqrt{11}\) were rational then \(\sqrt{11}=3-(3-\sqrt{11})\) would be rational, a contradiction. Numerically \(3-\sqrt{11}\approx-0.316\), so it is neither zero nor an integer. Exam tip: check whether the radical term cancels exactly; if it doesn't, the expression remains irrational.

(\sqrt{256}=16), but (257) is not a perfect square. So the first root is rational and the second is irrational.

\(\sqrt{242}=\sqrt{121\times2}=\sqrt{121}\times\sqrt{2}=11\sqrt{2}\). Thus the simplified form is \(11\sqrt{2}\). Option B is a common mistake: \(2\sqrt{11}=\sqrt{4\times11}=\sqrt{44}\), not \(\sqrt{242}\). Option C wrongly splits a sum under a root (\(\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}\)). Option D has an incorrect factor. Exam tip: always factor out the largest perfect square first (here 121).

The surd terms cancel: \((5+\sqrt{7})-(2+\sqrt{7})=5-2+(\sqrt{7}-\sqrt{7})=3\). The result is an integer, hence a rational number (\(3=3/1\)). Option B is incorrect because an irrational number cannot be written as a ratio of integers; here the result is an integer. Option C is incorrect because non‑real numbers have imaginary parts, which are absent. Option D is incorrect because the result is positive. Exam tip: combine like terms and cancel identical surds first — it quickly simplifies such expressions.

Like radical terms (same radicand) are combined by adding their coefficients. Both terms have \(\sqrt{3}\), so add coefficients: \(7+2=9\), giving \(9\sqrt{3}\). Option B is incorrect because it shows an incorrect coefficient (14); options C and D are wrong because they change the radicand (\(\sqrt{6}\) or \(\sqrt{30}\)), which cannot result from adding like radicals. Quick exam tip: combine radicals only when the radicands are identical; otherwise simplify each term first.