Irrational numbers and real numbers

The direct answer is option A: \\(\sqrt{43}\\) is irrational. The nearby perfect squares are \\(6^2=36\\) and \\(7^2=49\\), so 43 is not a perfect square. Therefore its square root is not an integer and cannot be expressed as a ratio of two integers; it is irrational. Option A is correct. Option B is wrong because 43 is not the square of an integer. Option C is wrong because \\(7^2=49\\), not 43, so \\(\sqrt{43}\\neq7\\). Option D is wrong because an irrational number has a non-terminating, non-repeating decimal expansion, not a terminating one. Also, \\(\sqrt{43}\\) is real because 43 is positive. Memory cue: check the nearest squares before deciding the type of a square root.

A square root of a positive integer is irrational when that integer is not a perfect square. Since 13 lies between the perfect squares 9 and 16, it cannot be written as the square of an integer or as a terminating fraction. Hence \(\sqrt{13}\) is irrational. It is nevertheless a real number because 13 is positive.

Putting a minus sign before an irrational real number changes its sign, not its number system or rationality. Thus \(-\sqrt{13}\) is still real and irrational. It is not rational, whole, or natural: whole and natural numbers are integers in the usual school classification, while this value is not an integer. Therefore, option A, irrational real number, is correct.

√72 = √(36×2) = √36 × √2 = 6√2. The largest perfect square factor of 72 is 36, so the simplified form is 6√2. Option B (−6√2) only flips the sign — the principal square root is non-negative. Options C (6√3) and D (12√2) are incorrect because squaring them gives 108 and 288 respectively, not 72. Exam tip: always factor the radicand into the largest perfect square times the remainder (4, 9, 16, 25, 36,...), then take the square root of that perfect square out front.

\(\sqrt{108}=\sqrt{36\times3}=\sqrt{36}\cdot\sqrt{3}=6\sqrt{3}\). Hence the correct simplified form is \(6\sqrt{3}\). Distractors like \(3\sqrt{6}\) or \(9\sqrt{2}\) give different numerical values — e.g. \(3\sqrt{6}\) would require \(\sqrt{6}=2\sqrt{3}\) to equal the correct answer, which is false. Exam tip: factor the number into the largest perfect square times a remainder, then take the square root of the perfect square outside the radical.

\(\sqrt{63}=\sqrt{9\times7}=3\sqrt{7}\) and \(\sqrt{28}=\sqrt{4\times7}=2\sqrt{7}\). These are like surd terms, so add coefficients: \(3\sqrt{7}+2\sqrt{7}=5\sqrt{7}\). Option C (\(\sqrt{91}\)) reflects the common mistake of adding under the radical (\(\sqrt{63+28}\)), which is invalid; you must simplify each radical first. Option B (\(11\sqrt{7}\)) is also incorrect — it results from an incorrect combination of coefficients. Exam tip: always factor each radicand to extract common square factors and then add only like surd terms (same \(\sqrt{\,}\) part).

Simplify each radical first: \(\sqrt{98}=\sqrt{49\cdot2}=7\sqrt{2}\) and \(\sqrt{50}=\sqrt{25\cdot2}=5\sqrt{2}\). Subtracting like surds gives \(7\sqrt{2}-5\sqrt{2}=2\sqrt{2}\). Option B (2) is incorrect because the factor \(\sqrt{2}\) must remain; option D (\(12\sqrt{2}\)) would result from wrongly adding the coefficients instead of subtracting. Exam tip: always factor out perfect squares from under the radical to combine surds correctly.

The direct answer is option A: 15. Use the rule \\(\sqrt a\,\sqrt b=\sqrt{ab}\\) for nonnegative numbers. Thus \\(\sqrt5\times\sqrt{45}=\sqrt{225}=15\\), because \\(15^2=225\\). Alternatively, \\(45=9\times5\\), so \\(\sqrt{45}=3\sqrt5\\), and the product is \\(3\sqrt5\times\sqrt5=3\times5=15\\). Option A is correct. Option B, \\(5\sqrt{45}\\), is not the value; it multiplies the second factor by 5 unnecessarily. Option C, \\(\sqrt{50}\\), would represent \\(\sqrt5\sqrt{10}\\), not the given product. Option D, \\(45\sqrt5\\), is also an unjustified multiplication and is much too large. Notice that two irrational numbers can have a rational product: here both roots combine to the perfect square 225. Memory cue: combine roots first, then check whether the product under the root is a perfect square.

9 is rational and \(\sqrt{17}\) is irrational. In general the sum of a rational and an irrational number is irrational because the irrational part cannot be expressed or cancelled by a rational number; hence \(9+\sqrt{17}\) is irrational. A common distractor is \(\sqrt{2}+(3-\sqrt{2})\) where the irrational parts cancel and the sum equals 3, which is rational. Exam tip: always simplify the expression first and check whether any irrational parts cancel out.

Since (25<30<36), (\sqrt{30}) lies between (5) and (6). (30) is not a perfect square so it is irrational.

\(\frac{7}{5}=1.4\) lies between 1 and 2 and is expressible as a ratio of integers, so it is rational. The other choices \(\sqrt{2},\,\sqrt{3}\) and \(\pi/2\) are irrational (they have non‑terminating, non‑repeating decimal expansions and cannot be written as a ratio of integers). Exam tip: to check rationality, try to express the number as a fraction of two integers; non‑perfect square roots and multiples of \(\pi\) are typically irrational.

In \(4.\overline{56}\) the block "56" repeats indefinitely, so it is a non-terminating repeating decimal (hence rational). Option B shows digits continuing without a fixed repeating block, so it is non-repeating; option C is a terminating decimal; option D equals \(\sqrt{4}=2\), an integer (terminating). Exam tip: look for a fixed group of digits that repeats periodically — that identifies a repeating decimal.

The decimal expansion of 7.03125 has a finite number of digits: it ends after five decimal places. Hence, it is a terminating decimal. It can be written as \(\frac{703125}{100000}\), where \(100000=10^5\). In option B, the block “03” repeats, so it is non-terminating recurring. Option C appears non-terminating and non-recurring, and \(\pi\) is also non-terminating and non-recurring. Exam tip: a terminating decimal always has a fixed, finite number of digits after the decimal point.

Since \((\sqrt{19})^2=19\), squaring removes the square root and yields the radicand. Option B is incorrect because it is the original \(\sqrt{19}\), option C is \(2\times19\) (wrong), and option D equals \(19^2\) (also wrong). Exam tip: For any nonnegative \(a\), \((\sqrt{a})^2=a\).

(\frac{5}{\sqrt{5}}=\sqrt{5}) because (5=\sqrt{5}\times\sqrt{5}). Learn to simplify denominators with roots.

Use the rule \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) for nonnegative \(a,b\). So \(\sqrt{\frac{49}{64}}=\frac{\sqrt{49}}{\sqrt{64}}=\frac{7}{8}\). The principal square root is nonnegative, so \(-\frac{7}{8}\) is not valid. The other distractors come from incorrect manipulation (for example taking root only of numerator or misplaced division). Exam tip: simplify perfect squares in numerator and denominator before taking the square root, and remember the principal root is positive.

\(\sqrt[3]{27}=3\) since \(3^3=27\). The result is an integer, and every integer is rational (can be written as a fraction, e.g. \(3=3/1\)). Option B is incorrect because irrational numbers cannot be expressed as a ratio of integers and have non-repeating, non-terminating decimals — that does not apply here. Option C is incorrect because the value is a real integer, not a non-real complex number. Option D is incorrect because the number is not a non-terminating non-repeating decimal but a terminating integer. Exam tip: first check if the radicand is a perfect power; the root of a perfect cube is an integer and hence rational.

The difference of a rational and an irrational number is irrational. This property helps identify mixed expressions.

The governing property is that the product of a non-zero rational number and an irrational number is irrational. Suppose, for contradiction, that rs were rational. Because r is non-zero and rational, its reciprocal 1/r is also rational. Multiplying the assumed rational number rs by 1/r would give s = (rs)(1/r), which would be rational. This contradicts the given fact that s is irrational. Therefore rs must be irrational. The condition r ≠ 0 is essential: if r were zero, the product would be 0, which is rational. It is not necessarily natural either, because the product may be negative or non-integral. Hence option A is correct.

\(\frac{4}{9}\) is a rational fraction and satisfies \(0<\frac{4}{9}<1\), so it lies between 0 and 1. The other choices evaluate to 1 (\(\frac{\sqrt{2}}{\sqrt{2}}=1\), \(\frac{\sqrt{5}}{\sqrt{5}}=1\), \(1.5/1.5=1\)), so they are not between 0 and 1. Exam tip: simplify expressions or convert to decimal form to quickly check whether a number lies between 0 and 1.

-3 is a real number but not a whole number because whole numbers are 0, 1, 2, ... (non‑negative integers). Options B (0) and C (1) are whole numbers. Option D (2/0) is undefined (division by zero) and therefore not a real number. Exam tip: first check if an expression is defined; then check whether it belongs to the whole-number set (non‑negative integers).

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.