Irrational numbers

The governing concept is solving an inequality over the integers and then counting the resulting set. Since x²<5, we have -√5<x<√5, so the possible integers are -2,-1,0,1,2. Their squares are 4,1,0,1,4, all less than 5. The next integers, -3 and 3, have square 9 and are excluded. Thus V has five elements, so C is correct.

To remove the square root from the denominator, multiply both numerator and denominator by \(\sqrt{5}\): \(\frac{4}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{4\sqrt{5}}{5}\), since \(\sqrt{5}\times\sqrt{5}=5\). Therefore, option A is correct. Option B is the reciprocal of the original fraction, while C and D are not equivalent to it. Exam tip: when a denominator contains a single square root, multiply the numerator and denominator by that same square root.

Since \(7^2=49<53<64=8^2\), it follows that \(7<\sqrt{53}<8\). Therefore, \(\sqrt{53}\) lies between 7 and 8, so option B is correct. Option C is incorrect because \(\sqrt{53}\) is less than 8, as \(8^2=64\). Exam tip: compare the number with the nearest consecutive perfect squares to locate its square root.

A real number is rational if its decimal expansion terminates or eventually repeats. The given expansion continues indefinitely and, as stated, has no fixed repeating block: the runs of digits keep changing. Therefore it cannot be represented as p/q with integers p and q, so it is irrational. Option B is correct; an integer is rational, and a terminating decimal also represents a rational number.

The governing concept is decimal representation of rational and irrational numbers. A rational number, when written in decimal form, either terminates or eventually repeats a fixed block of digits. For example, 0.333... repeats 3 and is rational. The given decimal 0.040040004... is stated to continue without termination and without any fixed repeating pattern. Therefore it cannot be expressed as a ratio of two integers and is irrational. It is still a real number because irrational numbers are included in the real-number system. It is not a whole or natural number, since those are non-negative integers. Hence option C is correct.

\(\sqrt{7}\times\sqrt{14}=\sqrt{98}=\sqrt{49\times2}=7\sqrt{2}\). Since \(\sqrt{2}\) is irrational, multiplying it by the nonzero rational number \(7\) still gives an irrational number. Therefore, the expression is an irrational real number, not a rational number, integer, or terminating decimal. Exam tip: combine the square roots first and extract perfect-square factors.

To rationalise the denominator, multiply both the numerator and denominator by \(\sqrt{7}\): \(\frac{6}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{6\sqrt{7}}{7}\), since \(\sqrt{7}\times\sqrt{7}=7\). Hence, option A is correct. Remember that the numerator must also be multiplied by \(\sqrt{7}\); changing only the denominator to 7 is not valid.

\(\sqrt{3}\times\sqrt{21}=\sqrt{63}=3\sqrt{7}\). Since 7 is not a perfect square, \(\sqrt{7}\) is irrational, and multiplying it by the non-zero rational number 3 remains irrational. Therefore, the given expression is an irrational real number. It is neither an integer, a rational number, nor a terminating decimal. Exam tip: the square root of a positive non-perfect square is irrational.

The governing classification theorem states that a real number is rational if and only if its decimal expansion terminates or eventually repeats. A decimal that continues forever without settling into any repeating block is non-terminating and non-repeating, so it cannot be written as p/q for integers p and q with q ≠ 0. Therefore the given number is irrational; because it is represented by a real decimal, it is an irrational real number. Option B is correct. Option A is ruled out by the absence of repetition, option C is ruled out because digits continue indefinitely, and option D is impossible because an integer has a terminating decimal representation such as 4.000.... The stated condition is decisive even if only an initial pattern is displayed.

The governing concept is simplification of like surds before addition. Rewrite each radical using 2 as the square-free factor: √2 remains √2, √8 = √(4×2) = 2√2, and √18 = √(9×2) = 3√2. These are now like surds, so their coefficients can be added: √2 + 2√2 + 3√2 = (1 + 2 + 3)√2 = 6√2. Thus option A is correct. Option B omits one coefficient, while option C adds incorrectly. Option D, √28, is actually equal to 2√7 and is not equal to the given sum. Radicals should be simplified before combining; unlike surds cannot be added by simply adding their radicands.

The number 3 is rational because it can be written as 3/1, while √2 is irrational; its decimal expansion is non-terminating and non-repeating. A fundamental property is that the sum of a rational number and an irrational number is irrational. To see why, suppose 3+√2 were rational. Subtracting the rational number 3 would then make √2 rational, contradicting the known irrationality of √2. Therefore 3+√2 is irrational, so option A is correct. It cannot be a rational number, integer, or whole number. The fact that the expression contains a radical does not by itself prove the result; the decisive reasoning is the rational-plus-irrational property and the contradiction argument. Thus the number is real, but among the listed choices its most specific correct classification is irrational.

An irrational number cannot be written as \(p/q\), so its decimal expansion is non-terminating and non-repeating. A repeating decimal is rational. Exam tip: look for both features together.

(\sqrt{18}=3\sqrt{2}) and (\sqrt{8}=2\sqrt{2}), so the sum is (5\sqrt{2}). Simplify radicals first.

An irrational number has a decimal expansion that never ends and never repeats a block of digits. Option C describes a rational number. Exam tip: non-terminating, non-repeating decimals are irrational.

Rationalising the denominator gives (\frac{1}{2+\sqrt{3}}=2-\sqrt{3}). Remember to multiply by the conjugate.

Writing a number in decimal form does not make it rational. The blocks of 1s grow as 1, 2, 3, ..., so no fixed repeating block exists. A non-terminating, non-repeating decimal is irrational. Exam tip: check for repetition.

To simplify a radical, separate any perfect-square factor from the number inside the root. Since 50 = 25 × 2 and \(\sqrt{25}=5\), we get \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). The original expression then becomes \(5\sqrt{2}-\sqrt{2}\).

These are like radicals because both terms contain the same \(\sqrt{2}\). Subtract their coefficients: \(5\sqrt{2}-1\sqrt{2}=(5-1)\sqrt{2}=4\sqrt{2}\). No further simplification is possible because 2 has no factor that is a perfect square greater than 1. Therefore option C is correct. The answer is not 5 or \(\sqrt{48}\); those forms do not represent the simplified subtraction as directly.

(\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the total is (6\sqrt{3}). First convert all terms into like radicals.

It has no fixed repeating block, so it is irrational. Avoid treating a non-repeating decimal as repeating.

For the quotient of square roots, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\). Thus, \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}=\sqrt{9}=3\), so option A is correct. The closest distractor, \(\sqrt{3}\), results from failing to simplify the quotient correctly. Exam tip: first divide the radicands when the denominator is nonzero, then simplify the resulting square root.

(\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the sum is (5\sqrt{5}). First take out perfect-square factors.

The governing concept is that the square-root function preserves order for non-negative numbers. Since 2 < 5 < 8, taking square roots gives √2 < √5 < √8, so option C is correct without needing decimal approximations. A numerical check confirms this: √2 is about 1.41, √5 is about 2.24, and √8 is about 2.83. Option A is 1, which is less than √2 because 1² < 2. Options B and D are greater than √8: 3² = 9 > 8 and 4² = 16 > 8. The important reasoning is to compare the radicands 2, 5, and 8, since all are non-negative; their square roots retain the same order.

The side of the square is \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\,\text{m}\). Since \(\sqrt{2}\) is irrational, \(3\sqrt{2}\) is also irrational. Option B has the correct length but the wrong classification. Exam tip: first extract perfect-square factors from a square root.

An irrational number has a decimal expansion that is non-terminating and non-repeating. In 0.101001000100001..., the blocks of zeros grow and no fixed digit pattern repeats indefinitely, so the expansion is non-terminating and non-recurring. Thus option A is irrational. Option B is recurring, option C terminates, and 11/13 is rational because it is a ratio of two integers.

In \(0.1010010001\ldots\), the zeros between consecutive 1s keep increasing, so no fixed block repeats. Hence it is irrational. Exam tip: identify whether an endless decimal has a repeating pattern.