Irrational numbers

5 is an integer, so it can be written as \(\frac{5}{1}\) and is therefore rational. In contrast, 11, 17, and 19 are not perfect squares, so \(\sqrt{11}\), \(\sqrt{17}\), and \(\sqrt{19}\) are irrational. Exam tip: The square root of a non-negative integer is rational only when the integer is a perfect square.

A decimal is terminating if its digits eventually stop, and it is repeating rational if a fixed finite block repeats forever. If the digits continue without stopping and no fixed block repeats indefinitely, the number is irrational. The displayed decimal follows the digit string formed by writing successive counting numbers: after the initial digits, it continues with 10, 11, 12 and so on. This pattern does not settle into a repeating cycle.

The decimal therefore has infinitely many digits and is non-terminating. Although it shows patterns in its construction, a visible pattern is not the same as a fixed repeating block. Since no finite block repeats forever, it is non-terminating and non-recurring, hence irrational. Thus option C is correct. It is neither a terminating rational decimal nor a negative integer.

Zeros between successive 1s increase as 1, 2, 3, …, so no block repeats. Non-termination alone is insufficient; a non-repeating decimal is irrational. Exam tip: check for repetition.

The number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Hence the decimal is irrational. Exam tip: only terminating or repeating decimals represent rational numbers.

The number of zeros between successive 1s increases as 1, 2, 3, 4, ..., so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check whether an infinite decimal repeats; infinity alone does not decide the type.

The number 3.14 is a terminating decimal, so it can be written as a fraction with integers: 3.14 = 314/100 = 157/50. Any number expressible as p/q, where p and q are integers and q is nonzero, is rational. Therefore, option B is correct. It is real as well, but it is not irrational and does not have an infinite decimal expansion.

\(0.333\ldots\) is non-terminating but repeating, and \(0.333\ldots=\frac{1}{3}\). Hence, it is rational. In contrast, \(0.1010010001\ldots\) is non-repeating. Exam tip: a decimal is irrational only when it is non-terminating and non-repeating.

\(0.\overline{3}=0.333\ldots=\frac{1}{3}\), so it is rational even though its decimal expansion never ends. \(\sqrt{2}\) and \(\pi\) are irrational. Exam tip: a non-terminating recurring decimal is always rational.

Option A is correct. The number of zeros between successive 1s is 1, 2, 3, 4, …, so no fixed block of digits repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: using only 0 and 1 does not make a number rational.

The number of zeros between successive 1s is 1, 2, 3, 4, ..., so no fixed repeating block occurs. A non-terminating, non-repeating decimal is irrational. Exam tip: only recurring decimals are rational.

Since \(12=4\times3\), \(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\). Therefore, \(\frac{\sqrt{12}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}=2\), because \(\sqrt{3}\neq0\). Hence, option A is correct, and the result is a rational number. Exam tip: When the numerator and denominator contain the same non-zero square-root factor, simplify and cancel that factor.

The zero blocks after 1 have lengths 1, 2, 3, 4, …, so no fixed block repeats. Thus it is non-terminating and non-repeating, making it irrational. Exam tip: check for a repeating block.

(\sqrt{2}) is irrational so (1+\sqrt{2}) is also irrational. Adding a rational number does not remove the irrational part.

In option C, the number of zeros between successive 1s keeps increasing, so there is no fixed repeating block of digits. It is non-terminating and non-repeating, hence it represents an irrational number. Options A and D repeat 6 and 7 respectively, so they are recurring decimals, while option B is terminating. Exam tip: A non-terminating decimal with no fixed repeating pattern is non-repeating.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), a rational number. Hence, the word “always” makes the statement false. Exam tip: test such claims using one counterexample.

A square root is rational when the number under the root is a perfect square, because its square root is an integer or another rational number. Since 64 = 8^2, \(\sqrt{64}=8\), which is rational and also an integer.

The number 65 is not a perfect square: it lies between 64 = 8^2 and 81 = 9^2. Therefore \(\sqrt{65}\) lies between 8 and 9 but is not an integer. The square root of a natural number that is not a perfect square is irrational, meaning it cannot be expressed as a ratio of integers and its decimal does not terminate or repeat. Hence option B, \(\sqrt{65}\), is the irrational number.

The governing definition is that an irrational number cannot be written as p/q, where p and q are integers and q is not zero. Since 23 is not a perfect square, √23 cannot be expressed as a ratio of integers. Its decimal expansion is non-terminating and non-repeating, so option D is irrational. The other choices are rational: 6/11 is already a quotient of integers, 0.875 terminates and equals 875/1000, and −3 is the rational number −3/1. The square-root symbol alone does not guarantee irrationality, because √25 = 5 is rational. Thus the deciding property is whether the number has a valid integer ratio, and only √23 fails that test.

The governing concept is the classification of numbers and the effect of multiplying a rational number by an irrational number. The factor 4 is a non-zero rational integer, whereas √3 is irrational because 3 is not a perfect square and √3 cannot be expressed as a ratio of integers. Thus 4√3 remains irrational. A direct contradiction also proves this: if 4√3 were rational, dividing it by the non-zero rational number 4 would make √3 rational, which is impossible. Therefore the part responsible for irrationality is √3, so option B is correct. The multiplication sign is merely an operation symbol, not a number, and the number 4 alone is rational. Option A therefore does not identify the source of irrationality.

Since \(\sqrt{81}=9\), the given number becomes \(9+\sqrt{2}\). Here, \(9\) is rational and \(\sqrt{2}\) is irrational. The sum of a rational number and an irrational number is always irrational, so the correct answer is irrational number. Therefore, it is neither an integer nor zero. Exam tip: simplify perfect-square roots first and remember that rational + irrational = irrational.

The governing concept is simplification of radicals by extracting perfect-square factors and then combining like radical terms. Since 27 = 9 × 3, √27 = √9 × √3 = 3√3. Similarly, 12 = 4 × 3, so √12 = √4 × √3 = 2√3. Therefore √27 − √12 = 3√3 − 2√3 = (3 − 2)√3 = √3, making option B correct. The subtraction is performed on the coefficients because both simplified terms contain the same radical √3. It is not valid to subtract the radicands directly to obtain √15. Option C leaves the original first term unchanged, and option D uses an incorrect coefficient. Since 3 is not a perfect square, √3 remains in the final simplified result.