Irrational numbers

The block 12 repeats continuously, so 0.121212... is a recurring decimal. Every recurring decimal can be written in the form \(\frac{p}{q}\); in fact, \(0.121212... = \frac{12}{99} = \frac{4}{33}\). Therefore, it is a rational number. Irrational numbers have non-terminating, non-recurring decimal expansions. Exam tip: terminating or recurring decimals are always rational.

The governing concept is the classification of real numbers. Since 13 is not a perfect square, √13 cannot be expressed as p/q, where p and q are integers and q is non-zero; therefore √13 is irrational. Multiplying an irrational number by −1 only changes its sign, not its irrational nature. Hence −√13 is also irrational, so option B is correct. It is not zero, because √13 is non-zero, and it is not a positive integer because it is negative and does not have an integral value. A negative sign does not make a number rational.

\(0.333\ldots=\frac{1}{3}\) is an infinite repeating decimal, so it is rational and disproves Reema’s claim. In contrast, \(\sqrt{2}\) is non-terminating and non-repeating. Exam tip: repeating decimals are rational.

The governing concept is the additive inverse property. The numbers √2 and −√2 are opposites, so their sum is zero: √2 + (−√2) = √2 − √2 = 0. Therefore option C is correct. This also illustrates that the sum of two irrational numbers need not be irrational; here, the irrational parts cancel exactly and produce the rational number zero. Options A and B do not represent the result of adding the two given terms; they introduce unrelated forms involving division or an unperformed expression. Option D is also incorrect because no operation in the original expression produces 2. Exact cancellation, rather than decimal approximation, gives the answer.

\(\sqrt{2}\times\sqrt{2}=(\sqrt{2})^2=2\), and 2 is an integer as well as a rational number. Therefore, option B is correct. Remember that the product of two irrational numbers is not always irrational; in this example, it is rational. Exam tip: use \(\sqrt{a}\times\sqrt{a}=a\) when \(a\) is positive.

(\sqrt{2}\times\sqrt{3}=\sqrt{6}) and (\sqrt{6}) is irrational. In multiplication multiply the numbers inside roots.

Adding irrational (\sqrt{5}) to rational (2) gives an irrational number. Identify the irrational part separately.

(\sqrt{7}) is irrational so (9-\sqrt{7}) is also irrational. Adding or subtracting a rational keeps irrationality.

(\sqrt{\frac{4}{9}}=\frac{2}{3}) so it is rational. Take square roots of numerator and denominator.

(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}) which is irrational. Be careful when the numerator is not a perfect square.

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.