Irrational numbers

The relevant concept is the classification of real numbers by their decimal representation and fractional form. π is irrational: it cannot be expressed exactly as p/q for integers p and q with q ≠ 0, and its decimal expansion is non-terminating and non-repeating. Thus option B is correct. The fraction 22/7 is a useful approximation to π, but it is not exactly equal to π, so option C is false. Since every integer is rational, π cannot be an integer either; option D is therefore false. Option A reverses the correct classification. Always distinguish an approximation symbol from exact equality in numerical questions.

A rational number has a terminating decimal when, in lowest terms, its denominator contains only the prime factors 2 and 5; otherwise its decimal is repeating. The numbers 1/4 = 0.25, 0.75 and 2.5 all have terminating decimal expansions, so they are rational. In contrast, 3 is not a perfect square, and √3 is irrational. Its decimal expansion therefore continues indefinitely without a repeating block. Hence option C is correct. The question tests a standard property of irrational numbers: their decimal expansions are non-terminating and non-repeating. Do not confuse a long decimal with an irrational one; the essential feature is the absence of termination and repetition.

The three dots indicate that the digit 3 repeats indefinitely: 0.333… = 0.333333… . Every repeating decimal represents a rational number because it can be converted into a fraction. For example, let x = 0.333…; then 10x = 3.333…, and subtracting gives 9x = 3, so x = 3/9 = 1/3. Therefore option B is correct. It is not irrational because its decimal digits repeat, and it is not an integer or natural number because 1/3 lies between 0 and 1. The important distinction is that non-terminating decimals may be rational when they repeat, or irrational when they do not repeat.

(5+\sqrt{2}) is irrational because adding a rational and an irrational gives an irrational. In such questions watch the irrational part.

(3\sqrt{5}) is irrational because multiplying by a non-zero rational keeps it irrational. Remember the multiplier must not be (0).

Both \(\sqrt{7}\) and \(-\sqrt{7}\) are irrational, but their sum is \(0\), which is rational. Hence the statement is false. Exam tip: one counterexample is enough to disprove an “always” statement.

(10) is not a perfect square so (\sqrt{10}) is irrational. Remembering perfect squares is useful.

\(49\) is a perfect square, and \(\sqrt{49}=7\), which is a rational integer. In contrast, 6, 15, and 30 are not perfect squares, so their square roots are irrational. Exam tip: the square root of a perfect square is rational.

An irrational number cannot be written as a ratio of two integers, and its decimal expansion is non-terminating and non-repeating. To identify the answer, each option must be checked both for its location and for its type. The number \(\sqrt{2}\) is approximately 1.414, so it lies strictly between 1 and 2.

Also, 2 is not a perfect square, so its square root is irrational. In contrast, \(\sqrt{4}=2\) is rational, while \(3/2\) and 1.5 are both equal to 1.5 and are rational. Therefore only option B satisfies both requirements: it lies between 1 and 2 and is irrational. The supplied answer and explanation are correct.

This decimal is non-terminating and non-repeating so it is irrational. Understand the pattern before calling it repeating.

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.