Irrational numbers

The groups of zeros after 1 have lengths 1, 2, 3, 4, …, so no fixed block repeats. The decimal is non-terminating and non-repeating, hence irrational. In exams, check whether a fixed repeating block exists.

Since \(2^2=4<5<9=3^2\), we get \(2<\sqrt{5}<3\). As 5 is not a perfect square, \(\sqrt{5}\) is irrational. The value 2.5 is only an approximation. Exam tip: compare with nearby perfect squares first.

The governing ideas are the comparison of square roots and the distinction between rational and irrational numbers. Since 1² = 1, 3² = 9, and 1 < 5 < 9, taking positive square roots gives 1 < √5 < 3. Also, 5 is not a perfect square, so √5 is irrational; it cannot be expressed as a ratio of integers and its decimal expansion is non-terminating and non-repeating. Thus option B satisfies both required conditions. The other choices simplify to rational numbers: √9/√9 = 3/3 = 1, 2/2 = 1, and 2.5/2.5 = 1. Each is rational and lies at the lower boundary rather than strictly between 1 and 3. Therefore √5 is the unique correct answer.

Subtracting irrational (\sqrt{2}) from rational (6) gives an irrational number. The irrational part does not disappear.

Both terms contain the same radical, so their coefficients are added: \(\sqrt{7}+\sqrt{7}=(1+1)\sqrt{7}=2\sqrt{7}\). \(\sqrt{14}\) is incorrect because radicands are not added directly when like radicals are combined. In an exam, treat radicals with the same radicand like like algebraic terms.

(\sqrt{12}=2\sqrt{3}) and (\sqrt{75}=5\sqrt{3}), so the sum is (7\sqrt{3}). Simplify both radicals first.

(\sqrt{80}=4\sqrt{5}) and (\sqrt{45}=3\sqrt{5}), so the difference is (\sqrt{5}). Subtract like radicals.

In 0.101001000100001…, the number of zeros between successive 1s keeps increasing, so no decimal block repeats. Its expansion is non-terminating and non-repeating, hence irrational. Exam tip: check for a repeating block, not merely the digits used.

Multiplying (\frac{1}{\sqrt{3}}) by (\frac{\sqrt{3}}{\sqrt{3}}) gives (\frac{\sqrt{3}}{3}). This is called rationalising the denominator.

(0.454545\ldots) is repeating so it is rational. A repeating decimal can be converted into a fraction.

The governing concept is that adding a rational number to an irrational number always gives an irrational number. Here π is irrational, while 2 is rational because it can be written as 2/1. Suppose π + 2 were rational. Subtracting the rational number 2 from it would then make π rational, contradicting the known property of π. Therefore π + 2 is irrational, so option B is correct. It is not an integer or a terminating decimal, because every integer and every terminating decimal is rational. The fraction 22/7 is only a rational approximation to π, not its exact value, so replacing π by 22/7 would incorrectly change the classification.

A rational decimal either terminates or eventually repeats a fixed finite block of digits. An irrational decimal does neither: it continues indefinitely without settling into a permanent repeating cycle. The displayed digits are arranged with growing groups of ones between zeros, as in 1, 01, 011, 0111, and so on. The gaps and blocks keep changing, so no fixed block can describe the tail forever.

The decimal does not end, because more digits continue after those shown. It also is not eventually periodic: the lengths of the runs change rather than repeating with one fixed period. Therefore the number is non-terminating and non-recurring, which makes it irrational. Option B is correct. It cannot be an integer or a terminating decimal, and merely seeing a pattern does not make a decimal rational unless that pattern eventually repeats exactly.

(\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}) and (\sqrt{7}) is irrational. Check perfect squares even in fractions.

The number of zeros between successive 1s is 1, 2, 3, 4, ... , so no fixed block of digits repeats. Its decimal expansion is non-terminating and non-recurring; therefore, it is irrational. Exam tip: check repetition in the decimal, not merely the digits used.

Since 11 is not a perfect square, \(\sqrt{11}\) is irrational. Multiplying an irrational number by the non-zero rational number \(-3\) gives an irrational result. Therefore, \(-3\sqrt{11}\) is irrational. It cannot be a natural number or zero because it is negative. Exam tip: The square root of a non-perfect square is irrational.

An irrational number has a decimal expansion that never ends and does not repeat a fixed block of digits. A non-terminating repeating decimal is rational. Exam tip: a repeating pattern indicates a rational number.

The governing concept is the reciprocal of a non-zero number. For any non-zero number x, its reciprocal is 1/x, because multiplying the two gives x × (1/x) = 1. Since √13 is positive and therefore non-zero, its reciprocal is 1/√13, so option B is correct. The answer can also be written in rationalised form as √13/13: multiply numerator and denominator of 1/√13 by √13 to obtain √13/(√13)² = √13/13. Both forms have exactly the same value. Option D is the original number, option C is its square, and option A is its negative. In fact, √13 × (−√13) = −13, not 1, so option A cannot be a reciprocal.

The (\sqrt{2}) terms cancel and the value is (1). Sometimes irrational terms can cancel to give a rational result.

This is ((a+b)(a-b)=a^2-b^2), so the value is (6-4=2). Conjugates can make the result rational.

The zero blocks between 1s have lengths 1, 2, 3, 4, …, so no fixed period exists. A non-terminating, non-repeating decimal is irrational. Exam tip: check whether a fixed block repeats.