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.

The direct answer is C, \(\sqrt{5}\). Simplify each radical by taking out the largest perfect-square factor: \(\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}\), and \(\sqrt{45}=\sqrt{9\times5}=3\sqrt{5}\). Therefore, \(\sqrt{80}-\sqrt{45}=4\sqrt{5}-3\sqrt{5}=\sqrt{5}\). Option A, \(\sqrt{35}\), is incorrect because subtraction of radicals is not done by subtracting the numbers inside the roots. Option B, \(5\sqrt{5}\), would result from adding 4 and 3, not subtracting them. Option C, \(\sqrt{5}\), correctly subtracts the coefficients of like radicals. Option D, \(4\sqrt{5}\), is only the simplified value of the first radical and ignores the second term. Like algebraic terms can be combined, but unlike radical parts cannot be combined directly.

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.

Direct answer: option D, 0.454545… . A rational number can be written as a fraction of integers. Every repeating decimal is rational because its repeating pattern can be converted into a fraction; here 0.454545… has the repeating block 45. For example, if x = 0.454545…, then 100x = 45.454545…, and subtracting gives 99x = 45, so x = 45/99 = 5/11. Option D is therefore rational. Option A, √29, is irrational because 29 is not a perfect square. Option B, √2 + 1, is irrational: adding the rational number 1 to irrational √2 remains irrational. Option C, π, is irrational by definition and cannot be expressed as a ratio of integers. Option D is the only repeating decimal. Exam cue: terminating and repeating decimals are rational; non-repeating infinite decimals are irrational.

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 direct answer is B, rational number \(1\). Treat the expression as \((\sqrt{2}+1)-\sqrt{2}\\). Rearranging the terms gives \(\sqrt{2}-\sqrt{2}+1=0+1=1\). The irrational terms cancel exactly, so the final result is rational. Option A, irrational \(\sqrt{2}\), wrongly keeps a term that has already cancelled. Option B, rational \(1\), is correct because 1 can be written as \(1/1\), a ratio of integers with a non-zero denominator. Option C, rational 2, is incorrect because no addition produces 2. Option D, irrational \(1+\sqrt{2}\), would be the result before subtracting the second \(\sqrt{2}\), so it ignores cancellation. The important lesson is that an expression containing an irrational number need not have an irrational final value.

Direct answer: option A, 2, is correct. The expression has the conjugate form \((a+b)(a-b)=a^2-b^2\). Take \(a=\sqrt6\) and \(b=2\): \((\sqrt6+2)(\sqrt6-2)=(\sqrt6)^2-2^2=6-4=2\). Option A matches this result. Option B, \(\sqrt6\), is only part of each factor, not the product. Option C, 10, would come from incorrectly adding or mishandling the terms and is not the difference of squares. Option D, \(4\sqrt6\), can arise from an incorrect cross-term calculation; in the correct expansion, the cross terms cancel: \(-2\sqrt6+2\sqrt6=0\). The result is rational even though the original factors contain an irrational number. Memory cue: whenever you see the same two terms with opposite signs, use “plus times minus equals square difference.”

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.

The governing concept is that the square-root function is increasing for non-negative numbers. Because 2 < 3, taking their principal square roots preserves the inequality, giving √2 < √3. This can also be checked by approximation: √2 is about 1.414 and √3 is about 1.732. Therefore option C is correct. The two numbers cannot be equal because equal non-negative square roots would have equal squares, whereas 2 and 3 are different. Option A reverses the valid inequality. Option D is also false because neither 2 nor 3 is a perfect square, so both √2 and √3 are irrational, although the question mainly asks about their order.

An irrational number has a non-terminating, non-repeating decimal expansion, such as \(\sqrt{2}\). Option C is repeating, so it is rational. Exam tip: a repeating decimal always represents a rational number.

The direct answer is option A: a right triangle whose two legs are each 1 unit. To construct \(\sqrt{2}\), first make two perpendicular sides of length 1. The side opposite the right angle is the hypotenuse. By Pythagoras, its square is \(1^2+1^2=1+1=2\), so its length is \(\sqrt{2}\). This length can then be transferred to the number line with a compass. Option A works because it gives exactly this calculation. Option B is an equilateral triangle of side 2; its relevant height or side does not give \(\sqrt{2}\). Option C only says two sides are 2 and does not specify a right angle or produce the required length. Option D is simply a segment of length 3, which is not \(\sqrt{2}\). Memory cue: for a right triangle, remember “sum of squares of the legs gives the square of the hypotenuse.”

The number line represents real numbers, including both rational and irrational numbers. Irrationals can be constructed geometrically.

\(0.333...\) is a recurring decimal and \(0.333...=\frac{1}{3}\), so it is rational. An infinite decimal is not always irrational; in exams, check whether the digits repeat.