Irrational numbers

This decimal never terminates, and the number of zeros increases as 1, 2, 3, 4, …, so no fixed digit block repeats. Hence it is irrational. Exam tip: a non-terminating, non-repeating decimal is irrational.

The governing concept is the distinction between rational and irrational terms in a sum. The number 4 is rational because it can be written as 4/1. The number √6 is irrational because 6 is not a perfect square, so its square root cannot be expressed as a ratio of integers. In the expression 4 + √6, the radical term is therefore the irrational part, making option D correct. Option A is the rational part, not the irrational part. Option B is only the radicand and is itself rational. Option C incorrectly multiplies the two terms; no such multiplication appears in the given expression. The answer must be identified from the terms actually present.

(\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so (2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}). Add and subtract like radicals.

In \(0.272727\ldots\), the block 27 repeats, so it is a recurring decimal and therefore rational. \(\sqrt{2}\) and \(\sqrt{5}\) are irrational. Exam tip: every recurring decimal represents a rational number.

The square root of a positive integer is irrational when it is not a perfect square. This is the main rule for identifying square roots.

The numbers of zeros between successive 1s are 1, 2, 3, ... , so no fixed digit block repeats. Hence the decimal is non-terminating and non-repeating, making it irrational. Exam tip: check repetition, not the digits used.

(\sqrt{2}) and (\sqrt{3}) are different irrational radicals and their sum is irrational. It is not written as (\sqrt{5}).

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: always check for a repeating block.

A terminating decimal ends after a finite number of digits, such as 0.25. A repeating decimal has a fixed block of digits that repeats forever, and it represents a rational number. A non-repeating infinite decimal does not repeat a fixed pattern and is generally irrational. These distinctions identify the correct option.

In option A, the block 123 repeats indefinitely: 0.123123123... Therefore the decimal is non-terminating and repeating. Option B does not show a fixed repeating block, option C is the non-terminating non-repeating decimal of \(\sqrt{2}\), and option D terminates. Hence option A is correct. The note that a repeating decimal is rational is also important: repeating does not mean irrational.

\(0.\overline{3}=\frac{1}{3}\), so it is non-terminating but repeating and rational. Irrational decimals are non-terminating and non-repeating. Exam tip: identify a recurring block to spot a rational number.

Use the distributive property: \(\sqrt{3}\times\sqrt{3}=3\) and \(\sqrt{3}\times\sqrt{12}=\sqrt{36}=6\). Hence, the expression equals \(3+6=9\). Option 6 is only the value of the second product and omits the first term, 3. Exam tip: when multiplying square roots, simplify using \(\sqrt{a}\sqrt{b}=\sqrt{ab}\).

C is correct. Zero blocks between 1s have lengths 1, 2, 3, 4, ..., so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not digits.

Using the distributive law, \(\sqrt{2}(3\sqrt{2}+1)=3\sqrt{2}\times\sqrt{2}+\sqrt{2}\). Since \(\sqrt{2}\times\sqrt{2}=2\), the simplified form is \(3\times2+\sqrt{2}=6+\sqrt{2}\). Option C incorrectly doubles the term containing \(\sqrt{2}\). Exam tip: use \(\sqrt{a}\times\sqrt{a}=a\) when multiplying identical square roots.

The direct answer is C, \(3\sqrt{2}-3\). To rationalise the denominator, multiply numerator and denominator by the conjugate of \(\sqrt{2}+1\), namely \(\sqrt{2}-1\): \(\frac{3}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{3(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\). The denominator is \(2-1=1\), so the result is \(3(\sqrt{2}-1)=3\sqrt{2}-3\). Option A, \(3(\sqrt{2}+1)\), uses the same sign and does not remove the surd from the denominator. Option B, \(3\sqrt{2}+3\), has the wrong sign. Option C is exactly the correct numerator after the denominator becomes 1. Option D still leaves a radical in the denominator and is not rationalised. Remember: for \(a+b\sqrt{x}\), multiply by its conjugate with the opposite sign.

Adding the two expressions gives \((7+\sqrt{13})+(7-\sqrt{13})=7+7+\sqrt{13}-\sqrt{13}=14\). The irrational terms \(\sqrt{13}\) and \(-\sqrt{13}\) cancel each other, so the result is the rational number 14. Exam tip: when adding conjugate forms \((a+b)\) and \((a-b)\), combine the matching terms first.

Option A is correct. Zeros between successive 1s are 1, 2, 3, 4, ...; no fixed block repeats. Thus this non-terminating decimal is irrational. Exam tip: check repetition, not digit types.

The direct answer is D: irrational. We have the number p = 4 + \(\sqrt{5}\). The number 4 is rational because it can be written as \(4/1\). The number \(\sqrt{5}\) is irrational because 5 is not a perfect square, so its decimal expansion does not terminate or repeat in a fixed pattern. A rational number plus an irrational number is irrational: if 4 + \(\sqrt{5}\) were rational, subtracting rational 4 would make \(\sqrt{5}\) rational, which is impossible. Option A, integer, is wrong because the expression is not a whole number. Option B, rational, is wrong for the reason just shown. Option C, non-terminating repeating, is wrong because irrational decimals are non-terminating and non-repeating, not repeating. Option D is correct because the irrational square-root part remains after adding 4. Memory cue: rational plus irrational is irrational, provided the irrational part is not cancelled.

(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4), which is rational. The product of two irrationals can be rational.

The side of a square is the square root of its area. Thus, side \(=\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\) units, so option C is correct. The distractor \(2\sqrt{3}\) is incorrect because its square is \(12\), not \(18\). Exam tip: factor the number under the square root into a perfect square and the remaining factor before simplifying.

A rational number has a terminating or recurring decimal expansion. Here, the zeros between successive 1s are 1, 2, 3, 4, …, so no repeating block occurs. In exams, check recurrence, not merely the digits used.

The direct answer is C, both are equal. Simplify each radical carefully. First, \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\), so \(\sqrt{3}+\sqrt{27}=\sqrt{3}+3\sqrt{3}=4\sqrt{3}\). Next, \(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\). Therefore both expressions have exactly the same value. Option A, “the first is greater,” is wrong because both simplify to the same expression. Option B, “the second is greater,” is also wrong for the same reason. Option C is correct because both equal \(4\sqrt{3}\). Option D, “both are rational,” is wrong: \(3\) is not a perfect square, so \(\sqrt{3}\) is irrational, and a non-zero rational multiple such as \(4\sqrt{3}\) remains irrational. The useful method is to take perfect-square factors out of each radical and then compare like radical terms. Memory cue: simplify first, compare later; \(\sqrt{27}=3\sqrt3\) and \(\sqrt{48}=4\sqrt3\).

(\sqrt{20}=2\sqrt{5}), so (2\sqrt{5}+6\sqrt{5}=8\sqrt{5}). Watch both coefficients and radicals carefully.

Here, a 1 appears after 1, 2, 3, 4, … zeros successively, so no fixed block of digits repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: digits 0 and 1 alone do not make a number rational.

\(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\). Therefore, \(\sqrt{2}+\sqrt{32}=\sqrt{2}+4\sqrt{2}=5\sqrt{2}\). Option D is incorrect because the sum of separate square roots cannot generally be written as \(\sqrt{34}\). Exam tip: simplify each radical first, then combine terms with the same radical part.

In \(0.333\ldots\), the digit 3 repeats, so \(0.333\ldots=\frac{1}{3}\), which is rational. Hence, an infinite decimal need not be irrational. Exam tip: check whether the decimal digits repeat.