Irrational numbers

A square root is rational when the number under the root is a perfect square, because its square root is an integer or another rational number. Since 64 = 8^2, \(\sqrt{64}=8\), which is rational and also an integer.

The number 65 is not a perfect square: it lies between 64 = 8^2 and 81 = 9^2. Therefore \(\sqrt{65}\) lies between 8 and 9 but is not an integer. The square root of a natural number that is not a perfect square is irrational, meaning it cannot be expressed as a ratio of integers and its decimal does not terminate or repeat. Hence option B, \(\sqrt{65}\), is the irrational number.

The governing definition is that an irrational number cannot be written as p/q, where p and q are integers and q is not zero. Since 23 is not a perfect square, √23 cannot be expressed as a ratio of integers. Its decimal expansion is non-terminating and non-repeating, so option D is irrational. The other choices are rational: 6/11 is already a quotient of integers, 0.875 terminates and equals 875/1000, and −3 is the rational number −3/1. The square-root symbol alone does not guarantee irrationality, because √25 = 5 is rational. Thus the deciding property is whether the number has a valid integer ratio, and only √23 fails that test.

The governing concept is the classification of numbers and the effect of multiplying a rational number by an irrational number. The factor 4 is a non-zero rational integer, whereas √3 is irrational because 3 is not a perfect square and √3 cannot be expressed as a ratio of integers. Thus 4√3 remains irrational. A direct contradiction also proves this: if 4√3 were rational, dividing it by the non-zero rational number 4 would make √3 rational, which is impossible. Therefore the part responsible for irrationality is √3, so option B is correct. The multiplication sign is merely an operation symbol, not a number, and the number 4 alone is rational. Option A therefore does not identify the source of irrationality.

Since \(\sqrt{81}=9\), the given number becomes \(9+\sqrt{2}\). Here, \(9\) is rational and \(\sqrt{2}\) is irrational. The sum of a rational number and an irrational number is always irrational, so the correct answer is irrational number. Therefore, it is neither an integer nor zero. Exam tip: simplify perfect-square roots first and remember that rational + irrational = irrational.

The governing concept is simplification of radicals by extracting perfect-square factors and then combining like radical terms. Since 27 = 9 × 3, √27 = √9 × √3 = 3√3. Similarly, 12 = 4 × 3, so √12 = √4 × √3 = 2√3. Therefore √27 − √12 = 3√3 − 2√3 = (3 − 2)√3 = √3, making option B correct. The subtraction is performed on the coefficients because both simplified terms contain the same radical √3. It is not valid to subtract the radicands directly to obtain √15. Option C leaves the original first term unchanged, and option D uses an incorrect coefficient. Since 3 is not a perfect square, √3 remains in the final simplified result.

The governing concept is the definition of an irrational number. An irrational number cannot be expressed as p/q, where p and q are integers and q is non-zero. Therefore, option D is correct. In decimal form, an irrational number has an expansion that is non-terminating and non-repeating; hence option A is false because terminating decimals are rational, and option B is false because repeating decimals are also rational. Option C gives the defining property of rational numbers, not irrational numbers. For example, √2 is irrational because no ratio of integers represents it, while 3/8 is rational because it is already in p/q form.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence, the word “always” makes the statement false. In exams, test such claims using a counterexample.

The zero gaps grow as 1, 2, 3, ..., so no fixed block repeats. Hence the number is irrational; a decimal is rational only if it terminates or repeats. Exam tip: check for a recurring block.

\(0.\overline{27}\) is a recurring decimal, so it is rational. Let \(x=0.\overline{27}\); then \(100x-x=27\), giving \(x=27/99=3/11\). Exam tip: non-terminating recurring decimals are rational, not irrational.

Since \(98=49\times2\) and \(49\) is a perfect square, \(\sqrt{98}=\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}=7\sqrt{2}\). Therefore, option A is correct. Option B, \(2\sqrt{7}\), is not equivalent to \(\sqrt{98}\), since its square is \(28\). Exam tip: To simplify a square root, factor out the greatest perfect-square factor.

Direct answer: option A, √6. Both terms contain the same radical √6, so they are like terms, just as 2x − x = x. Factor √6: 2√6 − √6 = (2 − 1)√6 = 1√6 = √6. Option A is correct. Option B, 2, would result only if the radicals somehow cancelled completely, but the common factor √6 remains. Option C, 6, incorrectly removes the square-root sign. Option D, 3√6, would be the result of adding √6 to 2√6 rather than subtracting it. The fact that √6 is irrational does not change the ordinary rule for subtracting like terms. Since 6 is not a perfect square, √6 itself cannot be simplified further. Memory cue: combine the coefficients of identical radicals and keep the radical.

If (a) is not a perfect square then (\sqrt{a}) can be irrational. In exams first identify perfect squares.

Statement I is false because \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. Statement II is true: if \(q\times x\) were rational, then \(x=(q\times x)/q\) would be rational, a contradiction. Exam tip: test “always” statements with a counterexample.

The perimeter is (4\sqrt{7}) and (\sqrt{7}) is irrational. A non-zero rational multiplier keeps irrationality.

The area of a square is given by side², so its side is \(\sqrt{11}\) units. Since 11 is not a perfect square, \(\sqrt{11}\) cannot be expressed as a ratio of two integers and is therefore irrational. Hence, option D is correct. Exam tip: The square root of a positive integer that is not a perfect square is irrational.

An irrational number has a non-terminating and non-repeating decimal. In exams check the repeating part carefully.

It has no fixed repetition so it is irrational. Do not mistake a non-repeating pattern for a repeating one.

Here, the number of zeros between successive 1s increases as 1, 2, 3, …, so no fixed repeating block is possible. Hence it is irrational. Exam tip: a non-terminating, non-repeating decimal is irrational.

\(64\) is a perfect square, and \(\sqrt{64}=8\), which is a rational number. Therefore, \(\sqrt{64}\) is not irrational. In contrast, 31, 40, and 72 are not perfect squares, so their square roots are irrational. Exam tip: The square root of a perfect square is always rational.

The governing concept is the closure property of rational multiplication with an irrational number. Substituting x = √6 gives 2x = 2√6. The number √6 is irrational because 6 is not a perfect square, so its decimal expansion is non-terminating and non-repeating. Multiplying an irrational number by a non-zero rational number such as 2 cannot make it rational: if 2√6 were rational, dividing it by 2 would make √6 rational, which is a contradiction. Therefore option C is correct. It is not an integer, rational number, or zero; the coefficient 2 only changes its magnitude.

Since \(0.333...=1/3\), it is non-terminating yet repeating, so it is rational. In contrast, \(\sqrt{2}\) and \(\pi\) are non-repeating. Exam tip: recurring decimals are rational.

Since (10<10.5<11), (\sqrt{10.5}) lies between them. Compare square roots using the numbers inside.

\(0.272727\ldots=27/99=3/11\), so it is rational despite being non-terminating. Exam tip: every non-terminating recurring decimal is rational.

49 is a perfect square because \(7 \times 7=49\). Hence, \(\sqrt{49}=7\), an integer and therefore rational; not every square root is irrational. Exam tip: check for perfect squares first.

The statement is true because an irrational number has a decimal expansion that neither terminates nor repeats in a fixed pattern. Exam tip: every terminating or recurring decimal represents a rational number.