Irrational numbers

A rational number can be written as a fraction of two integers with a non-zero denominator. Option A gives 5/2 = 2.5, and 2 < 2.5 < 3, so it satisfies both conditions. Hence option A is correct. The number √5 lies between 2 and 3 because 4 < 5 < 9, but it is irrational, not rational. Similarly, √6 is between 2 and 3, since 4 < 6 < 9, yet it is irrational because 6 is not a perfect square. The value π is approximately 3.14, so it is not even between 2 and 3, although π itself is irrational. The question requires checking location and number type together; being between the endpoints alone is not sufficient. The fraction form of 5/2 proves rationality directly.

Every number that can be placed on the real number line is real. An irrational number is real but cannot be expressed as p/q, where p and q are integers and q is nonzero. Since 18 = 9 × 2, we have √18 = √9 × √2 = 3√2. Because √2 is irrational and multiplying it by the nonzero rational number 3 does not make it rational, √18 is real and irrational. Thus option A is correct. Option B equals 1, option C equals 1, and option D equals √100 ÷ √100 = 10 ÷ 10 = 1. The last three values are rational integers, so none of them is irrational.

For non-negative real numbers, √a × √b = √(ab). Hence √7 × √11 = √77. The prime factorisation of 77 is 7 × 11, and neither factor is repeated, so 77 is not a perfect square. Therefore √77 cannot be expressed as a ratio of two integers and is irrational. It is certainly real because the radicand 77 is positive, but being real does not make it rational. Options B, C, and D are not suitable: an integer and a terminating decimal are rational, and this particular square root is not rational. Thus option A is correct.

The cube root of an integer is rational only when the number under the cube-root sign is a perfect cube of a rational number. The integer 9 is not a perfect cube: the nearby perfect cubes are 1³ = 1, 2³ = 8, and 3³ = 27. Therefore ∛9 cannot simplify to an integer. More formally, if ∛9 were rational, its reduced numerator and denominator would imply that 9 is a cube of an integer, which is impossible because its prime factorisation is 3² and the exponent 2 is not a multiple of 3. Hence ∛9 is irrational, so option A is correct. It is not 3 because 3³ = 27, and an irrational number cannot be rational or an integer.

Answer: A, 9. For non-negative numbers, the product rule for square roots is √a × √b = √(ab). Therefore √3 × √27 = √(3×27) = √81 = 9. We can also simplify first: √27=√(9×3)=3√3. Then √3×3√3=3(√3×√3)=3×3=9. Option A is correct. Option B, 3√3, is only the simplified value of √27; it does not include the additional multiplication by √3. Option C, √30, incorrectly adds the radicands 3 and 27 instead of multiplying them. Option D introduces an unsupported factor and is much larger than the actual product. Memory cue: multiply radicands under one square root, and simplify perfect-square factors.

For non-negative radicands, the product rule permits the radicals to be combined: √a × √b = √(ab). Hence √24 × √6 = √(24 × 6) = √144 = 12. We can verify this without combining first: √24 = √(4 × 6) = 2√6, and then (2√6)(√6) = 2 × 6 = 12. Thus option A is correct. Option B is not the value of the original product; it incorrectly leaves a factor of 6 outside √24. Option C does not follow the multiplication rule and gives √30, while option D adds an unnecessary factor and still contains an unsimplified radical. Since 144 is a perfect square, the final exact answer is the integer 12.

The governing classification rule is based on decimal expansions. Every rational number can be written as p/q, where p and q are integers and q is non-zero; its decimal expansion either terminates or eventually repeats a fixed block of digits. An irrational number cannot be expressed in that form, so its decimal expansion continues indefinitely without any repeating period. Therefore a non-terminating, non-repeating decimal is an irrational number, making option A correct. Option B is the opposite classification. An integer is a special kind of rational number, so option C cannot fit the description. A terminating decimal ends after finitely many digits, which directly contradicts the statement in option D.

The key criterion is that the square root of a positive integer is rational only when the integer is a perfect square. Since 47 is not a perfect square—it lies strictly between 36 = 6² and 49 = 7²—√47 is irrational. Negating a number changes its sign but does not change whether it is rational or irrational. Therefore −√47 is also irrational. It is still real because the square root of a positive real number is real, and the negative of a real number remains real. Hence option A is correct. It cannot be a rational number, whole number, or natural number, so options B, C, and D are excluded by the same reasoning.

The governing criterion is that the square root of a positive integer is rational exactly when that integer is a perfect square. Among the choices, 400 = 20², so √400 = 20, which is an integer and therefore rational. The other values, 401, 402, and 403, lie between consecutive perfect squares 20² = 400 and 21² = 441. Since none of them is a perfect square, their square roots are irrational. Therefore option A is the only valid choice. The question is not asking for the largest value or the value closest to a square; it specifically asks which value makes √m rational. The perfect-square test settles the answer immediately.

For a positive integer k, √k is rational if and only if k is a perfect square. Therefore √k is irrational whenever k is not a perfect square. The number 123 is not a perfect square; it lies between 11² = 121 and 12² = 144, so √123 cannot be simplified to an integer or any rational number. In contrast, 121 = 11², 144 = 12², and 169 = 13², giving rational square roots 11, 12, and 13 respectively. Thus option A satisfies the condition, while each of the other three options is excluded because it is a perfect square. The perfect-square criterion provides a complete and decisive justification.

The governing concept is the classification of square roots as rational or irrational numbers. A positive integer has a rational square root only when it is a perfect square. Since 59 lies between 7²=49 and 8²=64, it is not a perfect square, so √59 is irrational. The negative sign does not change irrationality: if −√59 were rational, multiplying it by −1 would make √59 rational, which is impossible. Because 59 is positive, √59 is real, and its negative is also real. Therefore −√59 is an irrational real number, so option A is correct. It is not rational; whole and natural numbers are integers and cannot equal this non-integer value.

The governing identity is the difference of squares: (a+b)(a−b)=a²−b². Here take a=√15 and b=√6. Then (√15+√6)(√15−√6)=(√15)²−(√6)²=15−6=9. If the expression is expanded instead, the two cross-products are √90 and −√90, so they cancel, leading to the same result. Hence option A is correct. Option B adds 15 and 6 rather than subtracting them. Option C leaves an uncancelled radical and therefore does not represent the product. Option D incorrectly retains a cross-term and also uses the wrong sign. The exact identity is more reliable than decimal approximation.