Irrational numbers

The governing concept is solving an inequality over the integers and then counting the resulting set. Since x²<5, we have -√5<x<√5, so the possible integers are -2,-1,0,1,2. Their squares are 4,1,0,1,4, all less than 5. The next integers, -3 and 3, have square 9 and are excluded. Thus V has five elements, so C is correct.

The governing concept is filtering a set by two conditions. The natural perfect squares below 25 are 1²=1, 2²=4, 3²=9, and 4²=16. The next square, 5²=25, is excluded because the inequality is strict. The additional condition x≠1 removes 1, leaving V={4,9,16}. Hence option A is correct; the other choices either retain 1 or include 25.

The governing concept is solving an inequality over the whole numbers. From 2x+1≤9, subtract 1 to obtain 2x≤8, and divide by 2 to get x≤4. Whole numbers are 0,1,2,..., so the members satisfying this condition are 0,1,2,3,4. Hence option B is correct. Options A and D omit zero, while C incorrectly omits the valid endpoint 4.

Since 7 is not a perfect square, \(\sqrt{7}\) cannot be expressed as a ratio of two integers; therefore, it is irrational. Because 7 is positive, \(\sqrt{7}\) is also a real number. Thus, option B is incorrect— not every real number is rational. Exam tip: The square root of a non-negative integer is rational only when the integer is a perfect square.

Since \(50=25\times2\) and 25 is a perfect square, \(\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}=5\sqrt{2}\). Option C is incorrect because \((2\sqrt{5})^2=20\), not 50. Exam tip: factor the radicand and take the largest perfect-square factor outside the square root.

A decimal expansion that is non-terminating and non-repeating represents an irrational number. Therefore, 1.414213... is an irrational real number. A rational number has a decimal expansion that either terminates or repeats, so option A is incorrect. Exam tip: Identify a non-terminating, non-repeating decimal as irrational.

\(\frac{3}{2}=1.5\), whereas \(\sqrt{2}\approx1.414\). Therefore, \(\frac{3}{2}\) is greater than \(\sqrt{2}\). This can also be checked without finding decimal values: both numbers are positive, and \(\left(\frac{3}{2}\right)^2=\frac{9}{4}=2.25>2=(\sqrt{2})^2\). Hence, option B is correct. Exam tip: for positive numbers, comparing their squares gives the same order.

\(5\) is rational because it can be written as \(\frac{5}{1}\). \(\sqrt{2}\) is irrational because 2 is not a perfect square, so its decimal expansion is non-terminating and non-repeating. Therefore, option A contains one rational and one irrational real number. In option C, both numbers are irrational, while both numbers in options B and D are rational. Exam tip: The square root of a perfect square is rational; the square root of a non-perfect square is irrational.

The governing concept is the decimal test for rational and irrational numbers. A real number is rational when it can be written as p/q, with integers p and q and q non-zero; its decimal expansion terminates or eventually repeats. The given expansion continues indefinitely and, as stated, has no repeating pattern. Therefore it cannot be represented as a ratio of integers and is irrational. It remains a real number because every such decimal denotes a point on the real number line. Option C is correct. It cannot be an integer or natural number, since those have whole-number values and terminating decimal forms, and it is not rational because no repetition occurs.

Since \(17\) is not a perfect square, \(\sqrt{17}\) cannot be expressed as the ratio of two integers, so it is irrational. It is also a real number because the square root of every positive number is real. Option A is a rational fraction, option B is an integer, and option D is a terminating decimal; all three are rational. Exam tip: The square root of a positive number that is not a perfect square is generally irrational.

\(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\) and \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\). Therefore, \(\sqrt{50}-\sqrt{18}=5\sqrt{2}-3\sqrt{2}=2\sqrt{2}\), so option A is correct. Exam tip: only surds with the same radicand can be subtracted like algebraic terms.

Both numbers are positive, so we can compare their squares. We have \(5^2=25\) and \((\sqrt{23})^2=23\). Since \(23<25\), it follows that \(\sqrt{23}<5\). Therefore, \(\sqrt{23}\) is the smaller number. Exam tip: Squaring is a useful method for comparing positive numbers involving square roots.

\(\sqrt{2}\) is irrational, and it lies between 1 and 2 because \(1^2<2<2^2\). Hence, it satisfies the required condition. \(\frac{3}{2}\) is rational, while 1 and 2 are the endpoints, not numbers strictly between them. Exam tip: Check both the type of number and its position in the given interval.

Since \(6=\sqrt{36}\) and \(35<36\), we get \(\sqrt{35}<\sqrt{36}=6\). Therefore, \(\sqrt{35}\) is smaller. Option A is incorrect because \(6\) is the larger number. In such questions, compare the number under the square root with nearby perfect squares.

The defining decimal criterion is that a rational number has a terminating or eventually repeating decimal expansion. A decimal that continues forever without any repeating block cannot be expressed as p/q, where p and q are integers and q is nonzero; it is therefore irrational. Since its decimal value is defined on the number line, it is still a real number. Thus the stated number is an irrational real number, making option B correct. It is not rational because the decimal neither terminates nor repeats. It is not a whole number, since whole numbers are 0, 1, 2, … and have no fractional part, and it is certainly not undefined merely because its expansion is infinite.

To remove the square root from the denominator, multiply both numerator and denominator by \(\sqrt{5}\): \(\frac{4}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{4\sqrt{5}}{5}\), since \(\sqrt{5}\times\sqrt{5}=5\). Therefore, option A is correct. Option B is the reciprocal of the original fraction, while C and D are not equivalent to it. Exam tip: when a denominator contains a single square root, multiply the numerator and denominator by that same square root.

Since \(7^2=49<53<64=8^2\), it follows that \(7<\sqrt{53}<8\). Therefore, \(\sqrt{53}\) lies between 7 and 8, so option B is correct. Option C is incorrect because \(\sqrt{53}\) is less than 8, as \(8^2=64\). Exam tip: compare the number with the nearest consecutive perfect squares to locate its square root.

A real number is rational if its decimal expansion terminates or eventually repeats. The given expansion continues indefinitely and, as stated, has no fixed repeating block: the runs of digits keep changing. Therefore it cannot be represented as p/q with integers p and q, so it is irrational. Option B is correct; an integer is rational, and a terminating decimal also represents a rational number.

The governing concept is decimal representation of rational and irrational numbers. A rational number, when written in decimal form, either terminates or eventually repeats a fixed block of digits. For example, 0.333... repeats 3 and is rational. The given decimal 0.040040004... is stated to continue without termination and without any fixed repeating pattern. Therefore it cannot be expressed as a ratio of two integers and is irrational. It is still a real number because irrational numbers are included in the real-number system. It is not a whole or natural number, since those are non-negative integers. Hence option C is correct.

The correct classification is an irrational real number, but it should be proved rather than inferred only from the fact that both terms are irrational. Suppose, for contradiction, that √2 + √3 = r where r is rational. Then √3 = r − √2. Squaring both sides gives 3 = r² + 2 − 2r√2, so 2r√2 = r² − 1. The value r is positive and nonzero, because the left side of the original equation is positive. Therefore √2 = (r²−1)/(2r) would be rational, contradicting the known irrationality of √2. Hence √2 + √3 is irrational. Since both summands are real, their sum is real as well, so option B is correct.