Irrational numbers

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.

The governing classification theorem says that a real number is rational exactly when its decimal expansion terminates or eventually repeats. The question explicitly states that the decimal expansion continues forever and has no repeating block. Therefore it cannot be written as a ratio of two integers and is irrational. It is still a real number because it is represented by a decimal expansion on the real number line. Thus option B, irrational real number, is correct. Option A would require termination or periodic repetition, option C contradicts the word non-terminating, and option D is impossible because an integer has a terminating decimal representation.

An irrational number has a decimal expansion that neither ends nor repeats in a fixed pattern. For example, \(\sqrt{2}=1.414213\ldots\). A repeating non-terminating decimal is rational. Exam tip: link “non-terminating, non-repeating” with irrational numbers.

(13) is not a perfect square, so ( \sqrt{13} ) is irrational. An irrational number cannot be written as ( \frac{p}{q} ).

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), a rational number. Hence, the word “always” makes the claim false. Exam tip: one counterexample disproves a universal statement.

(8) is not a perfect square, so ( \sqrt{8} ) is irrational. Check whether the number under the radical is a perfect square.

The governing definition is that a rational number can be written as p/q, where p and q are integers and q is not zero. Its decimal expansion either terminates or repeats. Thus 0.333… equals 1/3, 11/4 is already a ratio of integers, and −2 equals −2/1; all three are rational. In contrast, 15 is not a perfect square, so √15 cannot be expressed as a ratio of two integers. Its decimal expansion is non-terminating and non-repeating, which identifies it as irrational. Therefore option B is the only correct choice. The presence of a square-root sign alone is not enough; roots of perfect squares, such as √16, are rational.

The number of zeros between successive 1s increases as 1, 2, 3, 4..., so no repeating decimal block occurs. It is non-terminating and non-repeating, hence irrational. Exam tip: check repetition, not merely the digits used.

( \sqrt{6} ) is irrational, so its decimal is non-terminating non-recurring. An irrational decimal neither terminates nor repeats.

(50) is not a perfect square, so ( \sqrt{50} ) is irrational. Only the square root of a perfect square is an integer.

(2) and (3) are not perfect squares, so ( \sqrt{2} ) and ( \sqrt{3} ) are irrational. Square roots of non-perfect squares are irrational.

An irrational number has a non-terminating, non-recurring decimal expansion, so option C is correct. Terminating or recurring decimals are rational numbers. Exam tip: check whether any digit pattern repeats.