Irrational numbers

\(\sqrt{2}\) is irrational, while \(1\) is rational. The sum of an irrational number and a rational number is always irrational, so \(\sqrt{2}+1\) is irrational. The other sums are \(5\), \(0\), and \(1\), all of which are rational. Exam tip: identify whether each term is rational or irrational first; irrational + rational is always irrational.

\(\sqrt{10}\) is irrational because 10 is not a perfect square. Therefore, its decimal expansion is non-terminating and non-recurring. \(\frac{3}{4}=0.75\) and \(\frac{7}{8}=0.875\) terminate, whereas \(\frac{2}{11}=0.1818\ldots\) is recurring. Exam tip: The square root of a natural number that is not a perfect square is irrational.

The intended irrational endpoints are \(\sqrt{2}\) and \(\sqrt{5}\). Since \(\sqrt{2}\approx1.414\) and \(\sqrt{5}\approx2.236\), we have \(1.414<2<2.236\). Therefore, 2 is the correct option. \(\sqrt{2}\) is an endpoint, not a number strictly between them, while 1 and 3 lie outside this interval. Exam tip: Use approximate decimal values to compare square roots quickly.

If p+q were rational, then subtracting the rational number p from it would make q rational. This contradicts the fact that q is irrational. Therefore, p+q is irrational. Integers and natural numbers are types of rational numbers, so they cannot be correct here. Exam tip: Adding or subtracting a rational number does not change an irrational number into a rational number.

Since \(16<17<25\), we get \(4<\sqrt{17}<5\). Also, \(\sqrt{17}\approx 4.12\), which is about 0.12 away from 4 but about 0.88 away from 5. Therefore, 4 is the closest number. Although 5 is the next integer, it is farther away. Exam tip: estimate a square root by locating the nearest perfect squares on either side.

\(\sqrt{11}\) is irrational, while \(1\) is rational. The difference between an irrational number and a rational number remains irrational, so \(\sqrt{11}-1\) is irrational. In contrast, \(7-2=5\), \(5-5=0\), and \(\frac{9}{2}-\frac{1}{2}=4\) are all rational. Exam tip: the square root of a number that is not a perfect square is generally irrational.

Since \(48=16\times3\) and \(75=25\times3\), \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\). Therefore, \(\sqrt{48}+\sqrt{75}=4\sqrt{3}+5\sqrt{3}=9\sqrt{3}\). A close option such as \(8\sqrt{3}\) may result from simplifying one radical incorrectly. Exam tip: simplify each surd first, then add like surds.

Since \(49<50<64\), we get \(\sqrt{49}<\sqrt{50}<\sqrt{64}\), so \(7<\sqrt{50}<8\). Also, \(\sqrt{50}\approx 7.07\), which is much closer to 7 than to 8. Therefore, 7 is correct. Exam tip: Compare with nearby perfect squares to estimate a square root quickly.

If p=0, then pq=0, which is rational. However, if p is a non-zero rational number, for example p=2 and q=\(\sqrt{2}\), then pq=\(2\sqrt{2}\), which is irrational. Hence, the product can be either rational or irrational. “Always irrational” is true only when p is non-zero. Exam tip: Always check the special case p=0 in such questions.

Since \(98=49\times2\), \(a=\sqrt{98}=7\sqrt{2}\); and since \(50=25\times2\), \(b=\sqrt{50}=5\sqrt{2}\). Therefore, \(a-b=7\sqrt{2}-5\sqrt{2}=2\sqrt{2}\). \(\sqrt{2}\) would result from incorrectly subtracting the coefficients. Exam tip: simplify surds to like radical terms before subtracting them.

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.