Irrational numbers

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.

Subtracting like radicals gives (2\sqrt{6}-\sqrt{6}=\sqrt{6}). (\sqrt{6}) is an irrational number.

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.