Irrational numbers

Subtracting irrational (\sqrt{11}) from rational (7) gives an irrational number. The irrational part remains.

In a right triangle, the hypotenuse is (\sqrt{1^2+3^2}=\sqrt{10}). Pythagoras theorem is used in construction.

(\sqrt{6}\times\sqrt{24}=\sqrt{144}=12), which is rational. The product of two irrationals can be rational.

The zeros between successive 1s increase as 1, 2, 3, ...; therefore no block repeats and the number is irrational. Using only 0 and 1 does not make it rational. In exams, check whether the decimal repeats.

The area of a square is \(A=s^2\), so its side is \(s=\sqrt{A}=\sqrt{72}\). Since \(72=36\times2\), \(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\). Therefore, option C is correct. Exam tip: To find the side of a square from its area, take the square root of the area, not half of it.

Rationalisation requires multiplying by the conjugate of the denominator. The conjugate of 6 − √5 is 6 + √5, so multiply numerator and denominator by (6 + √5): 1/(6−√5) × (6+√5)/(6+√5). The denominator becomes a difference of squares, (6−√5)(6+√5) = 6^2 − (√5)^2 = 36−5 = 31. Consequently the rationalised expression is (6+√5)/31, so option A is correct. Option B keeps the wrong sign, C omits the denominator 31, and D merely changes the sign in the denominator while leaving it irrational. The conjugate is essential because it converts the denominator into a rational number.

(\sqrt{8}=2\sqrt{2}) and (\sqrt{200}=10\sqrt{2}), so the sum is (12\sqrt{2}). Simplify before comparing.

(\sqrt{44}=2\sqrt{11}), so (4\sqrt{11}+6\sqrt{11}=10\sqrt{11}). Watch both coefficients and radicals carefully.

\(0.\overline{3}=3/9=1/3\), so its decimal expansion is non-terminating but repeating, making it rational. \(\sqrt{2}\) is irrational. Exam tip: convert a repeating decimal into a fraction to check it.

\(0.\overline{27}=27/99=3/11\), so it is rational even though its decimal expansion never ends. \(\sqrt{5}\) and \(\pi\) are irrational. Exam tip: every repeating decimal represents a rational number.

The number of zeros between successive 1s is 1, 2, 3, 4, ..., so no repeating block can occur. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not merely which digits appear.

The governing concept is simplification and addition of like surds. A square factor can be taken outside a radical, so √12 = √(4×3) = 2√3. Substituting this into the expression gives x = 2√3 + 2√3. Because both terms contain the same irrational factor √3, their numerical coefficients may be added: 2 + 2 = 4. Therefore x = 4√3, making option D correct. Option A keeps only the first simplified term and ignores the second term. Option B uses an incorrect coefficient sum. Option C incorrectly treats the sum of two radical terms as one radical. Unlike ordinary multiplication, addition under radicals is not combined by adding radicands; only like surds can be combined directly.

18 is not a perfect square. \(\sqrt{18}=\sqrt{9\times2}=3\sqrt2\), and \(\sqrt2\) is irrational, so \(3\sqrt2\) is also irrational. Exam tip: factor the number into perfect-square factors before classifying its square root.

Multiplying by the conjugate gives (\frac{4(\sqrt{3}-1)}{2}=2\sqrt{3}-2). Simplify the whole fraction after rationalising.

The governing concept is reduction and subtraction of like surds. First extract the perfect-square factor from √8: √8 = √(4×2) = 2√2. Substituting this into the expression gives a = 2√2 − √2 = (2−1)√2 = √2. Therefore option B is correct. Option A adds the coefficients instead of subtracting them. Option C incorrectly combines the radicands 8 and 2 under one square root, even though subtraction of square roots does not work that way. Option D treats √2 as though it were an integer. Direct addition or subtraction is permitted only when the remaining radical parts are identical, as they are here.

(0.4141141114\ldots) has no fixed repetition so it is irrational. A repeating decimal is rational.

(\sqrt{98}=7\sqrt{2}) and (\sqrt{18}=3\sqrt{2}) so division gives (4). First convert the numerator into like radicals.

Option B is non-terminating and non-repeating, so it represents an irrational number. In contrast, 0.636363... repeats a fixed block and is rational. Exam tip: identify irrational decimals by checking for no repeating pattern.

A rational number has a terminating or repeating decimal expansion. Here, the number of zeros between successive 1s increases as 1, 2, 3, 4…, so no block repeats. Exam tip: check repetition, not merely the digits used.

Both \(\sqrt{7}\) and \(1-\sqrt{7}\) are irrational, but \(\sqrt{7}+1-\sqrt{7}=1\), which is rational. Hence the statement is false. Exam tip: check whether irrational terms cancel in a sum.