Irrational numbers

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

(\sqrt{13}) and (\sqrt{17}) are different irrational radicals and their sum is irrational. Different radicals are not added directly.

The governing method is to extract the largest perfect-square factor from under the radical. Factor 392 as 196 × 2, and observe that 196 = 14². Therefore √392 = √(196 × 2) = √196 × √2 = 14√2. Hence option A is the fully simplified form. Option C, 7√8, is numerically equivalent to 14√2, but it is not in simplest form because √8 = √(4 × 2) = 2√2, giving 7√8 = 14√2. Option B incorrectly doubles the required coefficient; its square is 28² × 2 = 1568, not 392. Option D mistakes the perfect-square factor 196 for its square root and is also far too large. Squaring 14√2 gives 196 × 2 = 392, confirming option A.

\(0.\overline{36}\) repeats, so it is rational: \(0.\overline{36}=\frac{36}{99}=\frac{4}{11}\). Unlike it, \(\sqrt{7}\) is non-repeating. Exam tip: repeating decimals are rational.

Subtracting coefficients of like radicals gives (7\sqrt{3}). It is irrational because (\sqrt{3}) remains.

(\sqrt{45}=3\sqrt{5}) and (\sqrt{80}=4\sqrt{5}), so the bracket is (7\sqrt{5}) and the product is (35). Simplify the bracket first.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence the word “always” makes Reena’s claim false. Exam tip: disprove universal statements using one counterexample.

The statement is true in some cases and false in others. For example, \(\sqrt{2}+\sqrt{3}\) is irrational, but \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. In exams, test “always” statements using a counterexample.

Multiplying by the conjugate gives (\frac{4(\sqrt{5}-1)}{5-1}=\sqrt{5}-1). Make the denominator rational.

\(0.272727\ldots\) is non-terminating but recurring. It can be written as \(27/99=3/11\), so it is rational. \(\sqrt{7}\), \(\pi\), and \(\sqrt{11}\) are irrational. Exam tip: every recurring decimal is rational.

The two expressions are conjugates. Using \((a+b)(a-b)=a^2-b^2\), we get \((10+\sqrt{29})(10-\sqrt{29})=10^2-(\sqrt{29})^2=100-29=71\). Therefore, option B is correct. Exam tip: when conjugate expressions are multiplied, apply the difference-of-squares identity directly instead of expanding every term.

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

The correct choice is A: a right triangle with legs 1 and 3 has hypotenuse \(\sqrt{10}\). The key idea is the Pythagorean theorem: in a right triangle, hypotenuse squared equals the sum of the squares of the two legs. Thus \(h^2=1^2+3^2=1+9=10\), so \(h=\sqrt{10}\). This length can be transferred to the number line with a compass. Option A works exactly because its squared legs add to 10. Option B gives \(h=\sqrt{2^2+2^2}=\sqrt8=2\sqrt2\), not \(\sqrt{10}\). Option C gives \(h=\sqrt{1^2+2^2}=\sqrt5\). Option D gives \(h=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2\). The side lengths need not themselves be irrational; ordinary lengths 1 and 3 are enough. The exam cue is to square each pair of proposed legs and choose the pair whose sum is 10.

(\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.

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.

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