Irrational numbers

Since (10<11<12), (\sqrt{11}) lies between them. For positive square roots compare the numbers inside.

\(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3\), which is rational. The quotient of two irrational numbers need not be irrational. Exam tip: combine the radicals before deciding the number type.

(\sqrt{125}=5\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (3\sqrt{5}). Dividing gives (3).

The square of a square root gives the number inside. So \(\left(\sqrt{7+\sqrt{10}}\right)^2=7+\sqrt{10}\).

If \(p+q\) were rational, then \(p=(p+q)-q\) would be the difference of two rational numbers and hence rational, a contradiction. Thus \(p+q\) is irrational. Exam tip: adding or subtracting a rational number from an irrational number remains irrational.

In option A, \(\sqrt{19}\) and \(\sqrt{19}\) are both irrational, and their difference is \(\sqrt{19}-\sqrt{19}=0\). Zero is a rational number. In option B, \(\sqrt{2}-\sqrt{3}\) is irrational; option C simplifies to \(\sqrt{5}-\sqrt{20}=-\sqrt{5}\), and option D to \(\sqrt{7}-\sqrt{28}=-\sqrt{7}\), both irrational. Exam tip: the difference of identical irrational numbers is zero, which is rational.

The governing concept is the difference-of-squares identity, (a+b)(a−b)=a²−b². Set a=√20 and b=√45. Then the product is (√20)²−(√45)² = 20−45 = −25, so option A is correct. The result can be checked by simplifying first: √20=2√5 and √45=3√5. The two factors become 5√5 and −√5, whose product is −5×5 = −25. Option B reverses the subtraction order. Option C adds the radicands instead of applying the identity. Option D incorrectly combines the radicands and does not represent the given conjugate product. Recognising the conjugate pattern avoids unnecessary expansion and also shows why the middle terms cancel.

The sum of two irrational numbers need not have a fixed type. For example, \(\sqrt{2}+(-\sqrt{2})=0\) is rational, whereas \(\sqrt{2}+\sqrt{3}\) is irrational. Hence, the sum can be rational or irrational. In exams, test claims using “always” with a counterexample.

Multiplying by the conjugate gives numerator (10+2\sqrt{21}) and denominator (4). So the simplified form is (\frac{5+\sqrt{21}}{2}).

\(\sqrt{7}\) is irrational. If \(5-\sqrt{7}\) were rational, subtracting it from 5 would make \(\sqrt{7}\) rational, a contradiction. Exam tip: rational ± irrational is always irrational.

(\sqrt{245}=7\sqrt{5}), (\sqrt{180}=6\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). So the result is (7\sqrt{5}-6\sqrt{5}+4\sqrt{5}=5\sqrt{5}).

The governing concept is rationalisation of a surd in the denominator. The conjugate of √11+3 is √11−3. Multiply numerator and denominator by this conjugate: r = (√11−3)/[(√11+3)(√11−3)] = (√11−3)/[(√11)²−3²] = (√11−3)/(11−9) = (√11−3)/2. Thus option A is correct. Option B has the correct numerator but loses the denominator 2. Option C retains the wrong sign and therefore uses the original expression rather than its conjugate. Option D is the negative of the numerator and is not equivalent to the original positive fraction. The denominator is nonzero, so multiplying by the conjugate preserves equality and produces a rational denominator.

(\sqrt{18}=3\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the sum is (8\sqrt{2}). Its square is (128).

Multiplying by the conjugate makes the denominator (11-7=4). So (\frac{2(\sqrt{11}-\sqrt{7})}{4}=\frac{\sqrt{11}-\sqrt{7}}{2}).

(\sqrt{98}=7\sqrt{2}) and (\sqrt{200}=10\sqrt{2}), so (s=17\sqrt{2}). Dividing by (\sqrt{2}) gives (17).

The governing concept is the conjugate-product identity (u+v)(u−v)=u²−v². Since multiplication is commutative, rewrite ab as (√5+√3)(√5−√3). Taking u=√5 and v=√3 gives ab=(√5)²−(√3)²=5−3=2. Therefore option A is correct. The cross terms cancel because one product is +√15 and the other is −√15. Option B has the wrong sign and would correspond to reversing the order of the squares. Option C comes from an incorrect expansion or addition. Option D multiplies the radicals but fails to include the cancellation of the two cross terms. The conjugate structure is the important observation, and it gives an exact result without decimal approximation.

The first, second, and fourth options give (5), (5), and (6) respectively. The third gives (\sqrt{5}), which is irrational.

Side = \(\sqrt{50}=\sqrt{25\times2}=5\sqrt2\) cm. Since \(\sqrt2\) is irrational, multiplying it by non-zero rational 5 keeps it irrational. Exam tip: take the square root of the area.

The governing concept is the conjugate-product identity (a+b)(a−b)=a²−b². Let a=√45 and b=√5. Then (√45+√5)(√45−√5)=(√45)²−(√5)²=45−5=40, so option A is correct. An independent check is √45=3√5; the factors then become 4√5 and 2√5, whose product is 8×5=40. Option B results from adding 45 and 5 rather than finding their difference. Option C reverses or mishandles the subtraction. Option D is not equivalent: √225=15, so 2√225=30, not 40. The conjugate pattern is preferable to full expansion because the positive and negative cross terms cancel immediately.

\(\sqrt{2}+(-\sqrt{2})=0\), and 0 is rational. Hence, the sum of two irrational numbers need not be irrational. In option B, the sum is \(3\sqrt{2}\), which is irrational. Exam tip: one counterexample is enough to disprove an “always” statement.