Irrational numbers

Step 1: (4<\sqrt{n}<5) means (16<n<25). Step 2: (20) lies in this interval and is not a perfect square, so (\sqrt{20}) is irrational. Step 3: In square-root inequalities, square the bounds to form the interval.

Step 1: Multiply numerator and denominator by (\sqrt{3}) to remove the square root from the denominator. Step 2: (\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}). Step 3: After rationalising, the denominator should not contain a square root.

Step 1: First simplify all products. Step 2: The first three produce inside numbers (144), (36), and (81), which are perfect squares; the fourth gives (\sqrt{10}). Step 3: After multiplication, check whether the inside number is a perfect square.

Step 1: (\sqrt{98}=7\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{18}=3\sqrt{2}). Step 2: (7\sqrt{2}+5\sqrt{2}-3\sqrt{2}=9\sqrt{2}). Step 3: Add or subtract only after converting all terms to like radicals.

Step 1: Adding a rational number cannot make an irrational number rational. Step 2: If (r+s) were rational, then (s=(r+s)-r) would be rational, which is a contradiction. Step 3: Such rules can also be checked by reverse reasoning.

Step 1: (\sqrt{a}\times\sqrt{b}=\sqrt{ab}). Step 2: (\sqrt{3b}=12), so (3b=144) and (b=48). Step 3: In square-root equations, squaring both sides is useful.

Step 1: Since (4<7<9), (\sqrt{7}) lies between (2) and (3). Step 2: (\sqrt{7}\approx2.646), so (2.65) is the closest. Step 3: In approximation, first set the range using perfect squares.

Step 1: (\sqrt{5}) and (\sqrt{5}) are both irrational. Step 2: Their difference is (0), which is rational. Step 3: The difference of equal irrational terms can be rational.

Step 1: (\sqrt{27}\times\sqrt{12}=\sqrt{324}). Step 2: (\sqrt{324}=18), so the result is rational. Step 3: When multiplying, multiply inside numbers and check for a perfect square.

Step 1: For (\frac{1}{2-\sqrt{3}}), the conjugate of the denominator is (2+\sqrt{3}). Step 2: Multiplying gives denominator (4-3=1), so the value is (2+\sqrt{3}). Step 3: Rationalising with the conjugate gives the answer quickly.

Step 1: (5) is rational and (\sqrt{2}) is irrational. Step 2: The sum of a rational and an irrational number is irrational. Step 3: Adding an integer does not remove the square-root part.

Step 1: (\sqrt{200}=10\sqrt{2}) and (\sqrt{72}=6\sqrt{2}). Step 2: (10\sqrt{2}-6\sqrt{2}=4\sqrt{2}). Step 3: Before subtracting radicals, write both terms in simplified form.

Step 1: (\sqrt{30}) is irrational because (30) is not a perfect square. Step 2: But (25<30<36), so (\sqrt{30}) is greater than (5); the other options are rational. Step 3: This means the option set has no valid answer, so such a question should not be used.

Step 1: This is of the form \((a-b)(a+b)=a^2-b^2\). Step 2: \((\sqrt{7})^2-(\sqrt{3})^2=7-3=4\). Step 3: In conjugate multiplication, directly use the difference of squares.

Step 1: The first option cancels to (3+\sqrt{2}). Step 2: (\frac{7}{3-\sqrt{2}}=\frac{7(3+\sqrt{2})}{9-2}=3+\sqrt{2}), so it is also equal; hence there is no incorrect option. Step 3: In equivalent-form questions, simplify every option.

Step 1: Multiply the given approximate value by (3). Step 2: (3\sqrt{2}\approx3\times1.414=4.242). Step 3: In approximation questions, directly use the given value.

Step 1: (3\sqrt{5}=\sqrt{9}\sqrt{5}). Step 2: This equals (\sqrt{45}). Step 3: An outside coefficient can be moved inside by squaring it.

Step 1: (\sqrt{8}=2\sqrt{2}). Step 2: (x=\sqrt{2}+2\sqrt{2}=3\sqrt{2}). Step 3: Before adding, convert radicals into like form.

Step 1: (\sqrt{3}\approx1.732) and (\sqrt{5}\approx2.236). Step 2: (2) lies between these two values. Step 3: Approximate values or squaring can both help in comparison.

Step 1: (\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}). Step 2: The total is (6\sqrt{3}), which is irrational. Step 3: Decide the nature of the number only after simplification.

Step 1: A non-zero rational multiplier does not remove irrationality. Step 2: For example, (4\sqrt{3}) remains irrational. Step 3: The non-zero condition is important because multiplying by (0) gives (0).

Step 1: (\sqrt{150}=5\sqrt{6}) and (\sqrt{24}=2\sqrt{6}). Step 2: (5\sqrt{6}+2\sqrt{6}=7\sqrt{6}). Step 3: Add radicals only after they become like radicals.

Step 1: (3) is a non-zero rational number and (\sqrt{7}) is irrational. Step 2: (3\sqrt{7}) remains irrational. Step 3: Multiplication by zero is a special case, so focus on non-zero rational factors.

Step 1: If (\sqrt{a}=7), square both sides. Step 2: (a=49), and (49) is a perfect square. Step 3: In square-root equations, square to find the original number.

Step 1: (\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}). Step 2: (4\sqrt{2}) is irrational, so it is not rational. Step 3: In options, simplify every result before deciding its nature.