Irrational numbers

Step 1: The square root of a perfect square is rational. Step 2: (52) is not a perfect square, so (\sqrt{52}) is irrational. Step 3: In such questions, eliminate perfect squares first.

Step 1: Simplify each option first. Step 2: (\sqrt{11}) is irrational and (2) is rational, so (\sqrt{11}+2) remains irrational. Step 3: Do not treat the sum of a rational and an irrational number as rational.

Step 1: (\sqrt{98}=7\sqrt{2}) and (\sqrt{32}=4\sqrt{2}). Step 2: (7\sqrt{2}-4\sqrt{2}=3\sqrt{2}). Step 3: Convert radicals into like radicals before subtracting.

Step 1: (x-4=(4+\sqrt{7})-4). Step 2: This leaves (\sqrt{7}), which is irrational. Step 3: First simplify the expression, then identify the nature of the number.

Step 1: Terminating decimals, fractions, and recurring decimals are rational. Step 2: (\sqrt{45}=3\sqrt{5}), and (\sqrt{5}) is irrational. Step 3: Simplify the square root to identify its nature.

Step 1: (\sqrt{20}=2\sqrt{5}), (\sqrt{45}=3\sqrt{5}), and (\sqrt{125}=5\sqrt{5}). Step 2: The sum is (2\sqrt{5}+3\sqrt{5}+5\sqrt{5}=10\sqrt{5}). Step 3: Once radicals are like terms, add only the coefficients.

Step 1: A non-zero rational multiplier does not remove irrationality. Step 2: For example, (5\sqrt{2}) remains irrational. Step 3: Testing always-type statements with examples is a good habit.

Step 1: Multiply numerator and denominator by (\sqrt{7}) to remove the root from the denominator. Step 2: (\frac{7}{\sqrt{7}}=\frac{7\sqrt{7}}{7}=\sqrt{7}). Step 3: Rationalisation helps when the denominator contains a square root.

Step 1: Squaring a square root gives the number inside. Step 2: ((\sqrt{b})^2=b), and if (b) is an integer, it is rational. Step 3: The square of an irrational square root can give a rational result.

Step 1: Since (81<85<100), (9<\sqrt{85}<10). Step 2: (85) is not a perfect square, so (\sqrt{85}) is irrational. Step 3: In interval questions, use nearby perfect squares.

Step 1: (\sqrt{75}=5\sqrt{3}), (\sqrt{300}=10\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Step 2: (5\sqrt{3}+10\sqrt{3}-4\sqrt{3}=11\sqrt{3}). Step 3: Simplify all radicals before addition and subtraction.

Step 1: This decimal has no fixed block repeating again and again. Step 2: It is non-terminating and non-recurring, so it is irrational. Step 3: Decide by checking whether there is a fixed repeating pattern.

Step 1: (\sqrt{5}) and (-\sqrt{5}) are both irrational. Step 2: Their sum is (0), which is rational. Step 3: Opposite irrational terms can give a rational sum.

Step 1: Apply distribution: (\sqrt{5}\times3+\sqrt{5}\times\sqrt{5}). Step 2: This becomes (3\sqrt{5}+5). Step 3: In such multiplication, remember (\sqrt{5}\times\sqrt{5}=5).

Step 1: (4\sqrt{3}=\sqrt{16}\sqrt{3}). Step 2: This equals (\sqrt{48}). Step 3: When moving an outside coefficient inside the root, multiply by its square.

Step 1: This is of the form \((a+b)(a-b)=a^2-b^2\). Step 2: \(3^2-(\sqrt{5})^2=9-5=4\). Step 3: In conjugate multiplication, directly use difference of squares.

Step 1: (\sqrt{196}=14). Step 2: (\sqrt{196}-5=9), which is rational. Step 3: When asked for not irrational, carefully check perfect-square options.

Step 1: (ab=(\sqrt{3}+2)(\sqrt{3}-2)). Step 2: Using difference of squares, ((\sqrt{3})^2-2^2=3-4=-1). Step 3: Recognising conjugate form makes the calculation shorter.

Step 1: The conjugate of (3+\sqrt{8}) is (3-\sqrt{8}). Step 2: (\frac{1}{3+\sqrt{8}}\times\frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}). Step 3: Use the conjugate of the denominator for rationalisation.

Step 1: (\sqrt{2}) and (\sqrt{2}) are both irrational. Step 2: Their product is (2), which is rational. Step 3: Test always-type statements with a counterexample.

Step 1: (\sqrt{18}=3\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{162}=9\sqrt{2}). Step 2: The sum is (3\sqrt{2}+6\sqrt{2}+9\sqrt{2}=18\sqrt{2}). Step 3: Simplify all radicals completely first.

Step 1: (x^2=(\sqrt{5})^2=5). Step 2: (3x=3\sqrt{5}), so (x^2+3x=5+3\sqrt{5}). Step 3: Simplify the square and the multiplication separately.

Step 1: (10<\sqrt{n}<11) means (100<n<121). Step 2: (105) lies in this interval and is not a perfect square, so (\sqrt{105}) is irrational. Step 3: Square both bounds to handle square-root ranges.

Step 1: A recurring decimal is rational. Step 2: In (0.272727\ldots), the block (27) repeats. Step 3: Identifying the repeating block in a decimal is important.

Step 1: Use \((a+b)^2=a^2+2ab+b^2\). Step 2: \((\sqrt{7})^2+2\sqrt{7}\times1+1^2=7+2\sqrt{7}+1=8+2\sqrt{7}\). Step 3: Forgetting the middle term \(2ab\) is a common mistake.