Irrational numbers

Step 1: The square root of a positive integer is rational only when the integer is a perfect square. Step 2: For example, (25) is a perfect square and (\sqrt{25}=5) is rational. Step 3: In square-root questions, identify perfect squares first.

Step 1: (\sqrt{2}\times\sqrt{8}=\sqrt{16}). Step 2: (\sqrt{16}=4), which is rational. Step 3: The product of two irrational numbers is not always irrational.

Step 1: (\sqrt{72}=6\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Step 2: (6\sqrt{2}-3\sqrt{2}=3\sqrt{2}). Step 3: Simplify both square roots before subtracting.

Step 1: (x-3=(3+\sqrt{5})-3). Step 2: This leaves (\sqrt{5}), which is irrational. Step 3: Simplify the expression before deciding the nature of the number.

Step 1: Terminating decimals and fractions are rational. Step 2: (\sqrt{18}=3\sqrt{2}), and (\sqrt{2}) is irrational. Step 3: Simplifying a square root often helps identify the number correctly.

Step 1: (\sqrt{12}=2\sqrt{3}), (\sqrt{27}=3\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Step 2: Adding gives (2\sqrt{3}+3\sqrt{3}+5\sqrt{3}=10\sqrt{3}). Step 3: Once radicals are like terms, add only the coefficients.

Step 1: Adding a rational number does not remove the irrational part. Step 2: For example, (2+\sqrt{3}) is irrational. Step 3: Be careful with always-type statements about sums or products of two irrational numbers.

Step 1: To simplify the denominator, multiply top and bottom by (\sqrt{5}). Step 2: (\frac{5}{\sqrt{5}}=\frac{5\sqrt{5}}{5}=\sqrt{5}). Step 3: Rationalising is useful when a square root appears in the denominator.

Step 1: Multiplying a square root by itself gives the number inside. Step 2: (\sqrt{a}\times\sqrt{a}=a), and if (a) is an integer, it is rational. Step 3: The square of an irrational number can be rational.

Step 1: Since (4<5<9), (2<\sqrt{5}<3). Step 2: (5) is not a perfect square, so (\sqrt{5}) is irrational. Step 3: Use nearby perfect squares to locate a square root.

Step 1: (\sqrt{45}=3\sqrt{5}), (\sqrt{80}=4\sqrt{5}), and (\sqrt{20}=2\sqrt{5}). Step 2: (3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}). Step 3: Convert all radicals to like form before adding or subtracting.

Step 1: This decimal has no fixed repeating block. Step 2: It is non-terminating and non-recurring, so it is irrational. Step 3: In long decimals, always check for a fixed repeating pattern.

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

Step 1: Use distribution: (\sqrt{3}\times2+\sqrt{3}\times\sqrt{3}). Step 2: This becomes (2\sqrt{3}+3). Step 3: In radical multiplication, remember (\sqrt{3}\times\sqrt{3}=3).

Step 1: Write (50=25\times2). Step 2: (\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}). Step 3: To identify an equivalent form, simplify the square root first.

Step 1: This is of the form \((a+b)(a-b)=a^2-b^2\). Step 2: \(2^2-(\sqrt{3})^2=4-3=1\). Step 3: For conjugate products, difference of squares gives the answer quickly.

Step 1: (\sqrt{121}=11). Step 2: (\sqrt{121}-2=9), which is rational. Step 3: In not-irrational questions, check perfect squares carefully.

Step 1: (ab=(\sqrt{2}+1)(\sqrt{2}-1)). Step 2: Using difference of squares, ((\sqrt{2})^2-1^2=2-1=1). Step 3: Learn to recognise conjugate forms like (a+b) and (a-b).

Step 1: The conjugate of the denominator is (2-\sqrt{3}). Step 2: (\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}). Step 3: For rationalisation, multiply by the conjugate.

Step 1: (\sqrt{2}) and (-\sqrt{2}) are both irrational. Step 2: Their sum is (0), which is rational. Step 3: Test always-type statements using a counterexample.

Step 1: (\sqrt{8}=2\sqrt{2}), (\sqrt{32}=4\sqrt{2}), and (\sqrt{128}=8\sqrt{2}). Step 2: The sum is (2\sqrt{2}+4\sqrt{2}+8\sqrt{2}=14\sqrt{2}). Step 3: With many radicals, simplify all of them first.

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

Step 1: Since (64<65<81), (8<\sqrt{65}<9). Step 2: (65) is not a perfect square, so (\sqrt{65}) is irrational. Step 3: The square root of a number between two perfect squares lies between their roots.

Step 1: A recurring decimal is rational. Step 2: In (0.141414\ldots), the block (14) repeats. Step 3: Identifying the repeating block is the key in decimal questions.

Step 1: Use \((a+b)^2=a^2+2ab+b^2\). Step 2: \((\sqrt{5})^2+2(\sqrt{5})(2)+2^2=5+4\sqrt{5}+4=9+4\sqrt{5}\). Step 3: Missing the middle term \(2ab\) is a common mistake.