Irrational numbers

Step 1: (\sqrt{49}=7). Step 2: (7) is rational, so saying (\sqrt{49}) is irrational is false. Step 3: While choosing a false statement, check perfect squares carefully.

Step 1: Write (150=25 \times 6). Step 2: (\sqrt{150}=\sqrt{25 \times 6}=5\sqrt{6}). Step 3: Take the perfect square factor outside and leave the remaining factor inside.

Step 1: (\sqrt{5}\times\sqrt{20}=\sqrt{100}). Step 2: (\sqrt{100}=10), which is rational. Step 3: In multiplication, first multiply the numbers inside the roots.

Step 1: Since (25<30<36), (5<\sqrt{30}<6). Step 2: (30) is not a perfect square, so (\sqrt{30}) is irrational. Step 3: Use nearby perfect squares to locate a square root.

Step 1: (200=100 \times 2). Step 2: (\sqrt{200}=\sqrt{100 \times 2}=10\sqrt{2}). Step 3: Recognising a large perfect square like (100) gives the answer quickly.

Step 1: Both terms have the same radical (\sqrt{3}). Step 2: (1\sqrt{3}+2\sqrt{3}=3\sqrt{3}). Step 3: For like radicals, add only the coefficients.

Step 1: The value of (\sqrt{2}) is about (1.414). Step 2: Among the given options, (1.41) is closest. Step 3: Remembering approximate values of common roots helps in comparison.

Step 1: (9<10<16). Step 2: So (\sqrt{9}<\sqrt{10}<\sqrt{16}), meaning (3<\sqrt{10}<4). Step 3: Use nearby perfect squares to find the range of a square root.

Step 1: (63=9 \times 7). Step 2: (\sqrt{63}=\sqrt{9 \times 7}=3\sqrt{7}). Step 3: Remember to take a perfect square like (9) outside the root.

Step 1: (y=\sqrt{11}). Step 2: (y^2=(\sqrt{11})^2=11). Step 3: Squaring a square root gives the original number inside it.

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

Step 1: Adding zero does not change a number. Step 2: (\sqrt{5}+0=\sqrt{5}), which is irrational. Step 3: Even with zero, understand the original nature of the number.

Step 1: (6) is not a perfect square. Step 2: So (\sqrt{6}) is irrational and its decimal is non-terminating and non-recurring. Step 3: An irrational number has no fixed repeating pattern.

Step 1: Write (90=9 \times 10). Step 2: (\sqrt{90}=\sqrt{9 \times 10}=3\sqrt{10}). Step 3: Take the perfect square outside and keep the remaining part inside.

Step 1: (12=4 \times 3). Step 2: (\sqrt{12}=2\sqrt{3}). Step 3: To identify an equivalent form, simplify the square root first.

Step 1: The two terms are opposites of each other. Step 2: (\sqrt{5}+(-\sqrt{5})=0). Step 3: The sum of opposite terms is always zero.

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

Step 1: (112=16 \times 7). Step 2: (\sqrt{112}=\sqrt{16 \times 7}=4\sqrt{7}). Step 3: After simplification, check that the remaining number has no perfect square factor.

Step 1: Recurring or terminating decimals are rational. Step 2: A non-terminating decimal without fixed repetition can be irrational. Step 3: Do not judge only by length; check for repetition.

Step 1: Write (147=49 \times 3). Step 2: (\sqrt{147}=\sqrt{49 \times 3}=7\sqrt{3}). Step 3: Recognising the perfect square (49) is the main step here.

Step 1: (\sqrt{2}\times\sqrt{32}=\sqrt{64}). Step 2: (\sqrt{64}=8), so the result is rational. Step 3: In multiplication, multiply the numbers inside the square roots.

Step 1: (\sqrt{3}) is approximately (1.732). Step 2: So (1.73) is the closest correct value. Step 3: For estimation, remember common square-root values.

Step 1: (24=4 \times 6). Step 2: (\sqrt{24}=\sqrt{4 \times 6}=2\sqrt{6}). Step 3: If (6) remains inside, it cannot be simplified further because it has no perfect square factor.

Step 1: (9<13<16). Step 2: Therefore, (3<\sqrt{13}<4). Step 3: Nearby perfect squares are very useful for comparing square roots.

Step 1: (\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}). Step 2: (3\sqrt{2}) is irrational. Step 3: In options, simplify the result before deciding its nature.