Irrational numbers

Step 1: (\sqrt{20}=2\sqrt{5}), so (x=3\sqrt{5}). Step 2: (x^2=(3\sqrt{5})^2=9\times5=45). Step 3: Simplify surd terms before squaring.

Step 1: (\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}). Step 2: The difference is (7\sqrt{2}-5\sqrt{2}=2\sqrt{2}), which is irrational. Step 3: For like surds, subtract only the coefficients.

Step 1: Assuming (\sqrt{3}=\frac{p}{q}) gives (p^2=3q^2). Step 2: This makes both (p) and (q) divisible by (3), contradicting that they are coprime. Step 3: In such proofs, finding a common factor creates the contradiction.

Step 1: (\sqrt{12}=2\sqrt{3}) and (\sqrt{3}) are both irrational. Step 2: (\frac{\sqrt{12}}{\sqrt{3}}=\sqrt{4}=2), which is rational. Step 3: In quotients, check whether the value inside the root becomes a perfect square.

Step 1: A non-terminating decimal can also be recurring. Step 2: For example, (0.\overline{6}) is non-terminating but rational. Step 3: For irrational decimals, non-repetition is also necessary.

Step 1: (\sqrt{8}=2\sqrt{2}) and (\sqrt{32}=4\sqrt{2}). Step 2: (y=6\sqrt{2}), so (\frac{y}{\sqrt{2}}=6). Step 3: First add like surds, then divide.

Step 1: (2\sqrt{15}=\sqrt{4}\sqrt{15}). Step 2: Therefore (2\sqrt{15}=\sqrt{60}). Step 3: When moving a coefficient inside a square root, its square goes inside.

Step 1: (\sqrt{52}=2\sqrt{13}). Step 2: (a=\sqrt{13}-2\sqrt{13}=-\sqrt{13}), which is irrational. Step 3: A negative sign does not change irrationality.

Step 1: ((\sqrt{11}+1)(\sqrt{11}-1)) is a conjugate product. Step 2: Its value is (11-1=10), which is rational. Step 3: In conjugate forms, irrational terms can cancel.

Step 1: (0) is rational. Step 2: (x+0=x), so the nature remains the same and it is irrational. Step 3: Adding zero does not change either the value or the type of a number.

Step 1: (\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}). Step 2: (\sqrt{5}+3\sqrt{5}-2\sqrt{5}=2\sqrt{5}). Step 3: In questions with many radicals, first convert all terms to like surds when possible.

Step 1: (\sqrt{19}) is irrational because (19) is not a perfect square. Step 2: ((\sqrt{19})^2=19), which is rational. Step 3: The square of an irrational number can sometimes be rational.

Step 1: An irrational decimal neither terminates nor has a fixed repeating block. Step 2: Digit groups with changing lengths do not form a fixed repetition. Step 3: Once a fixed repetition appears, the decimal becomes rational.

Step 1: (\sqrt{24}=2\sqrt{6}). Step 2: So (x=\sqrt{6}+2\sqrt{6}=3\sqrt{6}), which is irrational. Step 3: Simplify radicals to like terms before adding.

Step 1: This decimal does not terminate. Step 2: The number of (5)'s keeps increasing, so no fixed repeating block is formed. Step 3: A non-terminating non-recurring decimal is irrational.

Step 1: Square both sides. Step 2: (a=(5\sqrt{2})^2=25\times2=50). Step 3: Apply ((k\sqrt{m})^2=k^2m) correctly.

Step 1: (\sqrt{3}) and (2\sqrt{3}) are both irrational. Step 2: Their sum is (3\sqrt{3}), which is irrational. Step 3: In sum questions, identify whether like surds cancel or combine.

Step 1: The conjugate of the denominator in (\frac{1}{2-\sqrt{3}}) 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: Multiplying by the conjugate removes the radical from the denominator.

Step 1: (49), (81), and (121) are perfect squares. Step 2: (90) is not a perfect square, so (\sqrt{90}) is irrational. Step 3: To decide the nature of a square root, first check perfect squares.

Step 1: Use the distributive law. Step 2: (\sqrt{3}(2+\sqrt{3})=2\sqrt{3}+(\sqrt{3})^2=2\sqrt{3}+3). Step 3: Remember that (\sqrt{3}\times\sqrt{3}=3).