Irrational numbers

Step 1: (a) is rational and non-zero. Step 2: If (a\sqrt{3}) were rational then (\sqrt{3}) would also become rational which is false. Step 3: In exams always check the special case of multiplication by zero.

Step 1: Simplify (\sqrt{8}=2\sqrt{2}). Step 2: (\sqrt{2}+\sqrt{8}=3\sqrt{2}) and (\sqrt{2}) is irrational. Step 3: Do not choose the answer before simplifying square roots.

Step 1: The square root of a positive integer is rational only when the integer is a perfect square. Step 2: For example (\sqrt{25}=5). Step 3: The square root of a prime number is usually irrational.

Step 1: Suppose (\sqrt{2}+\sqrt{3}) is rational. Step 2: Squaring gives (5+2\sqrt{6}) so (\sqrt{6}) would be rational which is false. Step 3: Do not decide the sum of two different irrational numbers without reasoning.

Step 1: The square root of a perfect square is rational. Step 2: (\sqrt{49}=7) so it is rational. Step 3: First look for perfect-square factors inside the radical.

Step 1: From (\sqrt{2}=\frac{p}{q}) we get (p^2=2q^2). Step 2: This makes (p) even and then (q) even. Step 3: Coprime numbers cannot both be even so the assumption is false.

Step 1: (\sqrt{5}) is not the square root of a perfect square. Step 2: So it is irrational and its decimal expansion is non-terminating and non-recurring. Step 3: While using decimal form check whether repetition exists.

Step 1: (\frac{3}{5}) is a non-zero rational number. Step 2: Multiplying it by (\sqrt{13}) gives an irrational number. Step 3: In products check first whether square roots combine to a perfect square.

Step 1: Since (8=4\cdot 2) we have (\sqrt{8}=2\sqrt{2}). Step 2: (\sqrt{2}) is irrational and its double is also irrational. Step 3: Compare like radicals after simplifying them.

Step 1: Dividing by a non-zero rational number is the same as multiplying by its reciprocal. Step 2: An irrational number multiplied by a non-zero rational number remains irrational. Step 3: Convert division questions into multiplication for easier reasoning.

Step 1: (\sqrt{18}=3\sqrt{2}) and (\sqrt{8}=2\sqrt{2}). Step 2: The difference is (\sqrt{2}) which is irrational. Step 3: Do not subtract the numbers inside square roots directly.

Step 1: Rational numbers have terminating or recurring decimals. Step 2: Irrational numbers have non-terminating and non-recurring decimals. Step 3: If a repeating block is visible the number may be rational.

Step 1: (\sqrt{5}) is irrational. Step 2: (\sqrt{5}\cdot \sqrt{5}=5) which is rational. Step 3: The product of two identical irrational square roots becomes the number inside.

Step 1: (3) is rational and (\sqrt{2}) is irrational. Step 2: The sum of a rational and an irrational number is irrational. Step 3: Adding an irrational number to an integer does not give an integer.

Step 1: If (\frac{1}{\sqrt{3}}) were rational then its reciprocal (\sqrt{3}) would be rational. Step 2: (\sqrt{3}) is irrational so the given number is irrational. Step 3: A denominator with an irrational radical does not make the value rational automatically.

Step 1: (48=16\cdot 3). Step 2: (\sqrt{48}=4\sqrt{3}) and (\sqrt{3}) is irrational. Step 3: The square root of an even number need not be rational.

Step 1: (\sqrt{18}) and (\sqrt{2}) are both irrational. Step 2: (\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3) which is rational. Step 3: In quotients check whether the ratio inside the radical becomes a perfect square.

Step 1: Suppose (5-\sqrt{6}) is rational. Step 2: Then (\sqrt{6}=5-) that rational number so (\sqrt{6}) would be rational. Step 3: In contradiction proofs identify the result that clashes with a known fact.

Step 1: (4<5<9). Step 2: Therefore (2<\sqrt{5}<3) and since (5) is not a perfect square (\sqrt{5}) is irrational. Step 3: Use squares to locate irrational square roots between integers.

Step 1: Like terms give (\sqrt{a}+\sqrt{a}=2\sqrt{a}). Step 2: Since (a) is not a perfect square (\sqrt{a}) is irrational and its double is irrational. Step 3: Add like radicals like algebraic terms.