Irrational numbers

Step 1: Multiply numerator and denominator by (\sqrt{3}). Step 2: (\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}). Step 3: Rationalizing the denominator gives a cleaner exam answer.

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 integer does not remove the irrational nature of the surd.

Step 1: (0\times\sqrt{7}=0), (\sqrt{7}\times\sqrt{7}=7), and (\sqrt{28}\div\sqrt{7}=2) are rational. Step 2: (4\sqrt{7}) is a non-zero rational multiple of an irrational number, so it is irrational. Step 3: Quickly identify multiplication by zero as rational.

Step 1: A positive integer has a rational square root only when it is a perfect square. Step 2: For example, (\sqrt{16}=4), but (\sqrt{18}) is irrational. Step 3: Check perfect squares to decide the nature of a square root.

Step 1: In the proof for (\sqrt{2}), we assume (\sqrt{2}=\frac{a}{b}). Step 2: This gives (a^2=2b^2), so (a^2) is even and hence (a) is even. Step 3: This parity argument leads to a contradiction.

Step 1: (\sqrt{13}) is irrational. Step 2: Its square is ((\sqrt{13})^2=13), which is rational. Step 3: The square of an irrational number is not always irrational, so examine examples carefully.

Step 1: (72=36\times2). Step 2: (\sqrt{72}=\sqrt{36}\sqrt{2}=6\sqrt{2}), which is irrational. Step 3: Use the largest perfect square factor for quick simplification.

Step 1: The denominator has (\sqrt{5}), which is irrational. Step 2: Rationalizing gives (x=\frac{2\sqrt{5}}{5}), a non-zero rational multiple of an irrational number. Step 3: Rationalizing the denominator often reveals the number type clearly.

Step 1: First look for like irrational terms. Step 2: ((2+\sqrt{3})+(5-\sqrt{3})=7) because (\sqrt{3}) and (-\sqrt{3}) cancel. Step 3: Opposite irrational terms can produce a rational result.

Step 1: From (x^2=7), (x=\sqrt{7}) or (x=-\sqrt{7}). Step 2: Among the options, (\sqrt{7}) is present and it is irrational. Step 3: Remember both positive and negative roots, then match the given options.

Step 1: (\sqrt{9}=3) and (\sqrt{16}=4). Step 2: Their sum is (7), which is rational. Step 3: If both radicands are perfect squares, the sum is easily rational.

Step 1: (3) is rational and (\sqrt{2}) is irrational. Step 2: A rational number minus an irrational number remains irrational. Step 3: Do not classify the whole expression by looking only at the rational part.

Step 1: This is a product of conjugates. Step 2: ((1+\sqrt{2})(1-\sqrt{2})=1-(\sqrt{2})^2=1-2=-1). Step 3: In conjugate multiplication, the middle irrational terms cancel.

Step 1: In (0.123123123\ldots), the block (123) repeats. Step 2: A recurring decimal is rational. Step 3: Do not call a decimal irrational just because it is non-terminating; check repetition.

Step 1: The sum of two irrational numbers can be rational. Step 2: For example, (\sqrt{2}+(-\sqrt{2})=0). Therefore, saying (a+b) is always irrational is false. Step 3: Be careful with universal statements about two irrational numbers.

Step 1: (\sqrt{27}=3\sqrt{3}) and (\sqrt{12}=2\sqrt{3}). Step 2: The sum is (3\sqrt{3}+2\sqrt{3}=5\sqrt{3}), which is irrational. Step 3: Do not combine separate square roots as (\sqrt{39}).

Step 1: Coprime means (p) and (q) have no common factor except (1). Step 2: In the proof of (\sqrt{2}), both (p) and (q) turn out even, so they have common factor (2). Step 3: This contradiction proves that (\sqrt{2}) is not rational.

Step 1: (\sqrt{32}=4\sqrt{2}). Step 2: (\sqrt{32}-\sqrt{2}=4\sqrt{2}-\sqrt{2}=3\sqrt{2}), which is irrational. Step 3: For like surds, subtract only the coefficients.

Step 1: (\sqrt{44}=\sqrt{4\times11}=2\sqrt{11}). Step 2: Hence (x=\sqrt{11}+2\sqrt{11}=3\sqrt{11}), and (\sqrt{11}) is irrational. Step 3: For like surds, add only the coefficients, not the numbers inside the roots.

Step 1: A non-terminating decimal can still be rational if it is recurring. Step 2: In (0.37373737\ldots), the block (37) repeats, so it is rational. Step 3: Do not call a decimal irrational just because it is non-terminating; check for a repeating block.

Step 1: (a) and (b) are conjugates. Step 2: (ab=(\sqrt{3})^2-2^2=3-4=-1), which is rational and negative. Step 3: In conjugate multiplication, the middle irrational terms cancel.

Step 1: Simplify each radical first. Step 2: (\sqrt{75}=5\sqrt{3}), so (\sqrt{75}-4\sqrt{3}=\sqrt{3}), which is irrational. Step 3: Options where like terms cancel completely may give rational zero.

Step 1: If (k) is a perfect square, then (\sqrt{k}) is rational and the fraction becomes rational. Step 2: (18) is not a perfect square, so (\sqrt{18}) is irrational and (\frac{5}{\sqrt{18}}) remains irrational. Step 3: In such questions, first check whether (k) is a perfect square.

Step 1: (\sqrt{5}) and (\sqrt{20}=2\sqrt{5}) are both irrational and different. Step 2: (\frac{\sqrt{5}}{\sqrt{20}}=\frac{\sqrt{5}}{2\sqrt{5}}=\frac{1}{2}), which is rational. Step 3: A common irrational factor can cancel in a quotient.

Step 1: (x-4=(4+\sqrt{6})-4). Step 2: On simplifying, (x-4=\sqrt{6}), and since (6) is not a perfect square, (\sqrt{6}) is irrational. Step 3: When rational terms cancel, check the nature of the remaining radical.