Irrational numbers

Step 1: Multiplying an irrational number by a non-zero rational number keeps it irrational. Step 2: If (pq) were rational, then (q=\frac{pq}{p}) would be rational, which contradicts the given condition. Step 3: Always check that the rational multiplier is not zero.

Step 1: (\sqrt{8}=2\sqrt{2}). Step 2: So (a=\sqrt{2}+2\sqrt{2}=3\sqrt{2}), and (\sqrt{2}) is irrational. Step 3: Simplify like radical terms before deciding the type of number.

Step 1: The square root of a perfect square is an integer. Step 2: If (n) is not a perfect square, then (\sqrt{n}) is irrational. Step 3: Do not judge only by evenness or primality; check whether it is a perfect square.

Step 1: Simplify each surd carefully. Step 2: (\sqrt{50}-\sqrt{8}=3\sqrt{2}), (\sqrt{18}-\sqrt{2}=2\sqrt{2}), (\sqrt{27}-\sqrt{3}=2\sqrt{3}), and (\sqrt{3}+\sqrt{12}=3\sqrt{3}), all are irrational. Step 3: This item has no rational option, so it should be treated as an invalid question.

Step 1: (\sqrt{5}) is irrational and (2) is rational. Step 2: Subtracting a rational number from an irrational number gives an irrational number. Step 3: When an integer is subtracted from a surd, focus on the nature of the surd.

Step 1: Both numbers are individually irrational. Step 2: (\sqrt{5}\times\sqrt{20}=\sqrt{100}=10), which is rational. Step 3: Check whether the product inside the square root becomes a perfect square.

Step 1: (3) is rational. Step 2: If (x+3) were rational, then (x=(x+3)-3) would also be rational, contradicting the given fact. Step 3: The contradiction method is useful in such questions.

Step 1: (45=9\times5). Step 2: (\sqrt{45}=\sqrt{9}\sqrt{5}=3\sqrt{5}), and (\sqrt{5}) is irrational. Step 3: Separate the largest perfect square factor while simplifying surds.

Step 1: (\sqrt{5}) and (-\sqrt{5}) are both irrational. Step 2: Their sum is (0), which is rational. Step 3: Before applying a general rule for two irrationals, test possible counterexamples.

Step 1: The square root of a positive integer is rational when that integer is a perfect square. Step 2: Hence (mn) being a perfect square is sufficient for (\sqrt{mn}) to be rational. Step 3: Parity of the sum or difference does not decide the nature of the square root.

Step 1: ((\sqrt{2})^2=2), (\sqrt{2}\times\sqrt{8}=4), and (\sqrt{5}\times\sqrt{20}=10) are rational. Step 2: (\sqrt{3}+\sqrt{12}=3\sqrt{3}), which is irrational. Step 3: Treat addition and multiplication of surds differently.

Step 1: Assume (2+\sqrt{7}) is rational. Step 2: Then (\sqrt{7}=(2+\sqrt{7})-2) would be rational, but (\sqrt{7}) is irrational. Step 3: In contradiction proofs, isolate the surd using rational operations.

Step 1: Add the coefficients of like surds. Step 2: (\sqrt{3}+2\sqrt{3}=3\sqrt{3}), and (\sqrt{3}) is irrational. Step 3: Coefficients add; the number inside the radical remains unchanged.

Step 1: Non-terminating decimals can be recurring or non-recurring. Step 2: A non-terminating recurring decimal like (0.\overline{3}) is rational. Step 3: For an irrational decimal, it must be both non-terminating and non-recurring.

Step 1: A rational number minus an irrational number is irrational. Step 2: If (r-s) were rational, then (s=r-(r-s)) would be rational, which is impossible. Step 3: Use the same reasoning for subtraction as for addition.

Step 1: (\sqrt{2}) is irrational. Step 2: Multiplying the two irrational numbers gives (\sqrt{2}\times\sqrt{2}=2), which is rational. Step 3: One valid counterexample is enough to disprove a universal statement.

Step 1: The decimal does not terminate. Step 2: It has no fixed recurring block because the number of zeros keeps changing. Step 3: A non-terminating non-recurring decimal is irrational.

Step 1: The square of the square root of a positive number gives the number itself. Step 2: Hence ((\sqrt{5})^2=5), which is rational. Step 3: Do not mistake (\sqrt{5}+\sqrt{5}) for (10).

Step 1: Square both sides. Step 2: (a=(4\sqrt{3})^2=16\times3=48). Step 3: Remember that ((k\sqrt{n})^2=k^2n).

Step 1: (2\sqrt{6}) can be written as (\sqrt{4}\sqrt{6}). Step 2: Therefore (2\sqrt{6}=\sqrt{24}). Step 3: When moving a coefficient inside a square root, square the coefficient.