Irrational numbers

Step 1: Assume (\sqrt{2}+\sqrt{3}) is rational. Step 2: Squaring gives (5+2\sqrt{6}) rational, which would force (\sqrt{6}) to be rational, impossible. Step 3: Squaring is useful for sums of two different surds.

Step 1: (\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}). Step 2: Therefore (\frac{2}{x}=\sqrt{5}-\sqrt{3}). Step 3: (x-\frac{2}{x}=(\sqrt{5}+\sqrt{3})-(\sqrt{5}-\sqrt{3})=2\sqrt{3}).

Step 1: For (a=4), (\sqrt{4}=2). Step 2: For (b=18), (\sqrt{18}=3\sqrt{2}), which is irrational; so the sum (2+3\sqrt{2}) is irrational. Step 3: A rational plus an irrational remains irrational.

Step 1: (x^2=2+7+2\sqrt{14}=9+2\sqrt{14}). Step 2: Therefore (x^2-9=2\sqrt{14}), which is irrational. Step 3: Square first, then subtract the rational part.

Step 1: (12=4\times3). Step 2: (\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}). Step 3: Take the perfect square factor outside the radical.

Step 1: (x+1=\sqrt{3}). Step 2: Therefore ((x+1)^2=(\sqrt{3})^2=3). Step 3: Simplify the inner expression first, then square it.

Step 1: Both numbers are positive, so compare their squares. Step 2: ((2\sqrt{3})^2=12) and ((3\sqrt{2})^2=18), so (3\sqrt{2}) is greater. Step 3: Squaring is a safe method for comparing positive surds.

Step 1: (xy=2\sqrt{5}\times5\sqrt{2}=10\sqrt{10}). Step 2: (\sqrt{10}) is irrational, so (10\sqrt{10}) is irrational. Step 3: If the product inside the root is not a perfect square, the result may remain irrational.

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

Step 1: Write (x^2-4x=x(x-4)). Step 2: With (x=\sqrt{13}+2), (x-4=\sqrt{13}-2), so the product is (13-4=9). Step 3: A conjugate form may be hidden in such expressions.

Step 1: Multiply by (2+\sqrt{3}) to rationalize the denominator. Step 2: (\frac{(2+\sqrt{3})^2}{4-3}=4+4\sqrt{3}+3=7+4\sqrt{3}). Step 3: When multiplying by the conjugate, the numerator may become a full square.

Step 1: ((2+\sqrt{3})^2=4+4\sqrt{3}+3). Step 2: This equals (7+4\sqrt{3}). Step 3: In such questions, identify the form (m+n+2\sqrt{mn}).

Step 1: (\sqrt{80}=4\sqrt{5}), (\sqrt{45}=3\sqrt{5}), and (\sqrt{20}=2\sqrt{5}). Step 2: (4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}), so none of the listed options is correct. Step 3: In such questions, trust your simplification before matching options.

Step 1: (\sqrt{80}=4\sqrt{5}), (\sqrt{45}=3\sqrt{5}), and (\sqrt{20}=2\sqrt{5}). Step 2: (4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}), which is irrational. Step 3: Handle the signs carefully when three terms are involved.

Step 1: To make (x+\sqrt{2}) rational, (x) should contain a (-\sqrt{2}) part. Step 2: If (x=3-\sqrt{2}), then (x+\sqrt{2}=3), which is rational. Step 3: Look for cancellation of the irrational part.

Step 1: ((u+v)^2-(u-v)^2=4uv). Step 2: Here (u=\sqrt{7}) and (v=\sqrt{2}), so the value is (4\sqrt{14}). Step 3: Using the identity makes the expansion shorter.

Step 1: Rationalize the denominator of (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}). Step 2: The numerator becomes ((\sqrt{6}+\sqrt{2})^2=8+4\sqrt{3}), and the denominator is (6-2=4), so the value is (2+\sqrt{3}). Step 3: Multiplying by the conjugate is effective in such quotients.

Step 1: (\sqrt{11}) is about (3.316), and (2\sqrt{3}) is about (3.464). Step 2: (3.4) lies between them. Step 3: For close values, estimating to two decimal places is helpful.

Step 1: (x^2=2). Step 2: Therefore (x^4=(x^2)^2=4), and the value is (4-8+4=0). Step 3: For powers of a surd, first find (x^2).

Step 1: From (p^2=2q^2), first (p^2) is even and hence (p) is even. Step 2: After writing (p=2k), we get (q^2=2k^2), so (q) is even. Step 3: Skipping this order makes the proof incomplete.

Step 1: (a^2-2a=a(a-2)). Step 2: (a-2=\sqrt{5}-1), so (a(a-2)=(1+\sqrt{5})(\sqrt{5}-1)=4). Step 3: Recognizing the hidden conjugate form is a quick method.

Step 1: (\sqrt{48}=4\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Step 2: (4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}). Step 3: For like surds, work with the coefficients.

Step 1: (x-\sqrt{3}=\sqrt{2}) and (x-\sqrt{2}=\sqrt{3}). Step 2: Their product is (\sqrt{2}\times\sqrt{3}=\sqrt{6}). Step 3: Simplify the small brackets first.

Step 1: (\sqrt{8}=2\sqrt{2}), (\sqrt{18}=3\sqrt{2}), and (\sqrt{32}=4\sqrt{2}). Step 2: The total is (1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}=10\sqrt{2}). Step 3: In ordered surds, identify the coefficient pattern.

Step 1: First observe the common structure and take (a=\sqrt{7}+\sqrt{5}) and (b=\sqrt{7}-\sqrt{5}). Step 2: (\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}). Here (a^2+b^2=24) and (ab=2), so (x=12). Step 3: For fractions with conjugate surds, use substitution instead of expanding everything directly.