Irrational numbers

Step 1: (\sqrt{8}=2\sqrt{2}), (\sqrt{32}=4\sqrt{2}), and (\sqrt{128}=8\sqrt{2}). Step 2: The sum is ((1+2+4+8)\sqrt{2}=15\sqrt{2}). Step 3: Recognize the pattern of perfect-square factors.

Step 1: Assuming (\sqrt{5}=\frac{p}{q}) and squaring gives (p^2=5q^2). Step 2: This makes both (p) and (q) divisible by (5), contradicting that they are coprime. Step 3: Finding a common factor is the key contradiction in the proof.

Step 1: Both numbers are positive, so compare their squares. Step 2: ((4\sqrt{3})^2=48) and ((3\sqrt{5})^2=45), so (4\sqrt{3}) is greater. Step 3: Squaring is safe for comparing positive surds.

Step 1: (x^2=11+7+2\sqrt{77}=18+2\sqrt{77}). Step 2: Therefore (x^2-18=2\sqrt{77}), which is irrational. Step 3: In the square of a sum of different surds, the middle term is the key.

Step 1: (\sqrt{12}=2\sqrt{3}) and (\sqrt{3}) are both irrational. Step 2: Their product is (\sqrt{36}=6), which is rational, and their sum is (3\sqrt{3}), which is irrational. Step 3: Check the nature of the sum and product separately.

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

Step 1: The conjugate of the denominator is (\sqrt{5}-\sqrt{2}). Step 2: The denominator becomes (5-2=3), so the form is (\frac{\sqrt{5}-\sqrt{2}}{3}). Step 3: For a sum of two surds, the conjugate changes the sign between them.

Step 1: In the square of three terms, pairwise products appear along with individual squares. Step 2: Thus (a^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}). Step 3: While squaring a sum of many surds, write all pairwise products.

Step 1: (25) and (49) are both perfect squares. Step 2: (\sqrt{25}+\sqrt{49}=5+7=12), which is rational. Step 3: For a rational sum, check both square roots separately.

Step 1: (x^2=5+2\sqrt{6}). Step 2: Using the identity (x^2+\frac{1}{x^2}=10), we get (x^4-10x^2+1=0). Step 3: In such questions, recognize the relation between (x) and its conjugate reciprocal.

Step 1: (\sqrt{a}\times\sqrt{b}=\sqrt{ab}). Step 2: For (a=6,b=15), (ab=90), which is not a perfect square, so (\sqrt{90}) is irrational. Step 3: In multiplication, the key check is whether the product inside the root is a perfect square.

Step 1: ((5-\sqrt{24})(5+\sqrt{24})=25-24=1). Step 2: Therefore (5+\sqrt{24}) is the reciprocal of (5-\sqrt{24}). Step 3: If conjugates multiply to (1), the reciprocal is directly the conjugate.

Step 1: A non-zero rational multiplier does not remove irrationality. Step 2: If the product were rational, the irrational number would become rational, a contradiction. Step 3: Be careful with universal statements about two irrational numbers.

Step 1: (\sqrt{8}=2\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Step 2: (x=5\sqrt{2}), so (\frac{x}{\sqrt{2}}=5). Step 3: Division is easier after combining like surds.

Step 1: All terms are positive and (\sqrt{6}>0). Step 2: Since (\sqrt{12}=2\sqrt{3}) and (\sqrt{6}>\sqrt{3}), the sum (\sqrt{3}+\sqrt{6}) is greater than (2\sqrt{3}). Step 3: For comparison, convert what you can and use positivity.

Step 1: (a^2-6a=a(a-6)). Step 2: (a-6=\sqrt{5}-3), so (a(a-6)=(3+\sqrt{5})(\sqrt{5}-3)=5-9=-4). Step 3: Recognize the hidden conjugate form.

Step 1: Reappearance of a digit alone is not recurrence unless a fixed block repeats. Step 2: Here the number of zeros changes, so the decimal is non-recurring. Step 3: Identify a non-terminating non-recurring decimal as irrational.

Step 1: This is a conjugate product. Step 2: ((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=13-12=1). Step 3: In such forms, identify the difference of squares before expanding.

Step 1: (\sqrt{25}=5) and (\sqrt{9}=3). Step 2: The difference is (5-3=2), which is rational. Step 3: A simple way to get a rational difference is to use square roots of perfect squares.

Step 1: (x^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}). Step 2: Subtracting (2\sqrt{6}) leaves (5). Step 3: After squaring, cancel like irrational terms.