Irrational numbers

Step 1: Since (a) is not a perfect square, (\sqrt{a}) is irrational. Step 2: A sum of two positive square roots could become rational only if irrational parts cancel, but both terms are positive here. Step 3: Without opposite signs, irrational surd parts remain in the sum.

Step 1: (\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}). Step 2: The numerator becomes (5\sqrt{5}), so (\frac{5\sqrt{5}}{\sqrt{5}}=5). Step 3: Before division, convert the numerator surds into like terms.

Step 1: (\sqrt{28}=2\sqrt{7}), so (x=3\sqrt{7}). Step 2: (x^2=(3\sqrt{7})^2=63), hence (\frac{x^2}{7}=9). Step 3: Combine like surds before squaring.

Step 1: Write (\sqrt{20}=2\sqrt{5}) and (\sqrt{45}=3\sqrt{5}). Step 2: (\sqrt{5}+2\sqrt{5}-3\sqrt{5}=0), which is rational. Step 3: Terms that look irrational may cancel to give a rational result.

Step 1: In the proof, assume (\sqrt{p}=\frac{a}{b}) and square both sides. Step 2: From (a^2=pb^2), we get (p\mid a^2), so the idea (p\mid a) is used. Step 3: The prime factor argument leads to a contradiction.

Step 1: This is of the form ((u+v)(u-v)). Step 2: The value is (11-3=8), which is rational. Step 3: Multiplying conjugate surds often removes the irrational part.

Step 1: The conjugate of the denominator is (\sqrt{6}+\sqrt{5}). Step 2: The denominator becomes ((\sqrt{6})^2-(\sqrt{5})^2=6-5=1). Step 3: When the denominator is a difference of two surds, multiply by its conjugate.

Step 1: First check whether a fixed block of digits repeats. Step 2: In (0.120120012000120000\ldots), the number of zeros keeps changing, so there is no fixed repetition. Step 3: A non-terminating non-recurring decimal is irrational.

Step 1: The conjugate of (\sqrt{3}+\sqrt{2}) is (\sqrt{3}-\sqrt{2}). Step 2: The denominator becomes (3-2=1), so (\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}). Step 3: When the difference of the squared surds is (1), the result becomes very simple.

Step 1: (4+\sqrt{7}) and (4-\sqrt{7}) are both irrational. Step 2: Their sum is (8), which is rational. Step 3: In such examples, equal irrational parts cancel with opposite signs.

Step 1: (ab=(\sqrt{8})^2-(\sqrt{18})^2). Step 2: (ab=8-18=-10), which is rational. Step 3: In conjugate multiplication, you do not always need to simplify each radical first.

Step 1: (\sqrt{50}=5\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{98}=7\sqrt{2}). Step 2: (5\sqrt{2}+6\sqrt{2}-7\sqrt{2}=4\sqrt{2}). Step 3: Once all terms are like surds, add or subtract only the coefficients.

Step 1: (\sqrt{4}=2) and (\sqrt{9}=3). Step 2: Their sum is (2+3=5). Step 3: To get a rational integer sum, check perfect squares first.

Step 1: (\sqrt{3}) is about (1.732), so it lies between (1) and (2). Step 2: Since (3>2), (\sqrt{3}>\sqrt{2}). Step 3: For positive square roots, compare the numbers inside the roots.

Step 1: Use ((a-b)^2=a^2-2ab+b^2). Step 2: (x^2=10-2\sqrt{20}+2=12-4\sqrt{5}). Step 3: In the middle term (-2ab), write both the sign and the surd carefully.

Step 1: ((\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}), which does not match the given expression. Step 2: The expression (2\sqrt{3}+3\sqrt{2}) does not directly match any listed square form. Step 3: Always expand and match, not guess by appearance.

Step 1: ((\sqrt{3}+\sqrt{2})^2=3+2+2\sqrt{6}). Step 2: This equals (5+2\sqrt{6}). Step 3: When squaring a sum of two surds, the middle term becomes (2\sqrt{6}).

Step 1: (\sqrt{a}) is irrational only when (a) is not a perfect square. Step 2: (36) is a perfect square and (\sqrt{36}=6), so it is not suitable. Step 3: Quickly identify perfect squares among the options.

Step 1: (\sqrt{12}=2\sqrt{3}) is irrational. Step 2: Its reciprocal (\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}) is also irrational. Step 3: Do not assume the reciprocal of a non-zero irrational surd is rational.

Step 1: View the product as ((\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})). Step 2: It gives (5-2=3). Step 3: You can rearrange the order of addition to recognize a conjugate form.