Irrational numbers

Multiplying by the conjugate makes the denominator (6-1=5). So the form is (\frac{5(\sqrt{6}+1)}{5}), that is (\sqrt{6}+1).

(\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the sum is (5\sqrt{3}). Its square is (75).

A decimal represents an irrational number only when it is non-terminating and non-repeating. A terminating decimal ends after a finite number of digits, while a repeating decimal has a fixed block of digits that continues again and again. Both types represent rational numbers because they can be written as fractions.

The first decimal repeats 25, the second terminates, and the third repeats 251. The fourth decimal continues forever but does not have one fixed repeating block: its zero groups and digits do not follow a constant cycle. Therefore it can represent an irrational number. Hence option D is correct. The important test is non-termination together with non-repetition.

The student’s rule is false: a quotient of two irrational numbers need not be irrational. Here \\(\sqrt{18}/\sqrt2=\sqrt9=3\\), so the result is rational. Exam tip: combine square roots first, then simplify.

\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(2\sqrt{2}\). Multiplying by \(\sqrt{2}\) gives (4).

Multiplying by the conjugate makes the denominator (7-5=2). So (\frac{2(\sqrt{7}-\sqrt{5})}{2}=\sqrt{7}-\sqrt{5}).

The governing concept is simplifying a radical before performing subtraction. Since 12 = 4×3, √12 = √(4×3) = 2√3. Substituting this equivalent form into the expression gives x = 2√3 − 2√3. The two terms are identical and have opposite signs, so they cancel exactly: x = 0. Therefore option A is correct. Option B leaves one radical without justification, option C effectively adds the terms instead of subtracting them, and option D has no valid algebraic basis. The important step is not to treat √12 as an unrelated radical; it must first be rewritten using its perfect-square factor.

The expression contains the same square root multiplied by itself. For a nonnegative number, the principal square root satisfies \(\sqrt{x}\times\sqrt{x}=x\). Here the quantity inside the root is \(3+\sqrt{2}\), which is positive, so the rule applies directly. There is no need to expand the nested radical or approximate its value.

Applying the rule gives \(\sqrt{3+\sqrt{2}}\times\sqrt{3+\sqrt{2}}=3+\sqrt{2}\). Thus option A is correct. Option B incorrectly treats the expression as if the terms 3 and \(\sqrt{2}\) were separately squared and then added; that is not what the given product means. The result remains an exact expression, not a decimal approximation.

(\sqrt{98}=7\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{8}=2\sqrt{2}). Therefore the result is (11\sqrt{2}).

If \(x+r\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence A is correct. But \(x^2\) can be rational; for example, \((\sqrt{2})^2=2\). Exam tip: adding a rational number preserves irrationality.

(\sqrt{27}=3\sqrt{3}) and (\sqrt{12}=2\sqrt{3}), so (s=5\sqrt{3}). Dividing by (\sqrt{3}) gives (5).

Multiplying by the conjugate gives (\frac{(\sqrt{3}-1)^2}{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}). Rationalise the denominator.

\(\sqrt{3}\times\sqrt{12}=\sqrt{36}=6\), which is rational although both factors are irrational. Hence the claim is false. Exam tip: combine radicals first before deciding rationality.

To simplify a quotient of square roots with positive radicands, combine them as \(\sqrt{a}\div\sqrt{b}=\sqrt{a/b}\). Then check whether the resulting number is a perfect square or has a remaining square-free factor. A result is rational when the radical simplifies completely to an integer or rational number. It is irrational when a non-square factor remains inside the square root.

The first three results are \(\sqrt{32/2}=\sqrt{16}=4\), \(\sqrt{18/2}=\sqrt{9}=3\), and \(\sqrt{45/5}=\sqrt{9}=3\). The fourth gives \(\sqrt{20/2}=\sqrt{10}\), and 10 is not a perfect square, so \(\sqrt{10}\) is irrational. Therefore option D is the result that is not rational.

\(\sqrt{2}\) is irrational, but \((\sqrt{2})^2=2\), which is rational. Hence Riya’s statement is false. For instance, \((1+\sqrt{2})^2=3+2\sqrt{2}\) is still irrational, but one counterexample is enough to disprove a universal claim. In exams, test such claims with simple surds first.

Area is ((3+\sqrt{2})(3-\sqrt{2})=9-2=7). Multiplying conjugate dimensions can give a rational area.

(\sqrt{200}=10\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{18}=3\sqrt{2}). Therefore the result is (5\sqrt{2}).

The governing concept is rationalisation by using conjugate pairs. For the first fraction, multiply by √3 − √2. Its denominator becomes (√3 + √2)(√3 − √2) = 3 − 2 = 1, so the fraction equals √3 − √2. For the second fraction, multiply by √3 + √2; its denominator is again 3 − 2 = 1, so it equals √3 + √2. Adding gives (√3 − √2) + (√3 + √2) = 2√3 because the √2 terms cancel. Hence option A is correct. Option B retains the wrong radical, option C ignores the remaining terms, and option D incorrectly combines the radicands.

Since \(\sqrt{45}=3\sqrt{5}\), we get \(\sqrt{45}+\sqrt{5}=4\sqrt{5}\). As \(\sqrt{5}\) is irrational, its non-zero rational multiple is also irrational. Exam tip: never replace \(\sqrt{a}+\sqrt{b}\) with \(\sqrt{a+b}\).

If \(x+1\) were rational, subtracting 1 would make \(x\) rational, which is a contradiction. Hence A is irrational. \(x-x=0\), \(x/x=1\), and \(x^2\) can be rational. Exam tip: check claims containing “always”.