Irrational numbers and real numbers

A rational number can be expressed as a ratio of two integers. \(\frac{15}{2}=7.5\) lies between 7 and 8 and is explicitly a ratio of integers, so it is rational. The other choices are irrational: \(\sqrt{53}\) and \(\sqrt{60}\) are square roots of non-perfect squares, and \(\pi+4\) is irrational because \(\pi\) is irrational. The closest distractor is \(\sqrt{53}\) (≈7.2801) but it remains irrational. Exam tip: check if a number can be written as m/n with integers m,n; if not, test for perfect squares or known irrational constants to rule out rationality quickly.

Since \(2^2=4\) and \(3^2=9\), we have \(4<7<9\). Hence \(2<\sqrt{7}<3\), so \(\sqrt{7}\) lies between the integers 2 and 3. The closest distractor C (3, 4) is wrong because \(\sqrt{7}\) is less than 3; B and D are evidently incorrect. Exam tip: compare the given number with consecutive perfect squares to find the integer bounds for its square root quickly.

The square-root function is increasing for non‑negative real numbers: if \(a>b\ge0\) then \(\sqrt{a}>\sqrt{b}\). Since \(99>90\), we have \(\sqrt{99}>\sqrt{90}\). Numerically \(\sqrt{90}\approx9.4868\) and \(\sqrt{99}\approx9.9499\). Option B reverses the order, C is false because the radicands are different, and D is wrong because the comparison is definite. Exam tip: either compare the radicands directly or compute approximate square roots to decide quickly.

\(\sqrt{0.64}=0.8\) and \(\sqrt{0.09}=0.3\). Therefore the sum is \(0.8+0.3=1.1\), so 1.1 is correct. The closest distractor 0.89 is incorrect because it is 0.21 less than the correct sum. Exam tip: check decimal square roots by converting to fractions (e.g. \(0.64=64/100\)) or by remembering that \(0.8^2=0.64\) and \(0.3^2=0.09\) to avoid place‑value mistakes.

\(\frac{\sqrt{80}}{\sqrt{5}}=\sqrt{\frac{80}{5}}=\sqrt{16}=4\). Alternatively, since \(\sqrt{80}=4\sqrt{5}\), canceling \(\sqrt{5}\) gives \(\frac{4\sqrt{5}}{\sqrt{5}}=4\). Option C (\(2\sqrt{5}\)) equals \(\sqrt{80}\), not the quotient, so it is incorrect; B and D are also incorrect numerical values. Exam tip: simplify the expression inside the radical or factor out perfect squares before dividing to avoid mistakes.

(\frac{3}{\sqrt{3}}=\sqrt{3}) because (3=\sqrt{3}\times\sqrt{3}). Simplify when a root is in the denominator.

Distribute: \(\sqrt{3}(4+\sqrt{3})=4\sqrt{3}+\sqrt{3}\cdot\sqrt{3}=4\sqrt{3}+3\). So the simplified form is \(3+4\sqrt{3}\). Option (B) swaps the rational and irrational coefficients and is therefore incorrect; (C) and (D) are pure surd forms that would result from mistaken addition or multiplication of terms. Exam tip: multiply each term separately, simplify \(\sqrt{a}\cdot\sqrt{a}=a\), then combine rational and irrational parts.

\(2\sqrt{5}\times 3\sqrt{5}=(2\times3)\times(\sqrt{5}\times\sqrt{5})=6\times5=30\). Multiplying like radicals uses \(\sqrt{a}\cdot\sqrt{a}=a\), so the product is rational. The nearest distractor, \(6\sqrt{5}\), arises from multiplying coefficients only and forgetting to simplify the radical part. Exam tip: for products of the form \((a\sqrt{b})(c\sqrt{b})\), compute \(ac\) and use \(\sqrt{b}\cdot\sqrt{b}=b\) to simplify.

Like surds (same radicand) combine by adding their coefficients. The coefficients here are 5, 3 and −1, so 5+3−1 = 7, giving \\(7\sqrt{2}\\). Option B (\\(9\sqrt{2}\\)) would result from mistakenly treating −1 as +1. Option C (\\(7\sqrt{3}\\)) is wrong because the radicand changed to 3 — you cannot change the radicand when combining like surds. Option D (\\(\sqrt{14}\\)) reflects an incorrect multiplication of roots instead of combining coefficients. Exam tip: always check the radicands first — only combine terms with identical radicands by adding/subtracting their coefficients.

(\sqrt{50}=5\sqrt{2}), (\sqrt{72}=6\sqrt{2}), and (\sqrt{32}=4\sqrt{2}). The result is (7\sqrt{2}).

(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}). In simplest form no perfect square should remain inside the root.

Let \(x=0.\overline{123}\). Then \(1000x=123.\overline{123}\). Subtracting gives \(999x=123\), so \(x=\dfrac{123}{999}=\dfrac{41}{333}\). Since it equals a ratio of integers, it is rational. Option B is wrong because the decimal is repeating and converts to a fraction; C is wrong because the number is a real decimal; D is wrong because the value lies between 0 and 1, not an integer. Exam tip: For a repeating block of length n, use n nines in the denominator (e.g., 3-digit repeat → denominator 999).

In 3.010010001... the number of zeros between successive 1s increases each time (1, 2, 3, ...). Thus there is no fixed repeating block — after any chosen block length the pattern does not repeat. A rational number’s decimal expansion is either terminating or eventually periodic (repeating). Since this decimal is neither terminating nor periodic, it is irrational. The closest distractor C (a repeating decimal) is incorrect because a repeating decimal requires a fixed period of digits repeating indefinitely, which this number does not have. Exam tip: check whether a decimal has a fixed repeating block or terminates; if neither holds and the pattern keeps changing (like increasing zero runs), the number is irrational.

The square root of a positive integer is rational only when it is a perfect square. Apply this rule directly.

If a positive integer is not a perfect square its square root is irrational. So (k) is not a perfect square.

(\sqrt{289}=17), but (290) is not a perfect square. So the first root is rational and the second is irrational.

\(\sqrt[3]{125}=5\), which is an integer and can be written as \(5=5/1\); therefore it is a rational number. The cube root of a perfect cube is an integer (hence rational). The closest distractor B (irrational) is incorrect because irrational numbers have non‑terminating, non‑repeating decimal expansions, whereas 5 is a terminating decimal. C is wrong because the number is real, not non‑real. D is incorrect because a "non‑repeating decimal" typically describes irrationals, but 5 has a terminating decimal. Exam tip: first check whether the radicand is a perfect power—if so, the root will be an integer (rational).

If \(\sqrt[3]{12}\) were rational it could be written as a reduced fraction \(\dfrac{p}{q}\) with integers p,q; cubing gives \(p^3=12q^3\). That forces the prime exponents in \(p^3\) to match those of 12q^3, so the exponents of primes coming from 12 must be multiples of 3. But 12=2^2·3^1 has exponents 2 and 1, which are not multiples of 3, a contradiction. Hence \(\sqrt[3]{12}\) is irrational. Distractors fail: 3 and 4 are wrong because 3^3=27 and 4^3=64 (not 12); a terminating decimal would be rational. Exam tip: first check whether the radicand is a perfect cube; use prime-power exponents to test rationality of integer roots.

(45) is not a perfect square so (\sqrt{45}) is irrational. Adding rational (2) keeps it irrational.

(\sqrt{100}=10) is rational so (9-\sqrt{100}) will be rational. The root of a perfect square is rational.