Irrational numbers and real numbers

(\sqrt{19}) is irrational because (19) is not a perfect square. Subtracting an irrational from a rational gives an irrational result.

Simplify each term: \sqrt{125}=\sqrt{25\times5}=5\sqrt{5} and \sqrt{45}=\sqrt{9\times5}=3\sqrt{5}. Adding like surds gives 5\sqrt{5}+3\sqrt{5}=8\sqrt{5}, so A is correct. Option B (\sqrt{170}) reflects the incorrect assumption \sqrt{a}+\sqrt{b}=\sqrt{a+b}. Option D (10\sqrt{5}) is a mistaken addition of coefficients. Option C (6\sqrt{10}) arises from incorrect factorization or mixing radicands. Exam tip: always factor out perfect squares from each radicand first and then combine only like surds.

Simplify each radical: \(\sqrt{192}=\sqrt{64\times3}=8\sqrt{3}\) and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). So the difference is \(8\sqrt{3}-3\sqrt{3}=(8-3)\sqrt{3}=5\sqrt{3}\). The nearest distractor \(3\sqrt{3}\) equals \(\sqrt{27}\), not the difference; a common mistake is to confuse one term with the result. Exam tip: pull out perfect squares first and then combine like surd terms by adding/subtracting their coefficients.

The direct answer is A: \(\sqrt{10}\times\sqrt{40}=20\). For nonnegative numbers, \(\sqrt{x}\sqrt{y}=\sqrt{xy}\). Therefore \(\sqrt{10}\times\sqrt{40}=\sqrt{10\cdot40}=\sqrt{400}=20\), because \(20^2=400\). Option A is correct. Option B, \(\sqrt{50}\), is not equal to the product: it would correspond to combining the radicands by addition, but multiplication requires \(10\cdot40\), not \(10+40\). Option C, \(10\sqrt{40}\), incorrectly leaves the factor 10 outside the radical; the first factor is \(\sqrt{10}\), not 10. Option D similarly changes the expression into \(40\sqrt{10}\), which is not the original product. Both individual roots are irrational, but their product can be rational; here the product becomes the integer 20. A quick check is to square 20: its square is 400, exactly the product of the radicands. Remember that multiplication of square roots permits multiplication inside one radical, not addition.

(\sqrt{6}\times\sqrt{10}=\sqrt{60}=2\sqrt{15}). Since (15) is not a perfect square the result is irrational.

Add like terms: the rational parts 3 and 3 sum to 6, while the irrational parts \(+\sqrt{6}\) and \(-\sqrt{6}\) cancel. Therefore \(a+b=6\). Distractor D (\(6+2\sqrt{6}\)) is a common mistake resulting from adding the square-root terms instead of canceling them. Exam tip: look for conjugates — the ±√ terms cancel when summed, leaving twice the rational part.

The direct answer is A, \(uv=31\). The two expressions are conjugates: one has a plus sign and the other has a minus sign. Use the identity \((a+b)(a-b)=a^2-b^2\). Here \(a=6\) and \(b=\sqrt{5}\), so \(uv=(6+\sqrt5)(6-\sqrt5)=6^2-(\sqrt5)^2=36-5=31\). Option A is correct. Option B, \(36+\sqrt5\), results from failing to multiply the conjugate terms correctly and is not the product. Option C, \(12\sqrt5\), is only the result of the two middle terms being combined incorrectly; those middle terms cancel, rather than add. Option D, 41, would come from adding 36 and 5 instead of subtracting them. The product is rational because the irrational parts cancel. This is not a matter of approximating \(\sqrt5\); the identity gives the exact answer. Memory cue: conjugates with opposite signs turn a product into “first square minus second square.”

This is a product of conjugates; use the identity \((a+b)(a-b)=a^2-b^2\). With \(a=\sqrt{11}\) and \(b=2\), we get \((\sqrt{11})^2-2^2=11-4=7\). Option B (15) is a common mistake from adding 11 and 4 instead of subtracting. Exam tip: recognize conjugates quickly and apply \(a^2-b^2\) to avoid expanding unnecessarily.

Use the square identity \((a+b)^2=a^2+2ab+b^2\). With \(a=3\) and \(b=\sqrt{2}\) we get \((3+\sqrt{2})^2=3^2+2\cdot3\cdot\sqrt{2}+(\sqrt{2})^2=9+6\sqrt{2}+2=11+6\sqrt{2}\). Option B (\(9+6\sqrt{2}\)) omits the \(b^2\) term (+2) and is therefore incorrect. Exam tip: always write all three terms \(a^2,\;2ab,\;b^2\) when expanding a square of a binomial.

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.

The direct answer is option A: \\(\sqrt3\\). Start with \\(\frac3{\sqrt3}\\). Since \\(3=\sqrt3\times\sqrt3\\), replace the numerator: \\(\frac{\sqrt3\times\sqrt3}{\sqrt3}\\\\=\sqrt3\\), because the nonzero factor \\(\sqrt3\\) cancels. Equivalently, rationalising gives \\(\frac3{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{3\sqrt3}{3}=\sqrt3\\). Option A is correct. Option B, \\(3\sqrt3\\), is three times too large. Option C, 1, would be the result of incorrectly treating 3 and \\(\sqrt3\\) as equal. Option D, \\(1/3\\), does not follow from the division and is numerically different. The denominator is nonzero, so cancellation is valid. Memory cue: because \\(3=(\sqrt3)^2\\), dividing 3 by \\(\sqrt3\\) leaves one \\(\sqrt3\\).

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.

Direct answer: Option A, m must be a perfect square. Let \(\sqrt m=r\), where r is rational. For a positive integer m, a rational square root can occur only when m is a perfect square; then r is an integer such as 1, 2, 3 or 4. Examples are \(\sqrt1=1\), \(\sqrt4=2\), and \(\sqrt9=3\). If m is not a perfect square, its prime factorisation contains at least one prime with an odd exponent, so the square root retains that prime and is irrational. Option A is correct. Option B is wrong because a perfect square need not be prime: 4 and 9 are composite. Option C is wrong because perfect squares can be odd, such as 9, or even, such as 4. Option D is the opposite of the required condition. Exam cue: for a positive integer, rational square root means every prime exponent in m is even.

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

The direct answer is A: \(\sqrt{289}\) is rational and \(\sqrt{290}\) is irrational. Since \(289=17^2\), \(\sqrt{289}=17\), which is an integer and therefore rational. For 290, the nearby consecutive perfect squares are \(17^2=289\) and \(18^2=324\). Thus 290 is not a perfect square, so its square root cannot be rational and is irrational. Option A correctly states both classifications. Option B is wrong because \(\sqrt{290}\) is not rational. Option C is wrong because \(\sqrt{289}=17\) is clearly rational. Option D reverses the properties of the two roots. The important idea is that the number under the root, not merely the presence of a radical symbol, determines the nature of the answer. Check for a whole-number square root first. If the radicand is an exact square, the root is rational; if it lies between consecutive squares, the root is irrational. This simple comparison avoids guessing.