Irrational numbers and real numbers

The square root of a positive integer is rational only when the number is a perfect square. This is a direct exam rule.

Let \(x = 2.\overline{45}\). The repeating block has length 2, so multiply by 100: \(100x = 245.\overline{45}\). Subtracting gives \(99x = 243\), hence \(x = \frac{243}{99} = \frac{27}{11}\). This is a fraction, so the number is rational. Option B (irrational) is incorrect because irrational numbers do not have terminating or repeating decimal expansions; here the decimal repeats. Option C (non-real) is wrong because the number is real, and option D (whole number) is wrong because there is a nonzero fractional part. Exam tip: An overline means a repeating block — use multiplication by \(10^n\) (where n is block length) to convert to a fraction quickly.

\(\sqrt{162}=\sqrt{81\times2}=\sqrt{81}\cdot\sqrt{2}=9\sqrt{2}\). Therefore the correct simplified form is \(9\sqrt{2}\). Option B (\(6\sqrt{3}\)) is incorrect because \(6\sqrt{3}=\sqrt{36\times3}=\sqrt{108}\), which is not equal to \(\sqrt{162}\). Options C and D are also numerically different (\(18=\sqrt{324}\), and \(18\sqrt{2}\) is much larger). Exam tip: factor out the largest perfect square from under the root to simplify quickly.

Simplify first: \(\sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3}\). Thus \(4\sqrt{3}-\sqrt{27}=4\sqrt{3}-3\sqrt{3}=\sqrt{3}\), so option A is correct. Option B (1) is wrong because \(\sqrt{3}\approx1.732\), not 1. Option C (\(3\sqrt{3}\)) is incorrect because it treats the second term as if it should be added or left unsimplified; the correct simplification gives subtraction of like terms. Exam tip: always simplify radicals first and then combine only like surd terms (same radical part).

Simplify each radical: \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\). So \(\sqrt{5}+\sqrt{20}-\sqrt{45}=\sqrt{5}+2\sqrt{5}-3\sqrt{5}=(1+2-3)\sqrt{5}=0\). The closest distractor \(-\sqrt{5}\) would arise only from an arithmetic/sign error when combining coefficients; correct combination gives zero. Exam tip: factor out the common \(\sqrt{5}\) first to add coefficients quickly.

The correct answer is \(\sqrt{2}\). Reason: \(\frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}\). Option C, \(\dfrac{\sqrt{2}}{2}\), equals \(\dfrac{1}{\sqrt{2}}\) and is the reciprocal form, so it is incorrect; options B (1) and D (2) are also not equal to the given expression. Exam tip: multiply numerator and denominator by the square root in the denominator to remove the root and simplify quickly.

\(a-1=\sqrt{6}\). Since 6 is not a perfect square, \(\sqrt{6}\) cannot be written as a ratio \(p/q\) of integers, so it is irrational. Option B (rational) is incorrect because \(\sqrt{6}\) is not expressible as a fraction of integers; options C (integer) and D (zero) are incorrect because \(\sqrt{6}\) is neither an integer nor zero. Exam tip: simplify the expression first and check whether the square root is of a perfect square — if not, it is irrational.

Since \(1^2=1<2<4=2^2\), it follows that \(1<\sqrt{2}<2\). Therefore \(\sqrt{2}\) lies between the integers 1 and 2. The other choices are incorrect because they give integers that are too small (0 and 1) or too large (2 and 3; 3 and 4). Exam tip: to locate \(\sqrt{x}\) between integers compare squares of consecutive integers or use a quick decimal estimate (\(\sqrt{2}\approx1.414\)).

For positive numbers a larger number inside the root gives a larger square root. Since (72>50), (\sqrt{72}>\sqrt{50}).

The square root symbol denotes the principal (non-negative) root. Since \(0.81=\frac{81}{100}\), \(\sqrt{0.81}=\sqrt{\frac{81}{100}}=\frac{9}{10}=0.9\). Option B (−0.9) is a common trap because \((-0.9)^2=0.81\) as well, but the principal square root is positive. Options C and D are incorrect: \(0.09^2=0.0081\), and 0.81 is the original number, not its square root. Exam tip: convert decimals to fractions to simplify square-root calculations (e.g., \(0.81=81/100\)).

These are conjugates, so use the difference of squares identity: \((a+b)(a-b)=a^2-b^2\). With \(a=5\) and \(b=\sqrt{2}\), \((5+\sqrt{2})(5-\sqrt{2})=5^2-(\sqrt{2})^2=25-2=23\). Option B (27) is incorrect — it confuses adding instead of subtracting the square; options C and D give irrational forms that are not the product. Exam tip: recognize conjugates and apply the difference of squares to compute such products quickly and reliably.

\(\sqrt[3]{64}=4\) because 64 is a perfect cube (\(4^3\)). The result 4 is an integer, and every integer is rational (can be written as a fraction, e.g. \(4=4/1\)). Option B is incorrect since irrational numbers have non-terminating, non-repeating decimals, which does not apply to 4. Option D is incorrect because a non-repeating decimal implies irrationality, whereas 4 is a terminating decimal. Option C is wrong because non-real numbers involve imaginary parts; \(\sqrt[3]{64}\) is a real number. Exam tip: when evaluating nth roots, first check whether the radicand is a perfect power—if it is, the root is an integer (hence rational).

Prime factorization gives 16 = 2^4. A number is a perfect cube only if every prime exponent is a multiple of 3. Since 4 is not divisible by 3, \(\sqrt[3]{16}\) is not a perfect cube and therefore is irrational. Options B and C are incorrect because 4^3 = 64 and 8^3 = 512, not 16. Option D is wrong because an irrational number cannot be expressed as a terminating decimal. Exam tip: to test whether \(\sqrt[k]{N}\) is rational, factor N and check if every prime exponent is divisible by k.

The key idea is that the square root of a positive integer is rational when the integer is a perfect square, and irrational when it is not a perfect square. Adding the rational number 2 to an irrational number keeps the result irrational. Therefore, we must find the option whose square root cannot be written as a fraction or whole number.

For 18,
\(\sqrt{18}=3\sqrt{2}\), which is irrational because 2 is not a perfect square. In contrast, \(\sqrt{16}=4\), \(\sqrt{25}=5\), and \(\sqrt{36}=6\) are rational. Hence \(2+\sqrt{18}\) is irrational, so option A is correct. The other choices produce rational sums.

The direct answer is A: \(\sqrt{225}\) is rational and \(\sqrt{226}\) is irrational. A perfect square has an integer square root, while a positive integer that is not a perfect square has an irrational square root. Since \(225=15^2\), we get \(\sqrt{225}=15\), an integer and hence a rational number. The consecutive squares around 226 are \(15^2=225\) and \(16^2=256\). Therefore 226 is not a perfect square, so \(\sqrt{226}\) is irrational. Option A is correct because it states exactly these results. Option B is wrong: the second root cannot be rational because 226 is not a perfect square. Option C is wrong: the first root is 15, not irrational. Option D assigns the two properties in reverse order. Do not decide only from the fact that both expressions contain a radical; a radical can have a rational value when its radicand is a perfect square. The useful exam method is to compare the radicand with nearby squares.

The digit '2' appears at positions 1, 3, 6, 10, ... which are triangular numbers with general term \(T_n=\tfrac{n(n+1)}{2}\). The gaps between successive '2's increase, so there is no fixed repeating block (period). Any rational number's decimal expansion is either terminating or eventually periodic; this decimal is neither, therefore it is irrational. Exam tip: to test rationality, check whether the decimal eventually repeats or terminates — if not, it is irrational.

Direct answer: Option A, a rational number. A decimal that stops is called terminating; a decimal that continues in a repeating pattern is called recurring or repeating. Every terminating decimal can be converted into a fraction by using a denominator of 10, 100, 1000 and so on. Every repeating decimal also represents a fraction, although its conversion may require an algebraic step. Therefore both kinds are rational numbers. Option A is correct. Option B is wrong because irrational decimals are non-terminating and non-repeating. Option C is wrong because these decimals are real numbers; they are not non-real. Option D is too narrow: rational numbers include fractions, negative numbers and many non-natural numbers, not only natural numbers. For example, 0.5=1/2 is rational, although it is not a natural number. Memory cue: terminating or repeating decimal means rational; non-terminating and non-repeating means irrational.

Use the conjugate-product identity:

\((1+\sqrt{3})(1-\sqrt{3})=1^2-(\sqrt{3})^2=1-3=-2.\)

This follows from \((a+b)(a-b)=a^2-b^2\). The tempting wrong choice 4 comes from incorrectly adding 1 and 3 instead of subtracting. Exam tip: apply the conjugate formula and watch the signs carefully when multiplying surds.

(\sqrt{12}=2\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). The result is (4\sqrt{3}).

\(\frac{1}{2}\) is rational and \(\sqrt{2}\) is irrational. If their sum were rational, then \(\sqrt{2}=(\frac{1}{2}+\sqrt{2})-\frac{1}{2}\) would be a difference of rationals and thus rational, contradicting that \(\sqrt{2}\) is irrational. Hence \(\frac{1}{2}+\sqrt{2}\) is irrational. The closest distractor "rational" is incorrect for the reason above; "integer" and "terminating decimal" are specific types of rationals so they are also incorrect. Exam tip: memorize that rational + irrational = irrational (contradiction proof is quick and useful in exams).

\(\sqrt{300}=\sqrt{100\times3}=\sqrt{100}\cdot\sqrt{3}=10\sqrt{3}\). Hence option A is correct. The other choices are not equal to \(\sqrt{300}\): \(5\sqrt{6}\approx12.25\) (not 17.32), \(3\sqrt{100}=30\), and \(6\sqrt{50}=6\cdot5\sqrt{2}=30\sqrt{2}\). Exam tip: always extract the largest perfect square factor when simplifying square roots (here 100).

On subtracting, the (4) terms cancel and (\sqrt{7}-(-\sqrt{7})=2\sqrt{7}). Watch the signs carefully.

\(\sqrt{180}=\sqrt{36\times5}=\sqrt{36}\,\sqrt{5}=6\sqrt{5}\), so the simplified form is \(6\sqrt{5}\). Why other options are wrong: \(9\sqrt{2}\) squared is \(81\times2=162\) (not 180), \(5\sqrt{6}\) squared is \(25\times6=150\), and \(3\sqrt{12}=3\sqrt{4\times3}=6\sqrt{3}\), which is different from \(6\sqrt{5}\). Exam tip: always factor out the largest perfect square from under the radical to simplify quickly.

With \(x=\sqrt{17}\), we have \(x^2=(\sqrt{17})^2=17\). Therefore \(x^2-5=17-5=12\). The number 12 is an integer and can be written as \(12/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers, whereas 12 can. Option C (non-real) is wrong because 12 is a real number. Option D (non-repeating decimal) is wrong because 12 is a terminating decimal (12.0), not a non-repeating non-terminating decimal. Exam tip: simplify powers of radicals first — squaring a square root removes the root, then evaluate the remaining arithmetic to classify the result.

A rational number can be written as a fraction of integers, while an irrational number cannot be written in that form. The square root of a positive integer is irrational when the integer is not a perfect square. Since 13 is not a perfect square, \(\sqrt{13}\) is irrational. The number 4, however, is rational because it can be written as \(4/1\).

Adding a rational number to an irrational number always gives an irrational number. If the sum were rational, subtracting the rational number 4 would make \(\sqrt{13}\) rational, which is impossible. Therefore \(4+\sqrt{13}\) is irrational. It is consequently not an integer or a terminating decimal, so option A follows.