Irrational numbers and real numbers

(\sqrt{2}\times\sqrt{2}=2) is rational but (\sqrt{2}\times\sqrt{3}=\sqrt{6}) is irrational. So it is not always one type.

\(0.\overline{7}\) is a repeating decimal. Let \(x=0.\overline{7}\). Then \(10x-x=7\), so \(x=7/9\). Since it can be expressed as a fraction, it is rational. Option B (irrational) is incorrect because irrational numbers are non‑terminating, non‑repeating decimals; here the decimal repeats. Exam tip: convert a repeating decimal to a fraction to verify rationality quickly.

\(\frac{9}{3}=3\). The number 3 is a counting (natural) number and can be written as \(\frac{3}{1}\), so it is rational as well. Option C (\(-\frac{5}{2}\)) is rational but negative, so not a natural number. Options B (\(\sqrt{10}\)) and D (\(\pi\)) are irrational and not rational. Exam tip: natural numbers are positive integers (1,2,3,...) and a rational number can be expressed as \(\frac{p}{q}\) with integers p,q and \(q\neq0\).

\(\sqrt{8}=\sqrt{4\times2}=\sqrt{4}\,\sqrt{2}=2\sqrt{2}\), so option A is correct. The nearest distractor C (\(\sqrt{8}=\sqrt{2}\)) is wrong because 8 is not equal to 2. Options B and D give unnecessarily larger coefficients (4 or 8) and are therefore incorrect. Exam tip: factor out perfect square factors from under the radical to simplify quickly.

The square of a number means multiplying it by itself. The square-root symbol asks for the non-negative number whose square gives the number inside the symbol. Therefore, squaring a square root of a positive number returns the original number: \((\sqrt{a})^2=a\) for positive \(a\).

Here, \(x=\sqrt{3}\), so \(x^2=(\sqrt{3})^2=3\). The number 3 is rational because it can be written as \(3/1\), a quotient of integers with a nonzero denominator. It is also real, but the most specific listed choice is rational number. Thus option A is correct. This example also shows that an irrational number can have a rational square.

Direct answer: Option A, \(\sqrt6\) cannot be written as \(p/q\). The defining test is that p and q must be integers and q must not be zero. Since 6 is not a perfect square, \(\sqrt6\) is irrational; therefore no such fraction exists. Option A is correct. Option B, -4, is rational because it equals \((-4)/1\). Option C, 2.75, is a terminating decimal and equals \(275/100=11/4\), so it is rational. Option D, 0, is rational because it equals \(0/1\), and the denominator is non-zero. The word “cannot” is important: the question asks for the irrational number, not merely a number that is difficult to convert. Also, a negative integer remains rational. Memory cue: terminating decimals and integers always pass the p/q test; roots of non-perfect squares do not.

The decimal expansion of 4.125 has a finite number of digits, so it terminates. In fractional form, \(4.125=\frac{4125}{1000}=\frac{33}{8}\). Its simplified denominator is \(8=2^3\), so the decimal terminates. \(0.666...\) is repeating, \(\pi\) is irrational and non-terminating non-repeating, and option D is also infinite. Exam tip: a rational number has a terminating decimal exactly when its denominator in lowest terms contains only the prime factors 2 and/or 5.

Simplify each radical first: \(\sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}\) and \(\sqrt{45}=\sqrt{9\cdot5}=3\sqrt{5}\). Both are like terms with \(\sqrt{5}\), so add them: \(2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\). The common distractor \(\sqrt{65}\) is incorrect because in general \(\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}\). Exam tip: always factor out perfect squares first to combine like radicals easily.

All real numbers can be represented on the number line. This includes both rational and irrational numbers.

Since \(27=9\times3\), \(\sqrt{27}=\sqrt{9\times3}=\sqrt{9}\cdot\sqrt{3}=3\sqrt{3}\). Thus \(3\sqrt{3}\) is correct. \(\sqrt{9}\) equals 3, not \(3\sqrt{3}\); \(9\sqrt{3}\) and \(2\sqrt{3}\) have incorrect coefficients. Exam tip: Factor the radicand into the largest perfect square times the remainder, then take the square root of the perfect square outside the radical.

Any repeating (periodic) decimal represents a ratio of two integers, so it is rational. For example, if \(x=0.\overline{3}\) then \(10x-x=3\) which gives \(x=3/9=1/3\). A common mistake is to call every non-terminating decimal irrational; however irrationals are non-terminating and non-repeating. Option C (non-real) is incorrect because decimal expansions describe real numbers; option D (integer only) is wrong because integers have terminating decimal forms (e.g. 5 = 5.000...). Exam tip: spot the repeating block—if present, convert by the usual algebraic shift-and-subtract method to get a fraction.

Since \(\sqrt{8}=2\sqrt{2}\), we have \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Option C (\(2\sqrt{2}\)) omits the original \(\sqrt{2}\) term; option B (\(\sqrt{10}\)) incorrectly combines two surds under one radical. Exam tip: always simplify each radical first (extract perfect squares) and then add like surd terms.

(\frac{1}{7}) is rational and its decimal is repeating. Rational numbers give terminating or repeating decimals.

A rational number can be written as a fraction or integer; an irrational number has a non-terminating, non-repeating decimal expansion. Here \(4\) is rational and \(\sqrt{7}\) is irrational, so \(4+\sqrt{7}\) is the sum of a rational and an irrational number (and is itself irrational). Option B is the sum of two irrationals (not a rational plus an irrational). Options C and D are sums of rationals, so their sums are rational. Exam tip: check each term — if one term is irrational and the other clearly rational, the sum is irrational.

The real-number system is divided into two mutually exclusive classes: rational numbers and irrational numbers. Thus every irrational number belongs to the set of real numbers, so option A is always true. The converse is false because rational numbers such as 1/2, -3, and 0 are also real. Consequently, option B is false. Option C is false because a rational number cannot be irrational by definition, and option D is false because every integer can be written as a fraction, for example 3 = 3/1, so every integer is rational. The useful inclusion is irrational numbers ⊂ real numbers, while rational numbers ⊂ real numbers as well. Recognising these set relationships prevents the mistake of treating “real” and “irrational” as synonyms.

\(\sqrt{0.04}\)=\(0.2\)=\(\frac{1}{5}\). Since it is a terminating decimal and can be expressed as a fraction, it is a rational number. Option B is incorrect because irrational numbers have non-terminating, non-repeating decimals (e.g. \(\sqrt{2}\)). Option C is wrong because non‑real numbers have an imaginary part; \(0.2\) is a real number. Option D is wrong because natural numbers are positive integers and \(0.2\) is not an integer. Exam tip: terminating or repeating decimals always represent rational numbers.

Reason: \((\sqrt{2})^2=2\) and \(\sqrt{2}\) is irrational (standard proof). Option B, \(-\sqrt{2}\), also satisfies \(x^2=2\) but is negative — the question asks for a positive number. Option C gives \(2^2=4\) and option D gives \((3/2)^2=9/4\); hence both are incorrect. Exam tip: by convention \(\sqrt{\cdot}\) denotes the principal (positive) square root, so choose \(\sqrt{2}\).

Subtracting irrational (\sqrt{2}) from rational (7) gives an irrational number. Changing the sign does not change this property.

The direct answer is option A: \\(\sqrt n\\) will be irrational, assuming n is a positive integer that is not a perfect square. A perfect square has an integer square root, such as \\(\sqrt9=3\\). If n is not a perfect square, no integer squared gives n. The square root then cannot be written as \\(p/q\\) with integers p and q and nonzero q, so it is irrational. Option A is correct. Option B is wrong because a non-perfect-square root is not an integer. Option C is wrong because \\(\sqrt n=0\\) only when n=0, and 0 is a perfect square. Option D is wrong under the stated positive-integer assumption: the square root of every positive real number is real. For example, \\(\sqrt2\\) is real but irrational. Memory cue: perfect square means rational integer root; otherwise a positive integer root is irrational.

\(\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}=5\sqrt{2}\). Thus the correct simplified form is \(5\sqrt{2}\). The closest distractor \(2\sqrt{5}\) is incorrect because \(2\sqrt{5}=\sqrt{4\times5}=\sqrt{20}\), not \(\sqrt{50}\). Exam tip: always factor out the largest perfect square from under the radical (here 25) to simplify quickly.

(5) is a non zero rational number so (5\sqrt{3}) is irrational. Remember multiplication by (0) gives (0).

Here \(x-2=\sqrt{5}\). Since 5 is not a perfect square, \(\sqrt{5}\) is irrational and cannot be written as a ratio \(p/q\) of integers; hence \(x-2\) is irrational. The closest distractor, rational (B), is wrong because rationals can be expressed as \(p/q\), which \(\sqrt{5}\) cannot. Integer (C) and zero (D) are also incorrect because \(\sqrt{5}\) is neither an integer nor zero. Exam tip: always simplify the expression first and use the perfect-square test to check irrationality.

\(\sqrt{98}=\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}=7\sqrt{2}\), so the simplified form is \(7\sqrt{2}\). Option B (\(2\sqrt{7}\)) is incorrect because \((2\sqrt{7})^2=28\), not 98. Option C (14) is wrong since \(14^2=196\). Option D is the unsimplified radical. Exam tip: Factor the radicand into the largest perfect square times the remainder, take the square root of the perfect square outside the radical.

A terminating decimal can be written as an integer divided by a power of 10, so after reducing it can be expressed as \\(\frac{p}{q}\\). For example, \\(0.125=125/1000=1/8\\). Therefore a terminating decimal is rational and (since all rationals are real) also a real number. Option B is wrong because irrational numbers have nonterminating, nonrepeating decimals. Option C is incorrect because decimals represent real numbers. Option D is wrong since terminating decimals need not be natural numbers (only some, like 1.0, are). Exam tip: convert the decimal to a fraction or check if its decimal expansion terminates or repeats — terminating or repeating implies rationality.

(\sqrt{15}) is irrational because (15) is not a perfect square. Identifying perfect squares is very useful in exams.