Irrational numbers and real numbers

Real numbers have no imaginary part. In option A, -3 (integer), 0 and \(\sqrt{2}\) (irrational) are all real. Option B contains \(\sqrt{-1}=i\), which is imaginary; option C has \(\sqrt{-9}=3i\), imaginary; option D contains \(\sqrt{3}\) (real) but also \(\sqrt{-16}=4i\), so it is not all real. Thus A is the only choice with only real numbers. Exam tip: if you see \(\sqrt{\text{negative}}\) immediately mark it as non-real (imaginary).

(\sqrt{7}) is irrational and its negative is also irrational. The negative sign does not stop it from being real.

The number √5 is irrational because 5 is not a perfect square. Multiplying an irrational number by a non-zero rational number remains irrational. Here 3 is a non-zero rational number. For a short proof, suppose 3√5 were rational; dividing it by 3, a non-zero rational, would imply √5 is rational, which is impossible. Hence 3√5 is irrational, so option A is correct. It is not rational or an integer, and it is not zero because both 3 and √5 are positive. The common error is to think that multiplying by an integer automatically removes irrationality; that happens only in special expressions involving cancellation, not when a non-zero rational multiplies √5. The classification depends on the irrational factor and the non-zero multiplier.

Dividing (\sqrt{2}) by non zero rational (2) keeps it irrational. Check the nature of the root in such expressions.

(\sqrt{2}+(-\sqrt{2})=0) is rational but (\sqrt{2}+\sqrt{3}) is irrational. So there is no fixed always rule.

(\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.

(\sqrt{6}) is irrational so it cannot be written as (\frac{p}{q}). The other numbers are rational.

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.