Irrational numbers and real numbers

\(\sqrt{0.04}\)=\(0.2\)=\(\frac{1}{5}\). 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.

If positive integer (n) is not a perfect square then (\sqrt{n}) is irrational. This is an easy exam rule.

\(\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}=5\sqrt{2}\). 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.

In \(0.272727\ldots\), the block 27 repeats, so the number is rational. Let \(x=0.272727\ldots\); then \(100x-x=27\), giving \(x=\frac{27}{99}=\frac{3}{11}\). Exam tip: recurring decimals are rational, unlike non-recurring non-terminating ones.

A terminating decimal can be expressed as a ratio of two integers by writing it over an appropriate power of 10 (for example 0.75 = 75/100 = 3/4). Therefore such numbers are rational. Option B (irrational) is incorrect because irrational numbers have non-terminating, non-repeating decimal expansions (e.g. √2). Option C (non‑real) is wrong since decimal expansions represent real numbers. Option D (natural number only) is incorrect because terminating decimals include non-integer rationals (e.g. 0.5). Exam tip: To convert a terminating decimal to a fraction, write the decimal without the point as the numerator and use 10^n as denominator where n is number of decimal places, then simplify (e.g. 0.125 = 125/1000 = 1/8).

The block (27) repeats, so this is a recurring (repeating) decimal. Every repeating decimal is rational because it can be written as a fraction. For example let \\(x=3.2727\ldots\\). Then \\(100x=327.2727\ldots\\) and subtracting gives \\(99x=324\\), so \\(x=\frac{324}{99}=\frac{36}{11}\\). Thus the number is rational. The closest distractor, "irrational", is wrong because irrational numbers are non-terminating and non-repeating decimals. "Integer" is wrong since the value is not an integer, and "non-real" is wrong because this is a real number. Exam tip: convert repeating decimals to fractions by multiplying to align repeats and subtracting to eliminate the repeating part.

A decimal expansion that is non-terminating and non-repeating represents an irrational number. Rational numbers always have decimal expansions that either terminate or eventually repeat a fixed block. The given expansion 5.123123312333... shows no fixed repeating block, so it is irrational. The closest distractor, 'rational number', is incorrect because a rational decimal must exhibit a repeating pattern if it does not terminate. Exam tip: to test quickly, look for a consistent repeating block—if none exists, the number is irrational.

\(\sqrt{121}=11\). The number 11 is an integer and can be written as the fraction \(11/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers; 11 can. Option C (non‑real) is wrong because 11 is a real number. Option D (zero) is wrong because the value is 11, not 0. Exam tip: the square root of a perfect square integer is an integer and thus rational.

The square root of a whole number is rational only when the number under the root is a perfect square, such as \(1,4,9,16,25,36\), or another square of an integer. The number 37 lies between \(36=6^2\) and \(49=7^2\), so it is not a perfect square. Therefore \(\sqrt{37}\) cannot be written as an integer or as a terminating or repeating rational number; it is irrational.

Thus option A is correct. It is not a perfect square itself, because 37 is the radicand rather than a square number. Also, \(\sqrt{37}\) is slightly greater than 6, so it cannot equal 6; squaring 6 gives 36, not 37. It is not an integer for the same reason. The key test is that the square root of a positive integer is irrational when that integer is not a perfect square.

Integers and terminating decimals are rational. Options containing irrational roots or (\pi) are not only rational.

Real numbers form the large set of rational and irrational numbers. They can be represented on the number line.

You can write (-8) as \( -8=\frac{-8}{1}\), so it is a ratio of two integers and therefore rational. Every integer is also a real number, so (-8) is real as well.
Why other choices are wrong: B is incorrect because irrational numbers cannot be expressed as a ratio of integers (e.g. \(\sqrt{2}\)). C is wrong because natural numbers are positive integers (1,2,3,...); -8 is not natural. D is wrong because non-real (complex) numbers have a nonzero imaginary part (e.g. \(3+2i\)); -8 has no imaginary part.
Exam tip: To decide if a number is rational, try to express it as \(\frac{p}{q}\) with integers \(p,q\) and \(q\neq0\).

Adding irrational (\sqrt{11}) to rational (6) gives an irrational number. Do not mistake the root for a perfect square.