Real Numbers

( \frac{22}{7} ) is a ratio of two integers with non-zero denominator. So it is a rational real number.

( \frac{1}{2} ), ( \sqrt{3} ), and ( -5 ) are all real numbers. A fraction with zero denominator is not real.

The governing concept is the principal square root and the effect of an outside negative sign. The symbol √4 denotes the non-negative principal square root of 4, so √4 = 2. The minus sign is written before the radical and therefore changes the result after the square root is evaluated: −√4 = −(2) = −2. Thus option B is correct. Option A would be correct for √4 without the outside negative sign. Option C is the radicand, not its square root, while option D incorrectly treats the square root of 4 as 4 and also applies a negative sign. It is important not to confuse −√4 with √(−4), which is not a real number.

Absolute value represents a number’s distance from zero, so it is never negative. The distance of −7 from zero is 7; therefore, \(|-7|=7\). Option A is incorrect because it gives the signed value, not the absolute value. Exam tip: To find the absolute value of a negative number, remove its minus sign.

There are infinitely many real numbers between any two distinct real numbers. This is an important property of the number line.

\(\sqrt{36}=6\) and \(\sqrt{64}=8\), so \(6<8\), which gives \(\sqrt{36}<\sqrt{64}\). Therefore, option C is correct. Option A reverses the inequality, while option B incorrectly treats the two values as equal. Option D is also incorrect because 6 and 8 are both real and rational numbers. Exam tip: For perfect squares, calculate their square roots first and then compare the results.

The governing concept is the identity property of addition. An additive identity is a number that leaves every number unchanged when it is added to that number. For any real number x, the defining relation is x + 0 = 0 + x = x. Therefore, zero is the additive identity in the set of real numbers, and option C is correct. Option A is not an additive identity because x + 1 is generally greater than x. Similarly, adding −1 changes most numbers, so option B cannot be correct. Adding 2 also changes the original number. Notice that 1 is the multiplicative identity because x × 1 = x; this distinction helps avoid confusing the identities for addition and multiplication.

Multiplying any real number by (1) gives the same number. Hence (1) is the multiplicative identity.

The additive inverse of a number is the number that gives a sum of 0 when added to the original number. Here, \((-9)+9=0\), so the additive inverse of \((-9)\) is 9. \(\frac{1}{9}\) is its multiplicative inverse, not its additive inverse. Exam tip: to find an additive inverse, change the sign of the number.

The multiplicative inverse of a non-zero number is the number that gives product 1 when multiplied by the original number. For a non-zero fraction \\(\frac{a}{b}\\), the inverse is \\(\frac{b}{a}\\), because \\(\frac{a}{b}\times\frac{b}{a}=1\\). Here the numerator 4 and denominator 7 are interchanged, so the inverse of \\(\frac{4}{7}\\) is \\(\frac{7}{4}\\). Therefore, option A is correct.

We can verify the result directly: \\(\frac{4}{7}\times\frac{7}{4}=\frac{28}{28}=1\\). The negative choices are not correct because multiplying \\(\frac{4}{7}\\) by a negative version gives a negative product, not 1. The number \\(\frac{11}{7}\\) also does not produce 1. The fraction is non-zero, so its reciprocal exists.

Zero is neither positive nor negative. Positive numbers are greater than 0, such as 1 and \(\frac{1}{2}\), whereas negative numbers are less than 0, such as -1. Therefore, 0 is the correct answer. Exam tip: On the number line, numbers to the right of 0 are positive and those to the left are negative.

\(\sqrt{121}\) is the positive number whose square is 121. Since \(11^2=121\), we get \(\sqrt{121}=11\). Option 121 is the radicand, not its square root. Exam tip: for a positive number \(a\), \(\sqrt{a^2}=a\).

The governing concept is extracting a perfect-square factor from a radical. Factor 12 as 4 × 3, where 4 is a perfect square. Using √(ab) = √a × √b for non-negative factors, √12 = √(4 × 3) = √4 × √3 = 2√3. Therefore option B is correct. Option A, 6√2, would square to 72 and is too large. Option C, 3√2, would square to 18, not 12. Option D, 4√3, would square to 48. The expression 2√3 cannot be reduced further because 3 has no factor greater than 1 that is a perfect square. This method is preferable to using a decimal approximation because it gives the exact simplified radical.

For positive numbers, the square root of the greater number is also greater. Since (6>5), ( \sqrt{6}>\sqrt{5} ).

Adding the irrational ( \sqrt{2} ) to the rational number (3) generally gives an irrational number. It is also real.

Each point on the real number line represents a real number. Understanding the number line helps in comparison.

Direct answer: Option B, \(\frac{6}{0}\). Division by zero is undefined, so \(\frac{6}{0}\) is not a real number. A real number must have a definite value on the number line. There is no number \(x\) satisfying \(0\times x=6\), because multiplying zero by any number gives zero. Therefore the expression cannot be assigned an ordinary real value. Option A, -8, is an integer and therefore a real number. Option B is correct because its denominator is zero, making the fraction undefined. Option C, \(\sqrt{19}\), is irrational but still real: 19 is positive, so its square root exists on the real number line. Option D, 0.45, is a terminating decimal and equals \(\frac{45}{100}=\frac9{20}\), so it is rational and real. Do not confuse an irrational number with a non-real number. Irrational numbers are still real. Remember: denominator zero means undefined, not zero and not infinity as an ordinary real answer.

Since \(81=9^2\), \(\sqrt{81}=9\). The principal square root is the positive number whose square is 81. The squares of 8 and 18 are 64 and 324, respectively, so they are incorrect; 81 itself is the radicand, not its square root. Exam tip: Memorising common perfect squares helps solve such questions quickly.

An irrational number has a decimal expansion that is non-terminating and non-repeating, so option B is correct. Option C describes rational numbers, which can be written as the ratio of two integers. Option A applies to rational numbers with terminating decimal expansions. Exam tip: terminating or non-terminating repeating decimals are rational, whereas non-terminating non-repeating decimals are irrational.

The direct answer is B, rational number. A rational number is any number that can be written as \\(p/q\\), where \\(p\\) and \\(q\\) are integers and \\(q\\neq0\\). The decimal 0.125 has three digits after the decimal point, so write it as \\(125/1000\\). Simplifying by dividing numerator and denominator by 125 gives \\(1/8\\). Thus it can be expressed as a ratio of integers and is rational. Option A, irrational number, is wrong because an irrational decimal does not terminate and cannot be written as such a fraction; 0.125 terminates. Option B is correct. Option C, undefined number, is wrong because 0.125 has a clear finite value. Option D, only integer, is wrong because integers have no fractional part, while \\(0.125=1/8\\) is not an integer. The exam cue is that every terminating decimal is rational; repeating decimals are rational too. The supplied answer is correct.

( \frac{-11}{6} ) is a ratio of two integers with non-zero denominator. So it is a rational real number.

Since \(32=16\times2\) and \(16\) is a perfect square, \(\sqrt{32}=\sqrt{16\times2}=\sqrt{16}\times\sqrt{2}=4\sqrt{2}\). Therefore, option C is correct. Exam tip: To simplify a surd, factor the number using its largest perfect-square factor.

\(45=9\times5=3^2\times5\). Therefore, \(\sqrt{45}=\sqrt{3^2\times5}=3\sqrt{5}\), so option A is correct. Option B results from incorrectly treating 45 as \(5^2\times3\), but \(5^2\times3=75\). Exam tip: factor the number into the largest perfect square multiplied by the remaining factor, then take the square root of the perfect square outside the radical.

Since \(75=25\times3\) and \(25\) is a perfect square, \(\sqrt{75}=\sqrt{25\times3}=\sqrt{25}\times\sqrt{3}=5\sqrt{3}\). Option C, \(3\sqrt{5}\), results from separating the factors incorrectly. Exam tip: take the largest perfect-square factor outside the square root.

The governing concept is the product property of square roots for non-negative numbers: √a × √b = √(ab). Applying it gives √2 × √18 = √(2 × 18) = √36. The principal square root of 36 is 6 because 6² = 36 and the principal root is non-negative. Therefore option A is correct. Option B, √20, results from an incorrect operation and is not equal to the product. Option C simply rewrites part of the expression without using the correct product property, while option D is incorrect because 9² = 81, not 36. This example also illustrates that multiplying two irrational square roots can produce a rational integer.