Irrational numbers

\(\sqrt{7}\times\sqrt{14}=\sqrt{98}=\sqrt{49\times2}=7\sqrt{2}\). Since \(\sqrt{2}\) is irrational, multiplying it by the nonzero rational number \(7\) still gives an irrational number. Therefore, the expression is an irrational real number, not a rational number, integer, or terminating decimal. Exam tip: combine the square roots first and extract perfect-square factors.

To rationalise the denominator, multiply both the numerator and denominator by \(\sqrt{7}\): \(\frac{6}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{6\sqrt{7}}{7}\), since \(\sqrt{7}\times\sqrt{7}=7\). Hence, option A is correct. Remember that the numerator must also be multiplied by \(\sqrt{7}\); changing only the denominator to 7 is not valid.

\(\sqrt{3}\times\sqrt{21}=\sqrt{63}=3\sqrt{7}\). Since 7 is not a perfect square, \(\sqrt{7}\) is irrational, and multiplying it by the non-zero rational number 3 remains irrational. Therefore, the given expression is an irrational real number. It is neither an integer, a rational number, nor a terminating decimal. Exam tip: the square root of a positive non-perfect square is irrational.

The governing classification theorem states that a real number is rational if and only if its decimal expansion terminates or eventually repeats. A decimal that continues forever without settling into any repeating block is non-terminating and non-repeating, so it cannot be written as p/q for integers p and q with q ≠ 0. Therefore the given number is irrational; because it is represented by a real decimal, it is an irrational real number. Option B is correct. Option A is ruled out by the absence of repetition, option C is ruled out because digits continue indefinitely, and option D is impossible because an integer has a terminating decimal representation such as 4.000.... The stated condition is decisive even if only an initial pattern is displayed.

The governing concept is simplification of like surds before addition. Rewrite each radical using 2 as the square-free factor: √2 remains √2, √8 = √(4×2) = 2√2, and √18 = √(9×2) = 3√2. These are now like surds, so their coefficients can be added: √2 + 2√2 + 3√2 = (1 + 2 + 3)√2 = 6√2. Thus option A is correct. Option B omits one coefficient, while option C adds incorrectly. Option D, √28, is actually equal to 2√7 and is not equal to the given sum. Radicals should be simplified before combining; unlike surds cannot be added by simply adding their radicands.

The governing classification theorem says that a real number is rational exactly when its decimal expansion terminates or eventually repeats. The question explicitly states that the decimal expansion continues forever and has no repeating block. Therefore it cannot be written as a ratio of two integers and is irrational. It is still a real number because it is represented by a decimal expansion on the real number line. Thus option B, irrational real number, is correct. Option A would require termination or periodic repetition, option C contradicts the word non-terminating, and option D is impossible because an integer has a terminating decimal representation.

The governing concept is addition of surd fractions using conjugate denominators. The denominators 3+√8 and 3−√8 are conjugates, so their product is (3+√8)(3−√8)=3²−(√8)²=9−8=1. Taking the common denominator, the numerator becomes (3−√8)+(3+√8)=6 because the two radical terms cancel. Therefore the complete expression is 6/1=6, so option A is correct. Option B incorrectly retains only the rational part, option C mistakes the combined numerator for the answer, and option D gives the product of the conjugate denominators rather than the value of the sum. No decimal approximation is needed, and the cancellation is exact.

The number 3 is rational because it can be written as 3/1, while √2 is irrational; its decimal expansion is non-terminating and non-repeating. A fundamental property is that the sum of a rational number and an irrational number is irrational. To see why, suppose 3+√2 were rational. Subtracting the rational number 3 would then make √2 rational, contradicting the known irrationality of √2. Therefore 3+√2 is irrational, so option A is correct. It cannot be a rational number, integer, or whole number. The fact that the expression contains a radical does not by itself prove the result; the decisive reasoning is the rational-plus-irrational property and the contradiction argument. Thus the number is real, but among the listed choices its most specific correct classification is irrational.

\(\sqrt{3}\) is irrational, whereas 1 is rational. Adding or subtracting an irrational number and a rational number always gives an irrational result. Therefore, \(1-\sqrt{3}\) is irrational. It cannot be an integer or a natural number, since both of those are rational numbers. Exam tip: \(\sqrt{n}\) is irrational when \(n\) is not a perfect square.

11 is not a perfect square: no integer has square equal to 11. Therefore, \(\sqrt{11}\) cannot be expressed as a ratio of two integers, so x is irrational. Integers and whole numbers are all rational, so those options cannot be correct. Exam tip: the square root of a non-perfect-square integer is irrational.

\(\sqrt{3}\) and \(\sqrt{5}\) are both irrational because 3 and 5 are not perfect squares. In option A, \(\sqrt{9}=3\) is rational; in option C, \(\frac{2}{3}\) is rational; and in option D, \(\sqrt{16}=4\) and \(\sqrt{25}=5\) are both rational. Exam tip: the square root of a perfect square is an integer and hence rational.

\(\sqrt{13}\) is irrational because 13 is not a perfect square. The square root of a positive integer that is not a perfect square is irrational. In contrast, \(-2.5=-\frac{5}{2}\), \(\frac{13}{17}\), and \(0=\frac{0}{1}\) are all rational numbers. Exam tip: Check whether the integer inside a square root is a perfect square.

The sum of a rational number and an irrational number is always irrational. If \(a+b\) were rational, then \(b=(a+b)-a\) would also be rational because the difference of two rational numbers is rational. This contradicts the fact that \(b\) is irrational. Hence, the correct answer is irrational. An integer or a natural number is rational, so neither can be the sum. Exam tip: The sum and difference of a rational number and an irrational number are both irrational.

In \(0.123456789101112\ldots\), the digits of natural numbers are written consecutively, and no fixed block of digits repeats forever. It is an infinite non-repeating decimal, so it is irrational. In contrast, \(0.7777\ldots\), \(0.202020\ldots\), and \(0.123123123\ldots\) are recurring decimals and therefore rational. Exam tip: An infinite decimal is rational if it eventually repeats; if it is non-terminating and non-repeating, it is irrational.

Here \(a=\sqrt{2}\), so \(\frac{1}{a}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\). Therefore, \(a+\frac{1}{a}=\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\), which is irrational because a non-zero rational multiple of \(\sqrt{2}\) remains irrational. Hence, it cannot be an integer or a natural number. Exam tip: simplify surd expressions first and try to write them in the form \(k\sqrt{n}\).

\(\sqrt{14}\) is irrational because 14 is not a perfect square. The square root of an integer is rational only when that integer is a perfect square. \(\frac{17}{19}\) is rational, \(0.090909\ldots\) is a recurring decimal and hence rational, and \(-8\) is an integer, so it is rational. Exam tip: Recurring decimals are always rational, whereas the square root of a non-perfect square is irrational.

Here \(x=\sqrt{7}\), so \(\frac{1}{x}=\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}\). Hence, \(x+\frac{1}{x}=\sqrt{7}+\frac{\sqrt{7}}{7}=\frac{8\sqrt{7}}{7}\). The product of a non-zero rational number, \(\frac{8}{7}\), and the irrational number \(\sqrt{7}\) is irrational. Therefore, the expression is not an integer or a natural number. Exam tip: the square root of a non-perfect square is irrational.

In option C, the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Its decimal expansion is non-terminating and non-repeating; therefore, it is irrational. Options A and B are non-terminating recurring decimals, so they are rational, while option D is a terminating decimal and is also rational. Exam tip: Every terminating or recurring decimal represents a rational number.

13 is not a perfect square. The square root of a natural number that is not a perfect square is irrational. Therefore, \(x=\sqrt{13}\) is an irrational number. In contrast, the square root of a perfect square, such as \(\sqrt{16}=4\), is an integer. Exam tip: first check whether the number under the square root is a perfect square.

\(\sqrt{7}\) is irrational, while \(3\) is a non-zero rational number. The product of a non-zero rational number and an irrational number is irrational. Therefore, \(3\sqrt{7}\) is irrational. It cannot be an integer or a natural number. Exam tip: Multiplying an irrational number by zero is the exceptional case, as the result is the rational number 0.