Irrational numbers

Correct answer: B. Irrational real number

Explanation: The correct classification is an irrational real number, but it should be proved rather than inferred only from the fact that both terms are irrational. Suppose, for contradiction, that √2 + √3 = r where r is rational. Then √3 = r − √2. Squaring both sides gives 3 = r² + 2 − 2r√2, so 2r√2 = r² − 1. The value r is positive and nonzero, because the left side of the original equation is positive. Therefore √2 = (r²−1)/(2r) would be rational, contradicting the known irrationality of √2. Hence √2 + √3 is irrational. Since both summands are real, their sum is real as well, so option B is correct.

Correct answer: A. 6

Explanation: 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.

Correct answer: A. Irrational

Explanation: \(\sqrt{3}\) 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.

Correct answer: B. Irrational

Explanation: 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.

Correct answer: B. (\sqrt{3},\sqrt{5})

Explanation: \(\sqrt{3}\) 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.

Correct answer: C. \(\sqrt{13}\)

Explanation: \(\sqrt{13}\) 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.

Correct answer: B. Irrational

Explanation: 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.

Correct answer: D. \(0.123456789101112\ldots\)

Explanation: 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.

Correct answer: C. Irrational

Explanation: 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 \(k\sqrt{n}\).

Correct answer: C. \(\sqrt{7}\) and \(1-\sqrt{7}\)

Explanation: Both \(\sqrt{7}\) and \(1-\sqrt{7}\) are irrational, but \(\sqrt{7}+1-\sqrt{7}=1\), which is rational. Hence the statement is false. Exam tip: check whether irrational terms cancel in a sum.

Correct answer: A. (7+4\sqrt{3})

Explanation: Multiplying by the conjugate (2+\sqrt{3}) makes the denominator (1) and numerator ((2+\sqrt{3})^2). So the answer is (7+4\sqrt{3}).

Correct answer: D. \(x+\frac{1}{3}\)

Explanation: \(x+\frac{1}{3}\) must be irrational: if it were rational, subtracting the rational number \(\frac{1}{3}\) would make \(x\) rational. But \(x^2\) need not be irrational; for \(x=\sqrt{2}\), \(x^2=2\). Exam tip: adding a rational number to an irrational number keeps it irrational.

Correct answer: A. (\sqrt{5})

Explanation: (\sqrt{45}=3\sqrt{5}), (\sqrt{20}=2\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). Therefore the result is (\sqrt{5}).

Correct answer: A. \(\sqrt{2}+(-\sqrt{2})=0\)

Explanation: Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence, the sum of two irrational numbers need not be irrational. Exam tip: test “always” statements using a counterexample.

Correct answer: B. It is real and its square is (3+\sqrt{5})

Explanation: The square root of a positive number is real and its square gives the same number. Here (3+\sqrt{5}) is positive.

Correct answer: C. It is irrational because the groups of zeros between 1s keep increasing, so no repeating block is formed.

Explanation: This decimal is non-terminating and non-repeating: the number of zeros between successive 1s is 1, 2, 3, 4, ... . Hence no fixed period is possible. Exam tip: non-terminating recurring decimals are rational.

Correct answer: C. \(x^2\)

Explanation: Take \(x=\sqrt{2}\). It is irrational, but \(x^2=2\) is rational. If \(x+1\), \(2x\), or \(1/x\) were rational, then \(x\) would also be rational, a contradiction. Exam tip: test a claim using a counterexample.

Correct answer: D. \(x-x\)

Explanation: Since \(x-x=0\), and 0 is rational, option D is always rational. In contrast, \(x^2\) may be rational, as \((\sqrt{2})^2=2\), or irrational. Exam tip: first cancel identical terms before classifying a number.

Correct answer: A. (6\sqrt{3})

Explanation: (\sqrt{48}=4\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Therefore the result is (6\sqrt{3}).

Correct answer: A. (11)

Explanation: (\sqrt{72}=6\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the numerator is (11\sqrt{2}). Dividing by (\sqrt{2}) gives (11).

Correct answer: C. If \(x\) is irrational, then \(x^2\) is irrational.

Explanation: Option C is correct. For \(x=\sqrt{2}\), \(x\) is irrational but \(x^2=2\) is rational. Adding 2 or multiplying or dividing by non-zero rational 3 preserves irrationality. Exam tip: test “always” claims with a counterexample.

Correct answer: A. (4\sqrt{10})

Explanation: Option A is correct: the value is \(4\sqrt{10}\). Let \(a=\sqrt5\) and \(b=\sqrt2\). Use \((a+b)^2=a^2+2ab+b^2\) and \((a-b)^2=a^2-2ab+b^2\). Subtracting cancels \(a^2\\) and \(b^2\), leaving \(4ab\): \((a+b)^2-(a-b)^2=4ab=4\sqrt5\sqrt2=4\sqrt{10}\). Option A matches this identity. Option B, 7, would come from keeping only the square terms and wrongly ignoring the cross terms. Option C, \(2\sqrt{10}\), is half the correct cross-term difference; the two expansions together produce four products, not two. Option D, 14, has no valid calculation and could result from confusing the expression with a sum of squares. The reliable method is to use the difference-of-squares identity \((x+y)^2-(x-y)^2=4xy\). The memory cue is: opposite signs cancel the square terms and double the cross term twice.

Correct answer: A. (\frac{7+3\sqrt{5}}{2})

Explanation: Multiplying by the conjugate gives denominator (9-5=4) and numerator ((3+\sqrt{5})^2=14+6\sqrt{5}). So the answer is (\frac{7+3\sqrt{5}}{2}).

Correct answer: A. (5+2\sqrt{6})

Explanation: The conjugate of the denominator is (\sqrt{3}+\sqrt{2}) and the denominator becomes (1). So the value is ((\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}).

Correct answer: C. (\sqrt{7}+\sqrt{28})

Explanation: (\sqrt{7}+\sqrt{28}=3\sqrt{7}), which is irrational. In the other options the radicals simplify to rational values.