Irrational numbers

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.

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.

\(\sqrt{3}\) 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 \(k\sqrt{n}\).

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.

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}).

\(x+\frac{1}{3}\) 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.

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

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.

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

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.

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.

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.

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

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

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.

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.

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}).

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}).

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