Irrational numbers

To simplify a radical, separate any perfect-square factor from the number inside the root. Since 50 = 25 × 2 and \(\sqrt{25}=5\), we get \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). The original expression then becomes \(5\sqrt{2}-\sqrt{2}\).

These are like radicals because both terms contain the same \(\sqrt{2}\). Subtract their coefficients: \(5\sqrt{2}-1\sqrt{2}=(5-1)\sqrt{2}=4\sqrt{2}\). No further simplification is possible because 2 has no factor that is a perfect square greater than 1. Therefore option C is correct. The answer is not 5 or \(\sqrt{48}\); those forms do not represent the simplified subtraction as directly.

(\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the total is (6\sqrt{3}). First convert all terms into like radicals.

It has no fixed repeating block, so it is irrational. Avoid treating a non-repeating decimal as repeating.

For the quotient of square roots, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\). Thus, \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}=\sqrt{9}=3\), so option A is correct. The closest distractor, \(\sqrt{3}\), results from failing to simplify the quotient correctly. Exam tip: first divide the radicands when the denominator is nonzero, then simplify the resulting square root.

(\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the sum is (5\sqrt{5}). First take out perfect-square factors.

The governing concept is that the square-root function preserves order for non-negative numbers. Since 2 < 5 < 8, taking square roots gives √2 < √5 < √8, so option C is correct without needing decimal approximations. A numerical check confirms this: √2 is about 1.41, √5 is about 2.24, and √8 is about 2.83. Option A is 1, which is less than √2 because 1² < 2. Options B and D are greater than √8: 3² = 9 > 8 and 4² = 16 > 8. The important reasoning is to compare the radicands 2, 5, and 8, since all are non-negative; their square roots retain the same order.

The side of the square is \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\,\text{m}\). Since \(\sqrt{2}\) is irrational, \(3\sqrt{2}\) is also irrational. Option B has the correct length but the wrong classification. Exam tip: first extract perfect-square factors from a square root.

An irrational number has a decimal expansion that is non-terminating and non-repeating. In 0.101001000100001..., the blocks of zeros grow and no fixed digit pattern repeats indefinitely, so the expansion is non-terminating and non-recurring. Thus option A is irrational. Option B is recurring, option C terminates, and 11/13 is rational because it is a ratio of two integers.

In \(0.1010010001\ldots\), the zeros between consecutive 1s keep increasing, so no fixed block repeats. Hence it is irrational. Exam tip: identify whether an endless decimal has a repeating pattern.

This decimal never terminates, and the number of zeros increases as 1, 2, 3, 4, …, so no fixed digit block repeats. Hence it is irrational. Exam tip: a non-terminating, non-repeating decimal is irrational.

The governing concept is the distinction between rational and irrational terms in a sum. The number 4 is rational because it can be written as 4/1. The number √6 is irrational because 6 is not a perfect square, so its square root cannot be expressed as a ratio of integers. In the expression 4 + √6, the radical term is therefore the irrational part, making option D correct. Option A is the rational part, not the irrational part. Option B is only the radicand and is itself rational. Option C incorrectly multiplies the two terms; no such multiplication appears in the given expression. The answer must be identified from the terms actually present.

(\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so (2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}). Add and subtract like radicals.

In \(0.272727\ldots\), the block 27 repeats, so it is a recurring decimal and therefore rational. \(\sqrt{2}\) and \(\sqrt{5}\) are irrational. Exam tip: every recurring decimal represents a rational number.

The square root of a positive integer is irrational when it is not a perfect square. This is the main rule for identifying square roots.

The numbers of zeros between successive 1s are 1, 2, 3, ... , so no fixed digit block repeats. Hence the decimal is non-terminating and non-repeating, making it irrational. Exam tip: check repetition, not the digits used.

(\sqrt{2}) and (\sqrt{3}) are different irrational radicals and their sum is irrational. It is not written as (\sqrt{5}).

The number of zeros between successive 1s increases as 1, 2, 3, 4, …, so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: always check for a repeating block.

A terminating decimal ends after a finite number of digits, such as 0.25. A repeating decimal has a fixed block of digits that repeats forever, and it represents a rational number. A non-repeating infinite decimal does not repeat a fixed pattern and is generally irrational. These distinctions identify the correct option.

In option A, the block 123 repeats indefinitely: 0.123123123... Therefore the decimal is non-terminating and repeating. Option B does not show a fixed repeating block, option C is the non-terminating non-repeating decimal of \(\sqrt{2}\), and option D terminates. Hence option A is correct. The note that a repeating decimal is rational is also important: repeating does not mean irrational.

\(0.\overline{3}=\frac{1}{3}\), so it is non-terminating but repeating and rational. Irrational decimals are non-terminating and non-repeating. Exam tip: identify a recurring block to spot a rational number.

Use the distributive property: \(\sqrt{3}\times\sqrt{3}=3\) and \(\sqrt{3}\times\sqrt{12}=\sqrt{36}=6\). Hence, the expression equals \(3+6=9\). Option 6 is only the value of the second product and omits the first term, 3. Exam tip: when multiplying square roots, simplify using \(\sqrt{a}\sqrt{b}=\sqrt{ab}\).