Real Numbers

Since \(125=25\times5\) and \(25\) is a perfect square, \(\sqrt{125}=\sqrt{25\times5}=5\sqrt{5}\). Therefore, option A is correct. Option B has an extra factor of \(5\), while option C is not the simplified form and equals \(25\). Exam tip: factor the radicand using its largest perfect-square factor before simplifying the square root.

To simplify a square root, factor the radicand into the largest possible perfect-square factor and the remaining factor. Here 147 = 49 × 3 = 7² × 3. Therefore √147 = √(49 × 3) = √49 × √3 = 7√3. Thus option B is correct. Option C corresponds to √63 rather than √147, while option A incorrectly uses 21 as the coefficient and option D treats 49 itself as the coefficient after taking a square root. The simplified form should have no square factor remaining under the radical; since 3 has no perfect-square factor greater than 1, 7√3 is fully simplified.

To simplify a square root, factor the number inside it into a perfect square and another factor. A perfect-square factor can come outside the radical because \\(\sqrt{a^2b}=a\sqrt{b}\\) for positive values. For 192, a convenient factorisation is \\(192=64\times3\\), and 64 is the largest perfect-square factor shown among the options.

Therefore, \\(\sqrt{192}=\sqrt{64\times3}=\sqrt{64}\sqrt{3}=8\sqrt{3}\\). Hence option C is correct. Although \\(4\sqrt{12}\\) has the same value, it is not fully simplified because \\(\sqrt{12}=2\sqrt{3}\\), giving \\(8\sqrt{3}\\). The forms involving 12 or 16 do not equal the original radical.

( \sqrt{8}=2\sqrt{2} ) and ( \sqrt{18}=3\sqrt{2} ). So the sum is (6\sqrt{2}+6\sqrt{2}=12\sqrt{2}).

The area of a square equals side², so its side is \(\sqrt{2}\) metres. Since 2 is not a perfect square, \(\sqrt{2}\) cannot be expressed as a ratio of two integers and is therefore irrational. Option A is incorrect because a side of 2 metres would give an area of 4 square metres. Exam tip: To find the side of a square, take the square root of its area and check whether the area is a perfect square.

By distributive law, ( \sqrt{3}\times2+\sqrt{3}\times\sqrt{3}=2\sqrt{3}+3 ). Remember ( \sqrt{a}\sqrt{a}=a ) in surd multiplication.

Use the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(a=2\) and \(b=\sqrt{7}\), so \((2+\sqrt{7})^2=2^2+2(2)(\sqrt{7})+(\sqrt{7})^2=4+4\sqrt{7}+7=11+4\sqrt{7}\). Therefore, option A is correct. Remember that \((\sqrt{7})^2=7\), not \(\sqrt{2}\).

Using ( (a-b)^2=a^2-2ab+b^2 ), we get (25-10\sqrt{6}+6=31-10\sqrt{6}). Pay special attention to signs.

Since \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\), we get \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). This is not equal to \(\sqrt{10}\). The rule \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\) is not generally valid, so option D is incorrect. Exam tip: simplify square roots first and combine only like radical terms; do not add radicands directly.

The area of a square is (side)². Hence, side = \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\) cm. Its perimeter is 4 × side = \(4\times3\sqrt{2}=12\sqrt{2}\) cm, so option B is correct. In option D, \(\sqrt{18}\) cm is a correct form of the side, but the stated perimeter is incorrect. Exam tip: For a square, first find the side from the area and then multiply the side by 4 to obtain the perimeter.

Multiplying numerator and denominator by ( \sqrt{3} ) gives ( \frac{7\sqrt{3}}{6} ). The radical should be removed from the denominator.

Multiply by the conjugate ( \sqrt{6}-\sqrt{5} ) and the denominator becomes (6-5=1). So the answer is ( \sqrt{6}-\sqrt{5} ).

\(\sqrt{2}\) is an irrational number, so its decimal expansion is non-terminating and non-recurring. In contrast, \(0.125\) is terminating, \(0.\overline{27}\) is recurring, and \(\sqrt{81}=9\) is rational. Exam tip: the decimal expansion of every rational number is either terminating or recurring, whereas an irrational number has a non-terminating, non-recurring decimal expansion.

( \sqrt{45}=3\sqrt{5} ), ( \sqrt{80}=4\sqrt{5} ), and ( \sqrt{20}=2\sqrt{5} ). So the value is (3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}).

Convert all surds into terms of ( \sqrt{5} ). (3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}) is correct.

A rational number has a terminating or an eventually repeating decimal expansion. In the given number, the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Hence its decimal expansion is non-terminating and non-repeating, making the number irrational. The fact that only two digits are used does not determine rationality. Exam tip: Check whether the decimal terminates or repeats, not merely which digits appear.

An irrational number has a decimal expansion that is non-terminating and non-recurring, so option B is correct. Option A is incorrect because not every rational number has a terminating decimal expansion; for example,
has a non-terminating but recurring expansion. Integers and rational numbers are parts of the real number system, so options C and D are also false. Exam tip: a terminating or non-terminating recurring decimal represents a rational number, whereas a non-terminating non-recurring decimal represents an irrational number.

In option B, \(2-\sqrt{2}\) is irrational because subtracting an irrational number from a rational number gives an irrational result. However, \(\sqrt{2}+(2-\sqrt{2})=2\), which is rational. Thus, this example disproves the student's claim. Options A, C and D are irrational. Exam tip: The sum of two irrational numbers may be either rational or irrational.

Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but \(\sqrt{5}+(-\sqrt{5})=0\), and \(0\) is rational. Therefore, this example disproves the student's claim. In the other options, the sums remain irrational. Exam tip: The sum of two irrational numbers is not always irrational; opposite irrational numbers can add to zero.

Using the Pythagorean theorem for the square, \(d^2=1^2+1^2=2\), so \(d=\sqrt{2}\) metres. Since \(\sqrt{2}\) cannot be expressed as a ratio of two integers, it is irrational. Option B is incorrect because 2 is the value of the square of the diagonal, not the diagonal itself. Exam tip: the diagonal of a square with side \(a\) is \(a\sqrt{2}\).

Option A is correct. The decimal expansion of a rational number is either terminating or recurring. In this number, the number of zeros between successive 1s keeps increasing, so no fixed block repeats; hence the expansion is non-terminating and non-repeating. Using only the digits 0 and 1 does not make a number rational. Exam tip: For an infinite decimal, check whether a fixed repeating pattern exists before classifying it as rational.

The first term is ( \frac{3}{4} ) and the second is ( \frac{5}{6} ). The sum is ( \frac{9}{12}+\frac{10}{12}=\frac{19}{12} ).

( \sqrt{\frac{49}{100}}=\frac{7}{10} ) and ( \sqrt{\frac{4}{25}}=\frac{2}{5} ). The difference is ( \frac{3}{10} ).

Since (2^2<8<3^2), ( \sqrt{8} ) lies between (2) and (3). Since (8) is not a perfect square, it is irrational.

\(\sqrt{2}\) is irrational, but \(\sqrt{2}\times\sqrt{2}=2\), which is rational. Therefore, the product of two irrational numbers is not always irrational, so option A disproves the student's claim. In option B, the product is \(\sqrt{6}\), which is irrational; in C and D, the products \(2\sqrt{2}\) and \(3\sqrt{5}\) are also irrational. Exam tip: Never assume that multiplying two irrational numbers must give an irrational result; always simplify the product.