Real Numbers

( \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.

(5^2=25) and (6^2=36), so ( \sqrt{26} ) lies between (5) and (6). Compare nearby perfect squares.

( \sqrt{20}\approx4.47 ), so ( -\sqrt{20}\approx-4.47 ). ( -4.4 ) is closer to zero, so it is greater.

The governing concepts are the classification of real numbers and the fact that division by zero is undefined. Every rational and every irrational number belongs to the real-number system. Thus √2 is real even though it is irrational. The repeating decimal 0.333… equals 1/3, so it is rational and real. The integer −7 can be written as −7/1, making it rational and therefore real. However, 5/0 is not defined: if 5/0 were a number x, then multiplying by zero would require 0×x = 5, which is impossible because 0×x is always 0. Hence 5/0 is not a real number, and option C is the incorrect statement.

The diagonal of a square is side × \(\sqrt{2}\), so it is \(\sqrt{2}\times\sqrt{2}=2\) cm, which is rational. The perimeter is \(4\times\sqrt{2}=4\sqrt{2}\) cm, which is irrational because the product of a non-zero rational number and an irrational number is irrational. Therefore, option A is correct. Exam tip: Do not classify an expression merely by looking at an irrational term; simplify it first.

A rational number has a terminating decimal expansion when, after simplification, the denominator has no prime factors other than 2 and 5. This happens because powers of 2 and 5 can combine to make a power of 10. For \\(\frac{13}{40}\\), the numerator and denominator have no common factor, and \\(40=2^3\times5\\).

Since the denominator contains only the allowed prime factors 2 and 5, the decimal terminates. In fact, \\(\frac{13}{40}=\frac{13\times25}{40\times25}=\frac{325}{1000}=0.325\\). Therefore, option A is correct. It is not irrational or undefined, and it is not non-terminating non-repeating; those descriptions do not fit this rational fraction.

Factoring 18 gives \(18=9\times2\), so \(\sqrt{18}=3\sqrt{2}\). Since \(\sqrt{2}\) is irrational, multiplying it by the non-zero rational number 3 still gives an irrational number. Therefore, \(\sqrt{18}\) is irrational. Option A is wrong because the square root of every integer is not rational, and 18 is not a perfect square. Exam tip: Before classifying \(\sqrt{n}\), check whether \(n\) is a perfect square.

\(\sqrt{5}\times\sqrt{20}=\sqrt{5\times20}=\sqrt{100}=10\). The number 10 is an integer and a natural number, so it is also a rational real number. Therefore, option A is correct. It is not irrational because the product simplifies to a whole number. Exam tip: For positive radicands, use \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\), then check whether the resulting square root is a perfect square.

Conjugate multiplication gives (2^2-(\sqrt{13})^2=4-13=-9). Use the difference of squares rule here.

The governing concept is rationalising a denominator by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of 2 − √3 is 2 + √3. Thus, (2 + √3)/(2 − √3) = [(2 + √3)(2 + √3)]/[(2 − √3)(2 + √3)]. The numerator is (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3. The denominator is a difference of squares: 2² − (√3)² = 4 − 3 = 1. Consequently, the value is (7 + 4√3)/1 = 7 + 4√3, so option A is correct. Option B has the wrong sign in the irrational term. Option C ignores the numerator expansion and cannot result from cancellation because the numerator and denominator are not identical. Option D does not follow from either the square identity or the conjugate product.

Multiplying by the conjugate ( \sqrt{5}-1 ) makes the denominator (5-1=4). Use difference of squares in the conjugate method.

A rational number has a terminating or recurring decimal expansion. Here the gaps of zeros between 1s are 1, 2, 3, 4, ..., so no fixed repeating block exists. Exam tip: check the repetition pattern, not merely the digits used.

In 0.1010010001..., the number of zeros between successive 1s is 1, 2, 3, 4..., so no repeating block occurs. A non-terminating, non-repeating decimal is irrational. Exam tip: an infinite decimal can still be rational only if it repeats.

In real numbers, ( \sqrt{a^2}=|a| ) because the principal square root is non-negative. Pay attention to the sign.

Using the identity \(\sqrt{x^2}=|x|\), we get \(\sqrt{(-5)^2}=|-5|=5\). The principal square root is always non-negative, so \(-5\) is not correct. Exam tip: remember that \(\sqrt{x^2}=|x|\), not always \(x\).

Multiplying by the conjugate ( \sqrt{3}-\sqrt{2} ) makes the denominator (3-2=1). So the rationalised form is ( \sqrt{3}-\sqrt{2} ).

Since \(180=36\times5\) and \(36\) is the largest perfect-square factor, \(\sqrt{180}=\sqrt{36\times5}=\sqrt{36}\times\sqrt{5}=6\sqrt{5}\). Option B is the unsimplified form, while options C and D have incorrect coefficients. Exam tip: factor the radicand using its largest perfect-square factor before simplifying the surd.

\(245=49\times5=7^2\times5\). Therefore, \(\sqrt{245}=\sqrt{7^2\times5}=7\sqrt{5}\), so option B is correct. Option D is incorrect because the square root of the perfect-square factor \(49\) is \(7\), not \(49\). Exam tip: To simplify a radical, factor the radicand into the largest possible perfect square times the remaining factor.