Real Numbers

( \sqrt{2} ) and ( \sqrt{8}=2\sqrt{2} ) are both irrational. Square roots of perfect squares can be rational.

In 0.101001000100001…, the number of zeros between successive 1s keeps increasing, so the digits do not follow a fixed repeating pattern and the decimal expansion is non-terminating. Therefore, the number is irrational. Merely using the digits 0 and 1 does not make a number rational; a rational number has a terminating or repeating decimal expansion. Exam tip: a non-terminating, non-repeating decimal is irrational.

Multiply numerator and denominator by ( \sqrt{5} ). This gives ( \frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5} ).

To rationalise the denominator, multiply numerator and denominator by ( \sqrt{3} ). The answer is ( \frac{2\sqrt{3}}{3} ).

For positive numbers, the square root of the greater number is greater. Since (3>2), ( \sqrt{3}>\sqrt{2} ).

( \sqrt{5}\approx2.236 ), so ( -\sqrt{5}\approx -2.236 ). ( -2 ) is closer to zero, so it is greater.

sqrt{1}=1, whereas frac{3}{4}=0.75. Since 1>0.75, sqrt{1} is the greater number. As an exam tip, simplify the square root and convert the fraction to a decimal when comparing such numbers.

There are infinitely many rational numbers between two distinct real numbers. This is a simple form of the density property.

Since (2^2<6<3^2), ( \sqrt{6} ) lies between (2) and (3). Since (6) is not a perfect square, it is irrational.

A square root is the number whose square equals the given number. Since \(0.2^2=0.04\), \(\sqrt{0.04}=0.2\). The principal square root is always non-negative; checking the decimal places helps avoid choosing 0.02 or 0.4.

Since \(0.5 \times 0.5 = 0.25\), \(\sqrt{0.25}=0.5\). The principal square root is always non-negative, so \(-0.5\) is not taken as the answer. In an exam, verify a decimal square root by squaring the obtained value.

Taking the square root of the numerator and denominator gives \(\sqrt{\frac{49}{64}}=\frac{\sqrt{49}}{\sqrt{64}}=\frac{7}{8}\). The principal square root is taken as positive, so \(-\frac{7}{8}\) would not be the answer. In exams, simplify the numerator and denominator separately under the square root.

Using the square-root property, \(\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}\). Since \(\sqrt{16}=4\) and \(\sqrt{81}=9\), the value is \(\frac{4}{9}\). The principal square root is taken as positive, so \(-\frac{4}{9}\) is not the answer. Exam tip: For a fraction made of perfect squares, take the square root of the numerator and denominator separately.

Adding irrational ( \sqrt{3} ) to rational (2) gives an irrational number. It is also a real number.

Subtracting an irrational number from a rational number gives an irrational result. Thus (5-\sqrt{2}) is an irrational real number.

Multiplying the same square root by itself gives the number inside. Therefore ( \sqrt{7}\times\sqrt{7}=7 ).

The decimal expansion of 0.125 terminates. Writing it as \(125/1000\) and dividing by 125 gives \(1/8\), so it is rational. Option B is a misconception: terminating decimals are rational. Exam tip: a terminating or repeating decimal can be expressed as a fraction.

The number is irrational because its decimal expansion is non-terminating and does not repeat in a fixed pattern; the number of zeros between successive 1s keeps increasing. Merely being written with digits does not make a number rational. A rational number has a terminating or recurring decimal expansion. Exam tip: classify an infinite non-repeating decimal as irrational.

(3^2=9) and (4^2=16) so ( \sqrt{11} ) lies between (3) and (4). Use squares to locate square roots.

Since \(98=49\times2\) and 49 is a perfect square, \(\sqrt{98}=\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}=7\sqrt{2}\). Option C, \(2\sqrt{49}\), equals 14, not \(7\sqrt{2}\). To simplify a surd, factor out the largest perfect-square factor first.