Representing real numbers on the number line

Positive numbers lie to the right of (0) on the number line. The value increases as we move right.

Negative numbers are located to the left of 0 on the number line, so -4 is correct. The numbers 1, 2, and 5 are positive and lie to the right of 0. Exam tip: moving left on the number line decreases the value of a number.

The governing concept is the order of real numbers on a number line. A number lies between 2 and 3 when it is greater than 2 and less than 3, that is, when 2 < x < 3. For option A, 2 < 2.5 < 3, so 2.5 is located to the right of 2 and to the left of 3. Option B, 3.5, is greater than 3; option C, 1.5, is less than 2; and option D, 4.5, is also greater than 3. Therefore, only option A satisfies both required inequalities and is the unambiguous answer.

(\frac{1}{2}=0.5), so it lies between (0) and (1). Thinking of a fraction as a decimal helps locate it.

The distance between two numbers is the absolute value of their difference: \(|6-4|=2\) units. Therefore, option A is correct. The 10-unit distractor may result from adding the numbers, but distance is found by taking their difference. Exam tip: Distance on a number line is always positive.

\(\sqrt{2}\) is irrational because it cannot be written as \(\frac{p}{q}\), and its decimal expansion is non-terminating and non-repeating. In contrast, 0.375 is a terminating decimal, so it is rational. Exam tip: the square root of a non-perfect square is usually irrational.

The greater number lies to the right and (-2>-5). Among negative numbers, the one closer to (0) is greater.

The decimal number \\(1.25=\\frac{125}{100}=\\frac{5}{4}\\). Therefore, option A is correct. Option B, \\(\\frac{4}{5}\\), is equal to \\(0.8\\), not \\(1.25\\). Exam tip: When a decimal has two digits after the decimal point, write it over 100 and then reduce the fraction.

The denominator of (\frac{3}{4}) is (4), so divide (0) to (1) into (4) equal parts. Then move (3) parts to the right.

(\frac{7}{4}=1.75), so it lies between (1) and (2). Think of an improper fraction in mixed or decimal form.

Since 0.2 has one digit after the decimal point, it can be written as \(\frac{2}{10}\). Dividing the numerator and denominator by 2 gives \(\frac{2}{10}=\frac{1}{5}\), so option A is correct. In exams, count the decimal places to choose the denominator as 10, 100, and so on.

Converting the decimal to a fraction gives \(0.75=\frac{75}{100}=\frac{3}{4}\). Therefore, the point lies between 0 and 1, at a distance of \(\frac{3}{4}\) from 0. Option B, \(\frac{1}{4}\), represents 0.25, not 0.75. Exam tip: convert a terminating decimal into a fraction and reduce it to the simplest form.

Since \(1^2=1<2<4=2^2\), taking positive square roots gives \(1<\sqrt{2}<2\). Therefore, \(\sqrt{2}\) lies between 1 and 2 on the number line. Exam tip: locate a square root by comparing the number with consecutive perfect squares; hence 0 and 1 is not correct.

Since (1^2<3<2^2), (\sqrt{3}) lies between (1) and (2). For square roots, check nearby perfect squares.

Since (2^2<5<3^2), (\sqrt{5}) lies between (2) and (3). Perfect squares quickly give the interval.

Since \(9=3^2\), \(\sqrt{9}=3\), so it represents the point 3 on the number line. Although \((-3)^2=9\), the square-root symbol denotes the principal, non-negative root. Exam tip: remember that \(\sqrt{a^2}=|a|\), not always \(a\).

The governing concept is the principal square root. The symbol √16 denotes the non-negative number whose square is 16. Since 4 × 4 = 16, we have √16 = 4, so its position on the number line is the point 4. Although both 4 and −4 have square 16, only the principal, non-negative square root is written as √16; −4 would be a solution of the equation x² = 16, not the value of √16. Option B is 8, whose square is 64, and option C is 16, whose square is 256. Hence option A is correct.

Since (3^2<10<4^2), (\sqrt{10}) lies between (3) and (4). Always place (n) between nearby perfect squares.

The distance between \(a\) and \(b\) on the number line is \(|a-b|\). The absolute value ensures that the distance is never negative. Using only \(a-b\) may give a negative result, so remember to use the absolute value when finding distance.

The midpoint of two numbers is their average. Therefore, the midpoint is \(\frac{1+5}{2}=\frac{6}{2}=3\). The value 4 is not correct because it does not divide the interval from 1 to 5 into two equal parts. In an exam, use the formula \(\frac{a+b}{2}\) to find the midpoint.