Representing real numbers on the number line

Both (2.5) and (-2.5) are (2.5) units from (0). Opposite numbers are equally distant from the origin.

(0.333\ldots=\frac{1}{3}), so both are at the same point. Connect recurring decimals with fractions.

(\sqrt{20}) is about (4.47), so it is closer to (4). Place (\sqrt{20}) between (4) and (5) and compare distances.

The total length is (2) and there are (8) parts, so each part is (\frac{2}{8}=\frac{1}{4}). Divide total length by the number of parts.

The distance between two numbers is the absolute value of their difference: \(|0.3-(-1.2)|=|1.5|=1.5\). Since -1.2 and 0.3 lie on opposite sides of zero, their distances from zero, 1.2 and 0.3, are added to get 1.5. The distractor 0.9 results from incorrectly subtracting 0.3 from 1.2. Exam tip: Use \(|a-b|\) to find the distance between any two points on the number line.

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

(\frac{9}{10}=0.9), whose distance from (1) is (0.1), while (0.95) has distance (0.05). The smaller distance is closer.

Since \(\sqrt{a^2}=|a|\), substituting \(a=-4\) gives \(\sqrt{(-4)^2}=\sqrt{16}=4\). Therefore, the represented point is 4. Option A is incorrect because the principal square root is never negative, while 16 is only the value of \(a^2\). Exam tip: rewrite \(\sqrt{a^2}\) as \(|a|\) before substituting the value of \(a\).

(-\frac{3}{5}=-0.6), so the order is (-0.7<-0.6<-0.2). Among negatives, the farther left number is smaller.

Since (3^2<12<4^2), (\sqrt{12}) lies between (3) and (4). A square root can be much smaller than the number.

Writing the decimals with the same number of places gives 1.050, 1.500, and 1.005. Since 1.005 is the smallest value, it lies farthest to the left on the number line. Remember that adding a zero at the end of a decimal does not change its value, so 1.05 = 1.050.

Since \(3^2=9\) and \(8<9\), we get \(\sqrt{8}<\sqrt{9}=3\). Therefore, \(\sqrt{8}\) lies to the left of 3 on the number line. Option B is incorrect because \(\sqrt{8}\) is not equal to \(\sqrt{9}\). Exam tip: For positive numbers, compare their squares to determine their order.

The governing concept is the midpoint property: the midpoint of two numbers a and b is their arithmetic mean, (a + b)/2. Here the endpoints are −2 and 4, so M = (−2 + 4)/2 = 2/2 = 1. Geometrically, 1 is equally far from both endpoints: its distance from −2 is 3 units and its distance from 4 is also 3 units. Therefore option A is correct. Option B results from using only half of the positive endpoint, option C reflects an incorrect sign calculation, and option D is not equidistant from the two given points. Adding signed numbers carefully is important: −2 + 4 equals 2, not −6 or 6.

Between (0) and (1), there are infinitely many rational and irrational numbers. Between any two real numbers, more numbers can be found.

The total distance is (0.8-0.2=0.6) and (\frac{0.6}{3}=0.2). Divide total distance by the number of parts for equal sections.

(\frac{5}{6}\approx0.833) and (\frac{7}{8}=0.875), so (\frac{7}{8}) is to the right. Use decimals or cross multiplication to compare.

(AC=|0.75-(-0.5)|=1.25), which is the greatest distance. The farthest points are often the endpoints.

The midpoint is (\frac{\frac{2}{5}+\frac{4}{5}}{2}=\frac{3}{5}). To find the exact middle point, take the average of the two points.

Since (3^2=9) and (4^2=16), (\sqrt{10}) lies between (3) and (4). In exams, bracket square roots between perfect squares.

Since (\sqrt{18}) lies between (4) and (5), (-\sqrt{18}) lies between (-5) and (-4). In exams, the direction reverses for negative square roots.