Representing real numbers on the number line

Since \\(4<5<9\\), we have \\(2<\\sqrt{5}<3\\). Adding 2 to all parts gives \\(4<2+\\sqrt{5}<5\\). Therefore, \\(2+\\sqrt{5}\\) lies between 4 and 5. Exam tip: Compare the number under the square root with consecutive perfect squares instead of finding a decimal approximation first.

The total distance is (2), so each part is (\frac{2}{8}=\frac{1}{4}). Find the distance and divide by equal parts.

After the third step, the value is (-1+3\times0.25=-0.25). For equal steps, multiply step number by step length.

The governing concept is comparison of real numbers on a number line. Estimate the irrational endpoint using squares: 1.7²=2.89 and 1.8²=3.24, so √3 is approximately 1.732. Therefore the open interval between √3 and 2 is approximately (1.732,2). The number 1.8 lies in this interval, so option B is correct. In contrast, 1.6 is less than √3, while 2.1 and 3 are both greater than 2. Hence those values cannot lie strictly between the two endpoints. Comparing squares gives a dependable bound for √3 and avoids the incorrect assumption that the irrational number is equal to a nearby integer.

\(\sqrt{2}\) is irrational, but every real number has a unique point on the number line. Since \(1^2<2<2^2\), it lies between 1 and 2. Option B wrongly treats a non-terminating decimal as unrepresentable; use consecutive squares in exams.

The governing concept is the increasing nature of the square-root function for non-negative numbers. From 6<√n<7, square all three parts; because all quantities are positive, the inequality direction is preserved: 36<n<49. Now test the choices. Values 30 and 35 are below 36, while 50 is above 49. Only 40 lies strictly between 36 and 49, so n=40 is possible and option C is correct. It is not necessary that n be a perfect square. Indeed, √40 is approximately 6.325, which is genuinely between 6 and 7. Squaring the endpoint bounds is more reliable than checking only the nearby perfect squares.

Convert the decimal to a fraction: −1.25 = −125/100 = −5/4 after simplifying by 25. Therefore, option B represents the same point.

The midpoint is (\frac{\frac{5}{4}+\frac{9}{4}}{2}=\frac{7}{4}). The average of two fractions gives the midpoint.

By the Pythagorean theorem, the hypotenuse is \(c=\sqrt{3^2+4^2}=\sqrt{25}=5\) units. Therefore, the point representing 5 can be constructed on the number line by marking a distance of 5 units from 0. The values 6, 7, and 12 are not the hypotenuse of this triangle. Exam tip: For perpendicular sides \(a\) and \(b\), use \(c=\sqrt{a^2+b^2}\).

(\sqrt{11}) lies between (3) and (4), so (-\sqrt{11}) lies between (-4) and (-3). The negative sign changes the side.

(0.125=\frac{125}{1000}=\frac{1}{8}). Start with denominator (10), (100), or (1000), then simplify.

Distance from (0) is magnitude, so \(\left|-\frac{7}{3}\right|=\frac{7}{3}\). Distance is never negative.

(\sqrt{2}\approx1.41), so (-1.5<-\sqrt{2}<-1.3). For negative numbers, the farther left number is smaller.

The principal square root of 0.49 is 0.7 because 0.7 × 0.7 = 0.49. Therefore, √0.49 is located at 0.7 on the number line.

The symmetric point about (0) is the opposite number, so the answer is (-1.75). Opposite numbers are at equal distance.

The midpoint is found by taking the average of the two endpoints: \(\frac{0.6+1.4}{2}=\frac{2}{2}=1\). Therefore, the correct answer is 1. The value 0.8 results from adding only 0.2 to 0.6, whereas 1 is 0.4 away from both numbers. Exam tip: use \(\frac{\text{first value}+\text{second value}}{2}\) to find the midpoint.

(\frac{2}{3}\approx0.667) and (\frac{3}{4}=0.75), so (0.7) lies between them. Use decimal form for comparison.

\(\sqrt{7}\) is irrational because 7 is not a perfect square. Since \(2^2<7<3^2\), its point lies between 2 and 3 and can be constructed using a square-root method. Exam tip: first check whether the radicand is a perfect square.

The governing concept is locating a square root between consecutive integers and then comparing distances on the number line. Since 25 < 32 < 36, taking positive square roots gives 5 < √32 < 6. To decide which integer is nearer, estimate √32 as 4√2, and use √2 ≈ 1.414; hence √32 ≈ 5.656. Its distance from 5 is approximately 0.656, whereas its distance from 6 is approximately 0.344. Therefore √32 is closer to 6, so option C is correct. Option B is the other adjacent integer but is farther away. Integers 4 and 7 are still more distant, so they cannot be the nearest point.

The distance between two points on a number line is the absolute value of the difference of their coordinates. Thus, AB = |−0.75 − (−2.25)| = |1.5| = 1.5 units. The value 0.75 is only the distance of B from zero, not the distance between A and B. Exam tip: subtract the smaller coordinate from the larger coordinate to obtain the positive distance.

Since \(\sqrt{3}\approx1.732\), we get \(1-\sqrt{3}\approx1-1.732=-0.732\). This number is greater than \(-1\) and less than \(0\), so it lies between \(-1\) and \(0\). For such questions, first estimate the irrational number and then carefully determine the sign after subtraction.

When \(n\) is not a perfect square, \(\sqrt{n}\) is irrational. Its length can be obtained through a semicircle construction and marked from 0 on the number line. Exam tip: irrational does not mean unrepresentable.

\(\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\approx0.7071\), so its closest decimal estimate is \(0.707\). The square of 0.5 is only 0.25, whereas \(0.707^2\approx0.5\), confirming that 0.707 is correct. An exam tip is to square the options to verify a square-root estimate.

(-0.1) is less than zero, so it is not between (0) and (1). Compare with both bounds when checking between values.

(\frac{2}{5}=\frac{4}{10}), so the distance is (\frac{7}{10}-\frac{4}{10}=\frac{3}{10}). Use a common denominator before subtracting.