Representing real numbers on the number line

To locate a number between \sqrt{3} and 2, first estimate the irrational endpoint. Since 1.7^2=2.89 and 1.8^2=3.24, \sqrt{3} is approximately 1.732, so the required interval is about (1.732,2). The number 1.8 lies inside this interval, making option B correct. Although 1.6 is less than \sqrt{3}, 2.1 and 3 are both greater than 2. Thus the other options cannot lie strictly between the two endpoints. Comparing squares or using a reliable decimal approximation avoids treating \sqrt{3} as exactly 1.7 or 1.8.

\(\sqrt{2}\) 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.

Because the square-root function is increasing for non-negative numbers, 6<\sqrt{n}<7 can be squared without changing the inequality directions. This gives 36<n<49. Test the options against this interval: 30 and 35 are below 36, 50 is above 49, while 40 lies strictly between 36 and 49. Therefore n=40 is possible, so option C is correct. The question asks for a possible value, not necessarily a perfect square; indeed, \sqrt{40} is approximately 6.325 and is genuinely between 6 and 7. Squaring the endpoint bounds is the safest method.

(-1.25 = -\frac{125}{100} = -\frac{5}{4}), so the correct fraction is (-\frac{5}{4}). Option C, (\frac{5}{4}), is positive and lies to the right of 0, whereas (-1.25) lies to the left. Exam tip: Convert a terminating decimal to a fraction over 10, 100, or 1000, then reduce it to lowest terms.

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.

Since (0.7)^2=0.49, we have (\sqrt{0.49})=0.7, so the point lies at 0.7 on the number line. Option A is incorrect because (0.07)^2=0.0049. In an exam, check the number of decimal places when finding the square root of a decimal.

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 nearest-point comparison on the number line. Since 25<32<36, we know that 5<\sqrt{32}<6. More precisely, \sqrt{32}=4\sqrt{2}\approx4(1.414)=5.656. Its distance from 5 is about 0.656, while its distance from 6 is about 0.344, so 6 is nearer. Therefore option C is correct. Option B is the other adjacent integer but is farther away; 4 and 7 are even more distant. Bounding the square root between consecutive integers first makes the nearest-integer decision straightforward and prevents confusing the value with \sqrt{36} or \sqrt{25}.

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.

The governing concept is equal partition of a distance on the number line. The total distance between the points 2 and 5 is the absolute difference |5-2|=3 units. Dividing this total length into 6 equal parts gives 3/6=1/2 unit for each part. Hence option B is correct. Option A would result from dividing 3 by 9, option C would incorrectly use a unit length, and option D is the entire distance rather than one of six equal portions. Using the absolute difference is important because distance is always non-negative, regardless of the direction in which the points are named.