Number line

For rational numbers \(a<b\), \(a+\frac{b-a}{\sqrt2}\) is an irrational number between them since \(0<\frac1{\sqrt2}<1\). Exam tip: every interval on the number line contains both rational and irrational numbers.

The midpoint of two numbers on a number line is their average: \(\frac{-10+7}{2}=\frac{-3}{2}=-1.5\). Therefore, \(-1.5\) is correct. Although \(-2\) is close, it is not equally distant from both numbers. Exam tip: add the two numbers and divide by 2 to find their midpoint.

The distance between two numbers on a number line is the absolute value of their difference. Thus, \(\lvert -20-(-4)\rvert=\lvert -16\rvert=16\). Therefore, 16 is correct. A result of 15 can arise from an arithmetic error. Exam tip: always express distance as a positive value.

( -0.00009 ) is closer to zero, whereas ( -0.0009 ) is a negative number farther from zero. On the number line, the negative number farther to the right is greater; therefore, ( -0.00009 ) is greater. Choosing ( -0.0009 ) is a common mistake because, among negative numbers, the one with the larger magnitude is smaller. Exam tip: When comparing negative decimals, first identify which number is closer to zero.

We can write -3.6 as -3.60. Since -3.60 is more negative than -3.06, it lies farther to the left on the number line and is therefore smaller. The close distractor, -3.06, has a smaller negative magnitude. Exam tip: while comparing negative decimals, first write them with the same number of decimal places.

Every interval between two distinct real numbers contains infinitely many rational as well as irrational numbers. Thus, the “only rational” and “only irrational” statements are false. Exam tip: pay close attention to the word “always”.

By the Pythagorean theorem, the hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Since \(9<13<16\), we get \(3<\sqrt{13}<4\). Adding 3 and 2 is incorrect. Exam tip: bracket the number between perfect squares to locate its square root.

On a number line, the distance between two numbers is the absolute value of their difference: \(\lvert -30-(-10)\rvert=\lvert -20\rvert=20\). Therefore, the correct answer is 20. \(-20\) can be a signed difference, but distance can never be negative. Exam tip: always take the absolute value of the difference when finding distance.

Since \(2^2<5<3^2\), we get \(2<\sqrt{5}<3\). As 5 is not a perfect square, \(\sqrt{5}\) is irrational; it is not 2.5. Exam tip: compare with nearby perfect squares first.

The distance between two numbers on a number line is the absolute value of their difference: \(\lvert -40-(-16)\rvert=\lvert -24\rvert=24\). Therefore, the correct answer is 24. \(-24\) can be the signed difference, but a distance can never be negative. Exam tip: always take the absolute value when finding distance.

On a number line, the number farther to the left is smaller. Here, \(-0.3=-0.300\), and \(-0.333\) is more negative; therefore, \(-0.333 < -0.300\). Hence, \((-0.333)\) is the correct answer. Although \(-0.3\) is close, it is nearer to zero and is therefore greater. Exam tip: While comparing negative decimals, write equal decimal places; the number with the greater magnitude is smaller.

The distance between two numbers on a number line is the absolute value of their difference: \(\left|-15-(-1)\right|=|-14|=14\). Therefore, the correct answer is 14. A value such as 13 results from incorrect subtraction. Exam tip: a distance is always non-negative.

Since \(2^2=4\) and \(3^2=9\), \(\sqrt{7}\) lies between 2 and 3. Also, \(\sqrt{7}\approx2.646\), so its distance from 3 is about 0.354, less than 0.646 from 2. Thus option B is wrong. Exam tip: compare nearby perfect squares first.

The midpoint of two numbers on a number line is their average: \(\frac{-6+(-18)}{2}=\frac{-24}{2}=-12\). Therefore, the correct answer is \(-12\). Although \(-11\) may seem close, it is not equally distant from both numbers. Exam tip: include the signs of negative numbers while adding them for a midpoint.

The distance between two numbers on a number line is the absolute value of their difference. Thus, \(|10-(-50)|=|60|=60\). Although \(-60\) can arise as a signed difference in the reverse order, a distance can never be negative. Exam tip: always take the absolute value of the difference when finding distance.

For negative numbers, the number closer to zero is greater. Here, \(-0.00201 > -0.00210\) because \(-0.00201\) is closer to zero. Writing \(-0.0021\) as \(-0.00210\) makes the comparison clear. Exam tip: Add trailing zeros when needed to compare the same number of decimal places.

The midpoint of two numbers on a number line is their average: \(\frac{-4+20}{2}=\frac{16}{2}=8\). Therefore, the correct answer is \(8\). \(10\) would be the midpoint of 0 and 20, so it is not correct here. Exam tip: add the two numbers and divide by 2 to find their midpoint.

On a number line, the distance between two numbers is the absolute value of their difference: \(\lvert -25-(-5)\rvert=\lvert -20\rvert=20\). Therefore, the correct answer is 20. A value such as 15 can result from incorrect subtraction. Exam tip: distance is always non-negative.

Write \(-0.56\) as \(-0.560\) to compare the decimals easily. Since \(-0.506\) is closer to zero than \(-0.560\), it is greater; among negative numbers, the number closer to zero is greater. Therefore, \(-0.506\) is the correct answer. Exam tip: Make the number of decimal places equal before comparing decimals.

The midpoint of two numbers on a number line is their average: \(\frac{-9+5}{2}=\frac{-4}{2}=-2\). Therefore, \(-2\) is correct. \(-1\) is not correct because it is not at an equal distance from \(-9\) and \(5\). Exam tip: Add the two numbers and divide by 2 to find their midpoint.