Number line

\(\sqrt{5}\) is irrational because \(4<5<9\) and 5 is not a perfect square. Its decimal expansion is non-terminating and non-repeating. \(0.125\) terminates, so it is rational. Exam tip: the square root of a non-perfect-square natural number is irrational.

For rational \(r<s\), \(r+(s-r)\frac{\sqrt2}{n}\) with \(n\ge2\) lies between them and is irrational. Different values of \(n\) give infinitely many such numbers. Exam tip: every interval contains both rational and irrational numbers.

On a number line, the distance between two numbers is the absolute value of their difference: \(\lvert 8-(-1)\rvert=\lvert 9\rvert=9\). Hence, the correct answer is \(9\). Choosing \(7\) usually results from an error while subtracting a negative number. Exam tip: a distance is always non-negative.

Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between 2 and 3. Also, \(\left(\frac{11}{5}\right)^2=\frac{121}{25}<5\), so \(\frac{11}{5}<\sqrt{5}\). Thus option B reverses their order. Exam tip: for positive numbers, comparing squares is often quicker.

On the number line, the integers between -2 and 5 must be greater than -2 and less than 5: -1, 0, 1, 2, 3, and 4. Therefore, 1 is correct. -3 lies to the left of -2, while 5 and 6 are not less than 5. Exam tip: “Between” usually excludes the two endpoints.

On a number line, the distance between two numbers is the absolute value of their difference: \(|10-(-10)|=|20|=20\). Therefore, the correct answer is 20. The value 10 is the distance of either number from 0, not the distance between the two numbers. Exam tip: Always take the absolute value of the difference when finding distance.

( -3.001 ) is more negative than ( -3 ). On a number line, the more negative number lies farther to the left and is therefore smaller. Hence, ( -3.001 ) is correct; the two numbers are not equal because their decimal values differ. Exam tip: Among negative numbers, the number with the greater magnitude is the smaller number.

The absolute value of a number is its distance from zero on the number line, so it is never negative. Since \(-12\) is 12 units from zero, \(|-12|=12\). \(-12\) is the original number, not its absolute value. Exam tip: for a negative number, remove the minus sign to find its absolute value.

Write 1.2 as 1.20 so that both decimals have the same number of decimal places. Comparing 1.20 and 1.02, the tenths digit 2 is greater than 0, so 1.2 is greater. The 2 in the hundredths place of 1.02 does not make it equal to 1.2. Exam tip: add zeros to the right of decimals before comparing place values.

The midpoint of two numbers on a number line is their average: \(\frac{-5+5}{2}=\frac{0}{2}=0\). Therefore, \(0\) is at an equal distance from \(-5\) and \(5\). The numbers \(5\) and \(-5\) are the given endpoints, not the midpoint. Exam tip: The midpoint of opposite numbers \(a\) and \(-a\) is always \(0\).

The distance between two numbers on a number line is their absolute difference: \(\lvert -9-(-1)\rvert=\lvert -8\rvert=8\). Therefore, the correct answer is 8. The number 9 is only the absolute value of \(-9\), not the distance between the two points. Exam tip: a distance is always non-negative.

Both rational and irrational numbers correspond to distinct points on the real number line. For example, \(1<\sqrt{2}<2\), so \(\sqrt{2}\) has a position on the line. Exam tip: one point represents only one real number.

On the number line, a number between \(-7\) and \(2\) must lie to the right of \(-7\) and to the left of \(2\). Since \(-2\) is greater than \(-7\) and less than \(2\), it is the correct answer. \(-8\) is less than \(-7\), so it is not between them. Exam tip: For “between”, check that the number is greater than the lower number and less than the higher number.

Converting \(-\frac{3}{5}\) into a decimal gives \(-3 \div 5 = -0.6\). Thus, \(-\frac{3}{5}\) and \(-0.6\) represent the same point on the number line, so they are equal. Choosing either number as greater is incorrect because their values are identical. Exam tip: compare fractions by converting them to decimals or by using a common denominator.

The midpoint of two numbers on a number line is their average: \(\frac{-4+7}{2}=\frac{3}{2}=1.5\). Therefore, \(1.5\) is correct. \(2\) is not correct because it is not at an equal distance from \(-4\) and \(7\). Exam tip: To find a midpoint, add the two numbers and divide by 2.

On a number line, the distance between two numbers is the absolute value of their difference: \(\left|-16-(-4)\right|=\left|-12\right|=12\). Therefore, the correct answer is 12. \(20\) comes from adding the absolute values of the two numbers, which is not the correct method here. Exam tip: distance is always non-negative.

Between any two distinct real numbers, there are infinitely many rational as well as irrational numbers. The “only rational” choice is false. Exam tip: remember that real numbers are dense on the number line.

The midpoint of two numbers on a number line is their average: \(\frac{0+(-5)}{2}=\frac{-5}{2}=-2.5\). Therefore, \(-2.5\) is correct. \(2.5\) lies on the positive side, so it is not between 0 and −5. Exam tip: add the two numbers and divide by 2 to find their midpoint.

Since \(9<15<16\), we get \(3<\sqrt{15}<4\). Multiplying by \(-1\) reverses the order: \(-4<-\sqrt{15}<-3\). Also, \(\sqrt{15}\approx3.87\), so the point is nearer to \(-4\), not \(-3\). Exam tip: greater magnitude means farther left for negative numbers.

Irrational numbers occur between any two distinct rational numbers, so A is correct. For example, \(1^2<2<2^2\) shows that \(\sqrt{2}\) lies between 1 and 2. Exam tip: recurring decimals are rational, not irrational.