Representing real numbers on the number line

Since \(3^2=9\) and \(4^2=16\), we have \(9<10<16\). Taking square roots gives \(3<\sqrt{10}<4\), so \(\sqrt{10}\) lies between 3 and 4. Option B is incorrect because \(\sqrt{10}>\sqrt{9}=3\). Exam tip: To locate a square root, identify the consecutive perfect squares between which its radicand lies.

Since (\frac{-11}{5}=-2.2), it lies between (-3) and (-2). Be careful with position of negative decimals.

Distance from (0) is (|x|), so (|x|=3.5) gives (x=\pm3.5). Distance is always a positive measure.

Moving right means addition, so (-1+\frac{7}{3}=\frac{4}{3}). Decide addition or subtraction by direction.

Because ((\sqrt{2})^2+1^2=3), the hypotenuse is (\sqrt{3}). Successive square roots are constructed using Pythagoras.

(\frac{13}{8}=1.625), so both represent the same point. Convert the fraction into a decimal to identify position.

Since \(12=4\\times3\) and \(\\sqrt{4}=2\), we get \(\\sqrt{12}=\\sqrt{4\\times3}=2\\sqrt{3}\). Therefore, option A is correct. The squares of \(3\\sqrt{2}\), \(4\\sqrt{3}\), and \(\\sqrt{6}\) are 18, 48, and 6 respectively, so none of them equals \(\\sqrt{12}\). Exam tip: when simplifying a surd, first extract the greatest perfect-square factor from under the radical sign.

(\frac{\sqrt{2}}{2}) is irrational and its value lies between (0) and (1). An irrational divided by a non-zero rational remains irrational.

Since (-\sqrt{3}\approx-1.732) and (-\frac{3}{2}=-1.5), (-\sqrt{3}) is farther left. For negative numbers, larger magnitude means farther left.

Since (-2.75<-2.7), (A) is to the left on the number line. In negative decimals, the more negative number is farther left.

Points at the same distance from (0) occur on both sides, so the values are (\pm\sqrt{13}). In distance questions, check both directions.

Because (4^2=16) and (5^2=25), (4<\sqrt{20}<5). Perfect-square bounds quickly give the interval.

Let \(x=0.333\ldots\). Then \(10x=3.333\ldots\). Subtracting the first equation from the second gives \(9x=3\), so \(x=\frac{1}{3}\). Options B, C and D represent the terminating decimals \(0.3\), \(0.33\) and \(0.333\), respectively, not the infinitely repeating decimal. Exam tip: To convert a repeating decimal into a fraction, assign it to \(x\), multiply by the appropriate power of \(10\), and subtract.

(\sqrt{3}\approx1.732), so it lies near (1.732) on the number line. Remember approximations of common square roots.

The midpoint is (\frac{\frac{5}{6}+\frac{7}{6}}{2}=1). For fractions, the average method is safest.

Since \\(1<\\sqrt{3}<2\\), subtracting \\(\\sqrt{3}\\) from 2 gives \\(0<2-\\sqrt{3}<1\\). Therefore, the number lies in \\((0,1)\\). Its approximate value is 0.268, so the interval \\((1,2)\\) is not possible. Exam tip: estimate the irrational term first, then locate the resulting value.

Since \(1.25\) has two digits after the decimal point, \(1.25=\frac{125}{100}\). Dividing the numerator and denominator by 25 gives \(\frac{125}{100}=\frac{5}{4}\). Option B, \(\frac{4}{5}\), is the reciprocal of \(\frac{5}{4}\), not its equivalent fraction. Exam tip: convert a terminating decimal to a fraction with denominator \(10^n\), then reduce it to the lowest terms.

(\sqrt{6}\approx2.449), so (2.45) is closest. To test closeness, compare differences from the approximate value.

(\frac{9}{7}=1+\frac{2}{7}), so choose the second seventh part after (1). Converting an improper fraction into a mixed form is useful.

The midpoint is (\frac{-\frac{5}{2}-\frac{1}{2}}{2}=-\frac{3}{2}). Pay attention to signs in negative fractions.