Representation on number line

A number line is a straight line on which every point stands for a real number, with the integers marked at equal, fixed intervals. Rational numbers fill in the gaps between those integers.

Step 1 — find the bounding integers. For $\dfrac{p}{q}$ in standard form ($q>0$), the integer part of the number tells you which two consecutive integers it lies between. A positive number like $\dfrac34$ lies between $0$ and $1$; a number like $-1\dfrac23$ lies between $-1$ and $-2$.

Step 2 — subdivide into $q$ equal parts. Once you know the segment, divide it into $q$ equal parts, where $q$ is the denominator of the number in standard form. Every mark in that stretch is then a multiple of $\dfrac1q$.

Step 3 — count $p$ parts in the right direction. Starting from the nearer integer (usually $0$), count $p$ parts — rightward for a positive number, leftward for a negative one — and mark that point. The general position of $\dfrac{p}{q}$ on the line, measured as a signed distance from $0$, is simply the value of the fraction itself:

where: $x$ is the coordinate (position) of the point on the number line, $p$ is the numerator (with its sign), and $q>0$ is the denominator in standard form. This equation is really just a restatement of Steps 1–3: the coordinate is the number, and plotting is the physical act of locating that coordinate by equal subdivision.

Reading the line for comparisons. Numbers increase left to right, so if point $A$ is to the right of point $B$, then $A>B$. Every negative rational number is less than every positive one, and between two negative rationals, the one closer to $0$ (smaller absolute value) is the larger number — for example $-\dfrac14 > -\dfrac34$, because $-\dfrac14$ sits to the right.

Unlike denominators. To compare or jointly plot fractions with different denominators, first convert them to a common denominator (the LCM of the denominators), so every number you're placing lands on the same evenly-divided scale.

Absolute value as distance. It helps to think of the distance of a rational number from $0$ (ignoring sign) as its size, and the sign as pure direction — left for negative, right for positive. $\dfrac34$ and $-\dfrac34$ sit the same distance from $0$ on opposite sides. A quick way to plot a negative fraction is to first mark the positive version as a distance from $0$, then reflect it across $0$ to the correct side.

Why the unit segment always works. Whichever unit segment you subdivide — $0$ to $1$, $-1$ to $-2$, or any pair of consecutive integers — the logic is identical: split it into $q$ equal parts, and every mark in that stretch is a rational number with denominator $q$. This is exactly why picking the correct denominator (after reducing the fraction to standard form) is the entire trick of accurate plotting; there is nothing extra to memorise beyond Steps 1–3.

The goal is to decide which of two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is greater purely by arithmetic, using the same left-right logic that governs their positions on the number line, so that plotting is not needed for every comparison. Assume both denominators are written positive ($b>0$, $d>0$), as in standard form.

Step 1. Bring both fractions to the common denominator $bd$: $\dfrac{a}{b}=\dfrac{ad}{bd}$ and $\dfrac{c}{d}=\dfrac{cb}{bd}$. On the number line, this is exactly what happens when both numbers are placed on a scale subdivided into $bd$ equal parts instead of two separate scales.

Step 2. Since both fractions now share the positive denominator $bd$, their positions on the number line are ordered exactly as their numerators are ordered — a larger numerator over the same positive denominator sits further right.

Step 3. So $\dfrac{a}{b} > \dfrac{c}{d}$ exactly when $ad > bc$, and $\dfrac{a}{b} < \dfrac{c}{d}$ exactly when $ad < bc$. This is the familiar cross-multiplication test, and it is nothing more than the number-line comparison rule in disguise — no drawing required, only two multiplications.

Step 4. The rule works unchanged for negative numerators; the only condition is that both denominators be positive first, which standard form guarantees.

Locate $-\dfrac{5}{3}$ on the number line, then determine whether it is greater than or less than $-\dfrac{3}{2}$.

Step 1 — locate $-\dfrac53$. Since $-\dfrac53 = -1\dfrac23$, the number lies between $-1$ and $-2$. Divide that unit segment into $3$ equal parts and move $2$ parts from $-1$ towards $-2$; the mark falls closer to $-2$.

Step 2 — set up the comparison. Compare $-\dfrac53$ with $-\dfrac32$ using the cross-multiplication rule from the derivation, with $a=-5,\ b=3,\ c=-3,\ d=2$ (both $b,d>0$).

Step 3 — cross-multiply. $ad = (-5)(2) = -10$ and $bc = (3)(-3) = -9$. Since $-10 < -9$, the rule gives $\dfrac{a}{b} < \dfrac{c}{d}$.

Step 4 — check against the picture. $-\dfrac53 \approx -1.67$ lies between $-1$ and $-2$, closer to $-2$; $-\dfrac32 = -1.5$ lies exactly midway between $-1$ and $-2$, which is to the right of $-1.67$. Both the arithmetic and the picture agree.

$-\dfrac{5}{3} < -\dfrac{3}{2}$.

Competitive stretch

Order $\dfrac23, -\dfrac12, \dfrac56, -\dfrac16$ ascending. LCM of $3,2,6,6$ is $6$: $\dfrac23=\dfrac46,\ -\dfrac12=-\dfrac36,\ \dfrac56=\dfrac56,\ -\dfrac16=-\dfrac16$. Comparing numerators over the common denominator, $-3<-1<4<5$, so $-\dfrac12 < -\dfrac16 < \dfrac23 < \dfrac56$.

• Dividing the wrong segment — plotting $\dfrac53$ by subdividing $0$-to-$1$ instead of recognising it lies between $1$ and $2$.
• Thinking a negative number with a bigger digit is "bigger," e.g. believing $-5 > -2$; further left always means smaller.
• Comparing unlike denominators by numerators alone without converting to a common denominator first.
• Cross-multiplying with a negative denominator without flipping it to standard form first, which reverses the inequality silently.

Ask the tutor: Stuck on which segment to subdivide for a tricky negative fraction — ask the AI tutor to walk through your specific number step by step.