In short

A fraction \dfrac{p}{q} is a single number written as a ratio of two integers, and a decimal is the same number written in base ten. You can move freely between the two: long division turns a fraction into its decimal expansion, and a small algebraic trick turns a recurring decimal back into a fraction. The arithmetic of fractions has its own small ruleset — common denominators for adding, criss-cross for multiplying, flip-and-multiply for dividing — and every rule has a clean reason behind it. When an irrational like \sqrt{2} shows up in a denominator, you can almost always clean it up by rationalising — multiplying top and bottom by something that makes the bottom rational again.

Cut a chapati into 8 equal pieces. Eat 3 of them. The fraction \tfrac{3}{8} is doing exactly what its notation suggests: it is three out of eight equal parts of one whole. The number on the bottom says how many pieces the whole was divided into; the number on top says how many of those pieces you took.

But \tfrac{3}{8} is also a number in its own right. It is a single point on the number line, sitting between \tfrac{1}{4} and \tfrac{1}{2}. You can add it to other fractions, multiply it, compare it to other fractions, plot it. And if you actually carry out the division 3 \div 8, you get 0.375 — the same number, written in base ten instead of as a ratio. The "fraction" and the "decimal" are not two different things. They are two notations for the same point on the line, and being able to flip between them is one of the foundational skills of arithmetic.

This article is the operating manual for both notations. How to do arithmetic with fractions. How to convert between fractions and decimals in either direction. What to do with the strange decimals that go on forever without ever stopping. And how to clean up an expression where an irrational like \sqrt{2} has wandered into the denominator.

What a fraction is

A fraction \dfrac{p}{q} has two readings, and they give the same answer.

The first reading is parts of a whole. Take one whole thing — a chapati, a rupee, a metre — and divide it into q equal parts. Take p of them. That collection of p pieces is what the fraction \tfrac{p}{q} refers to. So \tfrac{3}{8} is "three of eight equal parts." This is the picture children meet first.

The second reading is the result of division. The fraction \tfrac{p}{q} is the number you get when you divide p by q. So \tfrac{3}{8} = 3 \div 8 = 0.375. This is the reading that lets you treat the fraction as a single number sitting at one specific point on the line.

The two readings are equivalent. If you share three chapatis equally among eight people, each person gets one of "three-out-of-eight" worth — which is also 0.375 of a chapati, by long division.

The fraction three-eighths shown on a number line and as a divided wholeTwo pictures of the same number. On the left, a horizontal bar divided into eight equal parts with the first three shaded in red, illustrating three out of eight equal parts. On the right, a number line from zero to one with major tick marks at zero, one quarter, three eighths, one half, and one. A red dot sits at the position of three eighths between one quarter and one half. The decimal value zero point three seven five is labelled below the dot.3 of 8 equal parts01/41/213/8= 0.375
The same number in two pictures. On the left, $\tfrac{3}{8}$ is "three of eight equal parts of one whole." On the right, it is a single point on the number line, sitting at $0.375$ — between $\tfrac{1}{4}$ and $\tfrac{1}{2}$. The two readings are interchangeable.

A fraction can be written in many different ways and still represent the same number: \tfrac{1}{2} = \tfrac{2}{4} = \tfrac{3}{6} = \tfrac{50}{100}, and so on. Multiplying the top and the bottom by the same nonzero number does not change the value, because you are really multiplying by \tfrac{n}{n} = 1, and multiplying by 1 never changes anything (the multiplicative identity rule from Operations and Properties). The form where the top and bottom share no common factor — like \tfrac{1}{2} for the family above — is called the lowest terms. Reducing a fraction to lowest terms is just dividing top and bottom by their highest common factor.

You can also feel the relationship between a denominator and the value of \tfrac{1}{q} directly. The figure below plots \tfrac{1}{x} as a smooth curve. Drag the red point along the curve. As x grows, the value 1/x shrinks toward zero. As x shrinks toward zero, 1/x shoots up. The hyperbola is the geometric face of "the bigger the denominator, the smaller the unit fraction."

Interactive graph of one over x with a draggable point on the curveA coordinate plane with a horizontal axis from zero point four to six and a vertical axis from zero to three. The curve y equals one over x is plotted as a smooth hyperbola descending from the upper left to the lower right. A draggable red point sits on the curve. A small readout panel shows the current value of x and the corresponding value of one over x, both updating as the reader drags.x1/x1234512↔ drag the red point
Drag the red point along the curve. The readout shows your $x$ value and the corresponding $1/x$ value. As $x$ doubles from $1$ to $2$, the unit fraction $1/x$ halves from $1$ to $\tfrac{1}{2}$. As $x$ grows to $5$, the unit fraction shrinks to $\tfrac{1}{5} = 0.2$. The curve is the visual face of "the bigger the denominator, the smaller the fraction."

Adding and subtracting

The rule for adding two fractions with the same denominator is the simplest in arithmetic. Just add the numerators:

\frac{2}{7} + \frac{3}{7} = \frac{5}{7}

The picture is exactly what you'd expect: two of seven equal slices, plus three of seven equal slices, gives five of seven equal slices.

Adding two sevenths and three sevenths to get five seventhsThree horizontal bars stacked, each divided into seven equal parts. The first bar has two parts shaded in red, representing two-sevenths. The second bar has three parts shaded in red, representing three-sevenths. The third bar has five parts shaded in red, representing five-sevenths. A plus sign sits between the first two bars and an equals sign before the third.2/7+3/7=5/7
Adding fractions with the same denominator is just counting equal-size pieces. Two pieces of size $\tfrac{1}{7}$ plus three pieces of size $\tfrac{1}{7}$ gives five pieces of size $\tfrac{1}{7}$. The denominator stays the same because the *piece size* hasn't changed.

The rule generalises in one obvious direction: subtraction works the same way. \tfrac{5}{7} - \tfrac{3}{7} = \tfrac{2}{7}. Same denominator, subtract the numerators.

Now what about adding two fractions with different denominators, like \tfrac{1}{3} + \tfrac{1}{4}? You cannot just add the numerators, because the pieces aren't the same size. A third of a chapati is bigger than a quarter of a chapati. You need to first re-cut both into pieces of the same size — and the natural common size is one where both fractions land on a clean number of pieces.

That common size is given by a common denominator. The simplest choice is the least common multiple (LCM) of the two denominators. For 3 and 4 the LCM is 12, because 12 is the smallest number that both 3 and 4 divide into. Rewrite each fraction over 12: \tfrac{1}{3} = \tfrac{4}{12} (multiply top and bottom by 4), and \tfrac{1}{4} = \tfrac{3}{12} (multiply top and bottom by 3). Now both fractions are in twelfths and the rule for the same denominator takes over:

\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}
Common denominator visual for one third plus one quarter equals seven twelfthsA horizontal rectangle divided into twelve equal cells in a single row. The first four cells are shaded in one colour representing one third equal to four twelfths. The next three cells are shaded in a different shade representing one quarter equal to three twelfths. The total shaded region is seven cells out of twelve, illustrating seven twelfths.1/3 = 4/121/4 = 3/12total: 7/12
One third plus one quarter, on a single bar cut into twelfths. The first four cells (light red) are $\tfrac{1}{3}$, the next three cells (dark red) are $\tfrac{1}{4}$, and the total shaded region is $7$ cells out of $12$ — that is, $\tfrac{7}{12}$. The common denominator works because once both fractions are written over $12$, the cell size is the same and you can simply count.

The same rule scales: to add three or more fractions, find a common multiple of all the denominators, rewrite each fraction over that, and add the numerators. The LCM is the most economical choice, but any common multiple works — multiplying all the denominators together always gives a valid common denominator, just not always the smallest.

Multiplying

The rule for multiplying fractions is even simpler than adding. Multiply the numerators together; multiply the denominators together:

\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}

No common denominators required. Just multiply across.

The reason this works is the most beautiful picture in fraction arithmetic. Take a unit square — a 1 \times 1 square representing one whole. To find \tfrac{2}{3} of \tfrac{4}{5}, first cut the square horizontally into 5 strips and shade 4 of them — that is \tfrac{4}{5} of the square. Then cut the square vertically into 3 columns and shade 2 of them — that is \tfrac{2}{3} of the square. The region shaded both ways is the product. Count the small cells: the square has been cut into 3 \times 5 = 15 small cells, and the doubly-shaded region is 2 \times 4 = 8 of them. So the product is \tfrac{8}{15}.

Geometric proof of two thirds times four fifths equals eight fifteenthsA square divided horizontally into five rows and vertically into three columns, making fifteen small cells in total. The bottom four rows are tinted in one shade to represent four fifths of the square. The leftmost two columns are tinted in another shade to represent two thirds of the square. The eight cells where the two shaded regions overlap are filled with a deeper colour, illustrating that two thirds of four fifths equals eight fifteenths.2/3 across4/5 down8 / 15
The unit square is cut into a $3 \times 5$ grid of small cells — fifteen in all. Shading $\tfrac{2}{3}$ of the columns and $\tfrac{4}{5}$ of the rows gives a doubly-shaded rectangle of $2 \times 4 = 8$ cells. The product $\tfrac{2}{3} \times \tfrac{4}{5}$ is therefore $\tfrac{8}{15}$ — and the picture is the proof. The general rule "numerators times numerators, denominators times denominators" is exactly the rule "small-cell columns times small-cell rows over total grid size."

The same picture explains a useful shortcut. When you have a common factor in the top and bottom — for example \tfrac{4}{9} \times \tfrac{3}{8} — you can cancel before multiplying. The 4 on top and the 8 on the bottom share a factor of 4; the 3 on top and the 9 on the bottom share a factor of 3. Cancelling first gives \tfrac{1}{3} \times \tfrac{1}{2} = \tfrac{1}{6} — much smaller numbers to handle than the \tfrac{12}{72} you get by multiplying naïvely.

Dividing

To divide one fraction by another, flip the second one and multiply:

\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}

The flipped fraction \tfrac{5}{4} is called the reciprocal of \tfrac{4}{5}. It is exactly the multiplicative inverse from Operations and Properties: the number that multiplies with \tfrac{4}{5} to give 1.

Why does flipping work? Two ways to see it.

The first is a meaning-of-division argument. Asking "what is \tfrac{2}{3} divided by \tfrac{4}{5}" is asking "how many copies of \tfrac{4}{5} fit into \tfrac{2}{3}?" When you divide a number by something less than one, more than one copy fits, so the answer should be bigger than the original. And indeed, \tfrac{5}{6} is bigger than \tfrac{2}{3} — by exactly the amount predicted by multiplying \tfrac{2}{3} by \tfrac{5}{4}.

The second is a clean algebraic argument using the identity rule. Write the division as a giant fraction:

\frac{2}{3} \div \frac{4}{5} = \frac{2/3}{4/5}

Multiply the top and bottom of this giant fraction by the reciprocal of the bottom, which is \tfrac{5}{4}. Since you are multiplying top and bottom by the same thing, the value is unchanged.

\frac{(2/3) \times (5/4)}{(4/5) \times (5/4)} = \frac{(2/3) \times (5/4)}{1} = \frac{2}{3} \times \frac{5}{4}

The bottom became 1 on purpose — that is what the reciprocal does — and the top is \tfrac{2}{3} \times \tfrac{5}{4}, which is the rule. So "flip and multiply" is a five-line consequence of the multiplicative-inverse property, not a magic trick.

Converting between fractions and decimals

Every fraction has a decimal expansion, and you find it by long division. Take \tfrac{3}{8}. Divide 3 by 8. You get 0 with remainder 3, then 30 \div 8 = 3 with remainder 6, then 60 \div 8 = 7 with remainder 4, then 40 \div 8 = 5 with remainder 0. The division has terminated, and the digits you wrote down — 0.375 — are the decimal expansion of \tfrac{3}{8}.

This direction is mechanical. Long division always works, and it always ends in one of two ways: the remainder hits 0 and the decimal terminates, or the remainder starts cycling and the decimal repeats forever. The full theorem — every rational has a terminating or recurring decimal expansion, and conversely — was proved in Number Systems.

Going the other direction takes a bit more thought.

Converting a terminating decimal to a fraction. A terminating decimal is, by definition, a sum of digits times powers of ten — and you can read it directly as a fraction over a power of ten. 0.375 has three decimal places, so it is \tfrac{375}{1000}. Then reduce to lowest terms: the highest common factor of 375 and 1000 is 125, and dividing both by 125 gives \tfrac{3}{8}. Done — and \tfrac{3}{8} is exactly the fraction the decimal came from.

Converting a recurring decimal to a fraction. This is the trickier case. The trick is to use a small piece of algebra to subtract the repeating tail away. Take the decimal 0.\overline{35} = 0.353535\dots and call it x. The repeating block has two digits, so multiply both sides by 10^2 = 100 to shift the digits two places to the left:

x = 0.353535\dots
100x = 35.353535\dots

Now subtract the first equation from the second. The repeating tails are identical, so they cancel, leaving:

100x - x = 35
99x = 35
x = \frac{35}{99}

So the recurring decimal 0.\overline{35} is equal to the fraction \tfrac{35}{99}. You can check it the other way by long-dividing 35 by 99 — the digits will be exactly 0.353535\dots.

The trick is general. If the repeating block is k digits long, multiply by 10^k first, subtract the original, and solve. The denominator that comes out is always 10^k - 1 — that is, 9, 99, 999, 9999, and so on, depending on the length of the repeating block. This is the engine that lets you write any recurring decimal as a fraction in two lines of algebra.

Rationalising denominators

Sometimes a fraction ends up with an irrational number — typically a square root — sitting in its denominator. For example, \dfrac{1}{\sqrt{2}}. This is a perfectly valid number (you met it on the number line in Number Systems), but it is in an awkward form: the denominator is irrational, which makes it hard to compare with other fractions and historically made it nearly impossible to work with by hand.

The fix is rationalising the denominator: multiply the top and the bottom by something that makes the bottom rational, without changing the value of the fraction. For \dfrac{1}{\sqrt{2}}, multiply top and bottom by \sqrt{2}:

\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

The denominator is now 2, a clean integer. And because you multiplied by \tfrac{\sqrt{2}}{\sqrt{2}} = 1, the value of the fraction did not change. Both \dfrac{1}{\sqrt{2}} and \dfrac{\sqrt{2}}{2} equal approximately 0.7071, exactly the same number — only the form is different.

Rationalising one over square root of two to get square root of two over twoTwo equivalent fractions side by side. On the left, one over square root of two. In the middle, the same fraction multiplied by square root of two over square root of two. On the right, the result square root of two over two. An arrow connects each form to the next, with the middle step labelled multiply by one. Below the three forms, the shared decimal value zero point seven zero seven one is shown.1√2×√2√2(this is just 1)=√22all three equal ≈ 0.7071
Rationalising $\dfrac{1}{\sqrt{2}}$ in one move. Multiply top and bottom by $\sqrt{2}$ — which is the same as multiplying by $1$, since $\tfrac{\sqrt{2}}{\sqrt{2}} = 1$ — and the denominator becomes the rational number $2$. The value of the fraction is unchanged; only the *form* changes.

For a more interesting case, take a denominator that is a sum containing a radical, like \dfrac{1}{1 + \sqrt{2}}. Multiplying top and bottom by \sqrt{2} does not work here — the denominator becomes \sqrt{2} + 2, which is still irrational. The right move is to multiply by the conjugate of the denominator, the same expression with the sign of the radical flipped. The conjugate of 1 + \sqrt{2} is 1 - \sqrt{2}. Why does this help? Because the product

(1 + \sqrt{2})(1 - \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1

is rational. The cross-terms cancel and the radical squares away. So:

\frac{1}{1 + \sqrt{2}} = \frac{1}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1

The conjugate trick works for any denominator of the form a + b\sqrt{c}, where multiplying by a - b\sqrt{c} gives the rational a^2 - b^2 c. It is one of the most useful tricks in school algebra and reappears everywhere — in coordinate geometry, in Quadratic Equations when simplifying roots, and again in complex numbers as the complex conjugate.

Two worked examples

Example 1: Compute $\dfrac{5}{6} + \dfrac{7}{9} - \dfrac{1}{4}$

Three fractions, three different denominators. The plan is to find a common denominator, rewrite each fraction over it, and then add and subtract numerators.

Step 1. Find the LCM of 6, 9, and 4.

The prime factorisations are 6 = 2 \times 3, 9 = 3^2, and 4 = 2^2. The LCM takes the highest power of each prime that appears: 2^2 \times 3^2 = 4 \times 9 = 36.

Why: 36 is the smallest number that all three denominators divide into, so it is the smallest common piece-size that works for all three fractions at once. You could also use 6 \times 9 \times 4 = 216, but you would just have to simplify back down at the end — the LCM saves you that work.

Step 2. Rewrite each fraction over 36.

\frac{5}{6} = \frac{5 \times 6}{6 \times 6} = \frac{30}{36}
\frac{7}{9} = \frac{7 \times 4}{9 \times 4} = \frac{28}{36}
\frac{1}{4} = \frac{1 \times 9}{4 \times 9} = \frac{9}{36}

Why: each fraction is multiplied top and bottom by whatever number is needed to turn its denominator into 36. Since multiplying top and bottom by the same thing is the same as multiplying by 1, the value of each fraction is unchanged.

Step 3. Add and subtract the numerators over the common denominator.

\frac{30}{36} + \frac{28}{36} - \frac{9}{36} = \frac{30 + 28 - 9}{36} = \frac{49}{36}

Step 4. Check whether the result reduces to lowest terms.

The numerator is 49 = 7^2 and the denominator is 36 = 2^2 \times 3^2. They share no common factor, so \tfrac{49}{36} is already in lowest terms.

Result. \dfrac{5}{6} + \dfrac{7}{9} - \dfrac{1}{4} = \dfrac{49}{36}, which is approximately 1.361.

Three fractions over thirty-six showing the addition and subtractionThree horizontal bars stacked, each representing the unit length divided into thirty-six equal cells. The first bar has thirty cells shaded representing thirty thirty-sixths or five sixths. The second bar has twenty-eight cells shaded representing twenty-eight thirty-sixths or seven ninths. The third bar has nine cells shaded representing nine thirty-sixths or one quarter, with the cells crossed out to indicate subtraction. Below, a result bar shows forty-nine cells shaded out of thirty-six (the bar extends past one whole) representing forty-nine thirty-sixths.5/6 = 30/367/9 = 28/36− 1/4 = 9/361 whole49/36 ≈ 1.361
The three fractions, rewritten over the common denominator $36$, lined up so the cell sizes match. The final bar is $49$ cells wide — slightly more than one whole, which has $36$ cells. The dashed line marks where one whole ends, and the spillover past it is the $\tfrac{13}{36}$ that makes $\tfrac{49}{36}$ greater than $1$.

Example 2: Rationalise and simplify $\dfrac{3}{2 + \sqrt{5}}$

The denominator has a sum with a radical, so the conjugate trick is the right tool.

Step 1. Identify the conjugate of the denominator.

The denominator is 2 + \sqrt{5}. Its conjugate is the same expression with the sign of the radical flipped: 2 - \sqrt{5}.

Step 2. Multiply top and bottom of the fraction by the conjugate.

\frac{3}{2 + \sqrt{5}} \times \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{3 \,(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})}

Why: multiplying top and bottom by the same thing is the same as multiplying by 1, so the value of the fraction is unchanged. The conjugate is the specific "1" that makes the new denominator rational.

Step 3. Expand the new denominator using the difference-of-squares pattern.

(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1

Why: (a + b)(a - b) = a^2 - b^2 for any a and b. Plugging a = 2 and b = \sqrt{5} kills the radical, because (\sqrt{5})^2 = 5, leaving only rational numbers behind. This is the entire reason the conjugate trick works.

Step 4. Expand the numerator and simplify.

\frac{3(2 - \sqrt{5})}{-1} = \frac{6 - 3\sqrt{5}}{-1} = -(6 - 3\sqrt{5}) = 3\sqrt{5} - 6

Result. \dfrac{3}{2 + \sqrt{5}} = 3\sqrt{5} - 6, approximately 0.708.

Step by step rationalisation of three over two plus root fiveFour boxed expressions stacked vertically connected by labelled arrows. The first box contains three over the quantity two plus root five. An arrow labelled multiply by conjugate over conjugate leads to a box containing three times the quantity two minus root five over the product of the conjugates. An arrow labelled difference of squares leads to a box with the simplified denominator negative one. A final arrow labelled distribute and simplify leads to the answer three root five minus six.3 / (2 + √5)× (2 − √5) / (2 − √5)3(2 − √5) / [(2+√5)(2−√5)]denominator: 4 − 5 = −13(2 − √5) / (−1)distribute, simplify sign3√5 − 6
Each step replaces the previous expression with an equal one until the radical has moved out of the denominator and the answer is a clean linear combination of $\sqrt{5}$ and integers. The decisive move is the second step, where the difference-of-squares pattern collapses the irrational denominator into the rational $-1$.

Common confusions

Going deeper

If you came here just to do arithmetic with fractions and decimals and to clean up the occasional radical denominator, you have everything you need. The rest of this section is for readers who want to know why the rules behave the way they do.

Why some fractions terminate and others don't

A terminating decimal is, by construction, a number of the form \dfrac{n}{10^k} for some integer n and some non-negative integer k. For example, 0.375 = \dfrac{375}{1000} = \dfrac{375}{10^3}. So if a fraction \dfrac{p}{q} in lowest terms terminates, it must be expressible as \dfrac{n}{10^k}, which means its denominator q has to divide 10^k for some k.

Now 10 = 2 \times 5, so 10^k = 2^k \times 5^k. The only divisors of 10^k are products of 2's and 5's. Therefore q must itself be a product of 2's and 5's — no other primes allowed. Conversely, if q is a product of 2's and 5's, you can always find a k large enough that q divides 10^k, and so the fraction terminates.

The full theorem: a fraction in lowest terms has a terminating decimal expansion if and only if its denominator's prime factorisation contains only 2's and 5's. Anything else — a 3, a 7, an 11, a 13 — forces the decimal to recur. So \tfrac{3}{8} terminates (8 = 2^3), \tfrac{7}{20} terminates (20 = 2^2 \times 5), and \tfrac{1}{6} does not (6 = 2 \times 3, and the factor of 3 blocks termination).

The conjugate trick is a small case of a much bigger pattern

The trick "multiply by a - b to kill a a + b" depends on the difference-of-squares identity (a + b)(a - b) = a^2 - b^2. The same idea works one level up: to rationalise a denominator with a cube root, you would multiply by a more complicated expression that is the analogue of the conjugate for cubes — specifically, (a + b)(a^2 - ab + b^2) = a^3 + b^3 does for cubes what difference-of-squares does for squares. Generally, for an n-th root, there is always some polynomial expression you can multiply by to kill the radical, related to the algebraic identity a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \dots + b^{n-1}).

This same trick reappears in the chapter on complex numbers, where multiplying by the complex conjugate turns the imaginary part into a real number using exactly the same difference-of-squares pattern: (a + bi)(a - bi) = a^2 + b^2.

LCMs save work, but any common multiple is valid

In Example 1 you used the LCM of 6, 9, 4, which was 36. You could just as well have used 6 \times 9 \times 4 = 216 as the common denominator. The arithmetic is uglier — every numerator gets six times bigger than it needs to be — but the final answer is the same. After adding, you simplify \tfrac{294}{216} down to \tfrac{49}{36}, and you land in the same place.

The reason the LCM works is that it is the smallest number that all the denominators divide into. Any common multiple of the denominators gives a valid common denominator; the LCM is just the most efficient choice. If you cannot find the LCM by inspection, the foolproof fallback is to multiply the denominators together — it always works, just sometimes wastefully.

Where this leads next

Fractions and decimals are the working currency of every later chapter.