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Equivalent Fractions and Simplifying

Learning goals

  • Build an equivalent fraction by scaling top and bottom alike
  • Explain why scaling by a nonzero number leaves the value alone
  • Simplify by dividing out a common factor, the rule in reverse
  • Reach lowest terms in one step by dividing by the GCF
  • Test two fractions for equality with cross products

When two fractions name the same amount

Two fractions are equivalent when they stand for the same number: the same point on the number line, the same amount of pizza. The clearest way to see it is with the bars you met in the last lesson. Draw one whole cut into 22 equal parts with 11 shaded, and below it the same whole cut into 44 equal parts with 22 shaded. The shaded region is identical in both: exactly half the bar is coloured.

One half and two fourths shade the same region of the same whole, so 1/2 and 2/4 are equal. Cutting each existing part into two does not change how much is shaded. Rectangular bars divided into equal parts, with some parts shaded to show a fraction. 1 2 2 4
One half and two fourths shade the same region of the same whole, so 1/2 and 2/4 are equal. Cutting each existing part into two does not change how much is shaded.

To go from the top bar to the bottom one, we cut each of the 22 parts into 22 smaller parts. That doubles the number of parts in the whole, so the denominator goes from 22 to 44. But that same cutting also doubles the number of shaded parts, because each shaded part got cut in two as well. So the numerator goes from 11 to 22. The shaded amount never moved; we only drew more cut lines. That is why 12\frac{1}{2} and 24\frac{2}{4} are the same number:

12=24.\frac{1}{2} = \frac{2}{4}.

You can keep going. Cut each part into three instead, and the same half of the bar is now 36\frac{3}{6}; into five, and it is 510\frac{5}{10}. Every one of these is just a finer description of the same shaded region:

12=24=36=510=50100.\frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \frac{5}{10} = \frac{50}{100}.

You can do that cutting yourself. In the figure below, the top bar is fixed at one half. The bottom bar is the same whole, and you choose how many pieces each of its parts is cut into.

Why a finer cut gives a new name and not a new amount

1/2 = 3/6. Each part is cut into three, so 3 of 6 parts are shaded. The shaded length has not moved. Two bars of the same width. The top bar is cut into 2 equal parts with 1 shaded. The bottom bar shows the same shaded length cut into finer parts. Use the controls below the figure to change how fine the cut is. 1 2 3 6
Cut each part into

1/2 = 3/6. Each part is cut into three, so 3 of 6 parts are shaded. The shaded length has not moved.

The same whole twice. The top bar is cut in half; you choose how finely the bottom bar is cut.

Watch the edge of the shaded region as you change the cut. It does not move. The numerator and the denominator both climb, so the fraction gets new names, but the amount of bar coloured in stays exactly where it started. That is what makes all these fractions equal.

Why multiplying top and bottom by the same number is allowed

The picture suggests a rule: to rewrite a fraction, multiply the numerator and the denominator by the same number.

Try that on 25\frac{2}{5} with the multiplier 33. Multiplying top and bottom by 33 turns it into 615\frac{6}{15}, and the bars say the same thing. Cut each of the 55 parts into 33 pieces and the whole holds 1515 pieces. Each of the 22 shaded parts becomes 33 pieces too, so 66 pieces are shaded. The extra cuts moved no shading, so 25\frac{2}{5} and 615\frac{6}{15} are the same number.

Now a different fraction and a different multiplier. Take 47\frac{4}{7}, a whole cut into 77 equal parts with 44 of them shaded, and cut every part in two. The whole was in 77 parts and each became 22 pieces, so it now holds 7×2=147 \times 2 = 14 pieces. The 44 shaded parts each became 22 pieces too, so 4×2=84 \times 2 = 8 pieces are shaded. You drew new lines but added and removed no area, so 47=814\frac{4}{7} = \frac{8}{14}.

Both times the multiplier did the same two jobs. It cut every part of the whole, and it cut every shaded part by the identical amount. That is why the top and the bottom both have to be multiplied, and any starting fraction with any nonzero multiplier behaves the same way.

Building rule. For any whole numbers aa and bb with bb not zero, and any nonzero whole number cc, ab=a×cb×c.\frac{a}{b} = \frac{a \times c}{b \times c}.

This is also why the multiplier has to hit both numbers. Change only one of them and the picture no longer matches the fraction you started with. The requirement that cc is not zero matters too: you cannot cut each part into 00 pieces, and a denominator of 00 is undefined.

Building an equivalent fraction

The most common job is to rewrite a fraction so it has a particular denominator you need. Rewriting to a required denominator is exactly what you will do when adding fractions in the next lesson.

To rewrite 23\frac{2}{3} with a denominator of 1212, ask ”33 times what gives 1212?” Since 3×4=123 \times 4 = 12, the multiplier is 44. Now multiply the numerator by that same 44:

23=2×43×4=812.\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}.

The rule guarantees these name the same number, because we multiplied top and bottom by the same 44. That gives the method: find what the old denominator was multiplied by, then multiply the numerator by that same number. The fixed idea is that whatever you do to the bottom, you must do to the top. Cutting every part into the same number of equal pieces is what keeps the value the same.

Check your understanding

Rewrite 35\frac{3}{5} as an equivalent fraction with denominator 2020. What is the new numerator?

Answer choices

Simplifying: the rule run in reverse

The building rule works in both directions. Building up cut each part into smaller pieces; simplifying runs the same picture backwards, grouping the small pieces back into larger ones.

Take 68\frac{6}{8}. Both 66 and 88 are divisible by 22, so divide each by 22:

68=6÷28÷2=34.\frac{6}{8} = \frac{6 \div 2}{8 \div 2} = \frac{3}{4}.

Look at what that division does to the picture. The whole was cut into 88 parts, and gathering those parts two at a time leaves 44 larger parts. The 66 shaded pieces gather into 33 shaded larger parts. No area was added or removed, only cut lines erased, so the shaded amount is unchanged.

Six eighths and three fourths shade the same region. Merging the eight parts into four, two small parts per larger part, turns 6/8 into 3/4 without changing the shaded amount. Rectangular bars divided into equal parts, with some parts shaded to show a fraction. 6 8 3 4
Six eighths and three fourths shade the same region. Merging the eight parts into four, two small parts per larger part, turns 6/8 into 3/4 without changing the shaded amount.

That same gathering works for any number that divides both the numerator and the denominator, which gives the rule.

Simplifying rule. If a whole number cc divides both aa and bb, then ab=a÷cb÷c.\frac{a}{b} = \frac{a \div c}{b \div c}.

The number cc you divide by must be a common factor of the numerator and denominator, a factor of both. That requirement is exactly the idea from the factors chapter. Dividing by a common factor is the only way to make a fraction simpler while keeping its value.

Lowest terms and the one-step shortcut

A fraction is in lowest terms (also called simplest form) when the numerator and denominator have no common factor larger than 11. At that point there is nothing left to divide out, so the fraction cannot be written with smaller whole numbers. For example 34\frac{3}{4} is in lowest terms, because the only factor 33 and 44 share is 11.

You could simplify 1824\frac{18}{24} by dividing out common factors one at a time: divide by 22 to get 912\frac{9}{12}, then by 33 to get 34\frac{3}{4}. That works, but it takes several steps and you have to keep checking whether you are done. There is a faster way that always finishes in a single division, and it uses the greatest common factor from the previous chapter.

For 1824\frac{18}{24}, the greatest common factor of 1818 and 2424 is 66, so

1824=18÷624÷6=34,\frac{18}{24} = \frac{18 \div 6}{24 \div 6} = \frac{3}{4},

and you are guaranteed to be finished. If 33 and 44 still shared a factor, that factor times 66 would divide both 1818 and 2424. But 66 is already the biggest number that does. If you instead divide by a common factor that is not the greatest, you simplify the fraction but you are not done yet.

So the shortcut is: divide the numerator and the denominator by their GCF, once.

Check your understanding

What is the greatest common factor of 2020 and 3030, and what is 2030\frac{20}{30} in lowest terms?

Answer choices

Checking whether two fractions are equal

Sometimes you are handed two fractions and asked whether they are equal, with no obvious cutting between them. One reliable test is to simplify both to lowest terms. Compare 69\frac{6}{9} and 1015\frac{10}{15}. The first simplifies by 33 to 23\frac{2}{3}, and the second simplifies by 55 to 23\frac{2}{3}. Same simplest form, so they are equal. That test works because equivalent fractions have the same lowest-terms form. If the simplified versions match the fractions are equal, and if they differ the fractions are not.

A second test avoids simplifying entirely. To compare 46\frac{4}{6} and 69\frac{6}{9}, multiply the numerator of each by the denominator of the other:

4×9=36and6×6=36.4 \times 9 = 36 \qquad\text{and}\qquad 6 \times 6 = 36.

Those two products are equal, and so are the fractions. The reason is that both of them, rewritten over the shared denominator 6×96 \times 9, become 4×96×9\frac{4 \times 9}{6 \times 9} and 6×69×6\frac{6 \times 6}{9 \times 6}. With equal denominators, the two fractions are equal exactly when the numerators are equal, and those numerators are the two products just multiplied out. In general, ab\frac{a}{b} and cd\frac{c}{d} are equal exactly when the cross products match, that is when a×d=b×ca \times d = b \times c. This test is especially handy with larger numbers, where simplifying both might take longer than one multiplication on each side.

Worked example 1 Write 34\frac{3}{4} with a denominator of 2828

Find the multiplier by asking what turns the old denominator into the new one. The denominator goes from 44 to 2828, and

4×7=28,4 \times 7 = 28,

so the multiplier is 77. The building rule says multiply the numerator by that same 77:

34=3×74×7=2128.\frac{3}{4} = \frac{3 \times 7}{4 \times 7} = \frac{21}{28}.

Both fractions name the same number, because we multiplied the top and bottom by the same 77. That multiplier just cuts each of the four parts into seven smaller pieces.

Worked example 2 Simplify 2436\frac{24}{36} to lowest terms

Use the one-step shortcut: divide by the greatest common factor of 2424 and 3636. List their factors and find the largest they share.

24:  1,2,3,4,6,8,12,2436:  1,2,3,4,6,9,12,18,3624: \; 1, 2, 3, 4, 6, 8, 12, 24 \qquad 36: \; 1, 2, 3, 4, 6, 9, 12, 18, 36

The common factors are 1,2,3,4,6,121, 2, 3, 4, 6, 12, and the greatest is 1212, so GCF(24,36)=12\operatorname{GCF}(24, 36) = 12. Divide the numerator and denominator by 1212:

2436=24÷1236÷12=23.\frac{24}{36} = \frac{24 \div 12}{36 \div 12} = \frac{2}{3}.

Now check that you are finished: 22 and 33 share no factor larger than 11, so 23\frac{2}{3} is in lowest terms. Because we divided by the greatest common factor, one division was enough.

Worked example 3 Are 812\frac{8}{12} and 1015\frac{10}{15} equivalent?

Use the cross-product test. Multiply the numerator of each by the denominator of the other:

8×15=120and12×10=120.8 \times 15 = 120 \qquad\text{and}\qquad 12 \times 10 = 120.

The two products are both 120120, so the fractions are equivalent:

812=1015.\frac{8}{12} = \frac{10}{15}.

You can confirm it by simplifying: 812\frac{8}{12} divides by 44 to give 23\frac{2}{3}, and 1015\frac{10}{15} divides by 55 to give 23\frac{2}{3}. Same lowest-terms form, which agrees with the cross-product test.

Check your understanding

Which of these fractions is not equal to 23\frac{2}{3}?

Answer choices

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Free Response Questions (FRQ)

Longer questions in parts, to be worked out on paper. Progressive hints, the answer on its own so you can check yourself and try again, then the full worked solution, plus a rubric to mark your own work against.

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Extra sets, as hard as the Challenge set. Each one opens on its own page.

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Other explanations of this lesson, if you want a second take.

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Why cutting every part the same way leaves the fraction alone

The lesson cuts 25\frac{2}{5} into fifteenths and 47\frac{4}{7} into fourteenths, and the shading never moves either time. This settles it for every fraction and every nonzero multiplier at once, then checks the same rule a second way by reading the bar as a division.

Why ab=a×cb×c\frac{a}{b} = \frac{a \times c}{b \times c}#

Take 38\frac{3}{8} first, with the multiplier 55: a whole cut into 88 equal parts, with 33 of them shaded. Cut each of those 88 parts into 55 equal pieces. The whole was in 88 parts and every part became 55 pieces, so the whole now holds 8×5=408 \times 5 = 40 pieces. Each of the 33 shaded parts also became 55 pieces, so the number of shaded pieces is 3×5=153 \times 5 = 15. The new lines added and removed no area, so the shaded amount is exactly what it was. That same amount, described with the smaller pieces, is 1540\frac{15}{40}, so 38=1540\frac{3}{8} = \frac{15}{40}.

Now the same argument in letters. Start from what ab\frac{a}{b} means: a whole cut into bb equal parts, with aa of them shaded. Now cut each of those bb parts into cc equal smaller pieces. The whole was in bb parts and every part became cc pieces, so the whole now holds b×cb \times c equal pieces. Each of the aa shaded parts also became cc pieces, so the number of shaded pieces is a×ca \times c. You drew new cut lines but added and removed no area, so the shaded amount is exactly what it was. That same amount, described with the smaller pieces, is a×ca \times c shaded out of b×cb \times c, which is a×cb×c\frac{a \times c}{b \times c}. Since the two fractions describe the identical shaded amount,

ab=a×cb×c.\frac{a}{b} = \frac{a \times c}{b \times c}.

Nothing in that argument depended on the values 33, 88 and 55 the first pass used.

The division reading agrees and uses only whole-number arithmetic. Suppose the division a÷ba \div b comes out to a whole number qq, meaning a=q×ba = q \times b. Multiplying both sides by cc gives a×c=q×(b×c)a \times c = q \times (b \times c), so (a×c)÷(b×c)(a \times c) \div (b \times c) is the same qq.

Why dividing by the GCF always finishes the job

The lesson divides 1824\frac{18}{24} by 66 and finds nothing left to cancel. This shows the same thing holds for every fraction, by working out what a leftover shared factor would have to be.

Dividing by the GCF gives lowest terms in one step#

Take 4270\frac{42}{70} first. The greatest common factor of 4242 and 7070 is 1414, and dividing both by 1414 gives 35\frac{3}{5}. Suppose 33 and 55 still shared some factor kk bigger than 11. Then 14×k14 \times k would divide both 4242 and 7070, and 14×k14 \times k is bigger than 1414, so 1414 would not be the greatest common factor. It is, so no such kk exists and 35\frac{3}{5} is in lowest terms. A smaller common factor is what leaves something behind: dividing 4270\frac{42}{70} by 77 gives 610\frac{6}{10}, whose terms still share 22. Those leftovers point straight back at the factor that was missed, because 7×2=147 \times 2 = 14.

Now the same argument in letters. Let g=GCF(a,b)g = \operatorname{GCF}(a, b) be the greatest common factor of the numerator and denominator. Because gg divides both, the simplifying rule lets us write

ab=a÷gb÷g.\frac{a}{b} = \frac{a \div g}{b \div g}.

The claim is that a÷gb÷g\frac{a \div g}{b \div g} is already in lowest terms, with nothing left to cancel. Suppose it were not. Then a÷ga \div g and b÷gb \div g would share some common factor kk bigger than 11. But if kk divides both a÷ga \div g and b÷gb \div g, then g×kg \times k would divide both aa and bb. Since g×kg \times k divides both, it is a common factor of aa and bb, and k>1k > 1 makes it larger than gg. That contradicts gg being the greatest common factor. So no such kk exists, which means a÷ga \div g and b÷gb \div g share no factor above 11. The fraction is in lowest terms, and nothing in the argument depended on the values 4242, 7070 and 1414 the first pass used.

A bit of history (Optional)

For centuries a page of music named the notes but said nothing about their lengths. A singer picked up the timing by ear. You learned a tune from somebody who already knew it. If two singers disagreed, the page could not settle the argument.

Around 1280 a music teacher named Franco, working in what is now Germany, wrote down a fix. The shape of a note, he said, should announce how long to hold it. Lengths could then be measured against each other. Musicians eventually settled on halving.

That is why the notes carry fraction names today. There is a whole note, a half note, a quarter note and an eighth note. Hold one half note and you have filled the time of two quarter notes. Four eighth notes fill it too. Splitting a long note into shorter ones changes the counting and never the music.

That is the building rule of this lesson, performed out loud. One half, two fourths and four eighths are a single duration under three names.