Free math worksheets

- What fraction represents one of the \(12\) equal parts? - Think of the numerator as a count of equal unit-fraction pieces. - Check that the number of addends matches the numerator.

1. One of \(12\) equal parts is the unit fraction \(\frac{1}{12}\). 2. Five equal parts are five copies of \(\frac{1}{12}\): \(\frac{5}{12}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\). 3. The numerator \(5\) counts the number of \(\frac{1}{12}\) unit fractions in \(\frac{5}{12}\).

a) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\) b) The numerator \(5\) means there are five copies of \(\frac{1}{12}\).

- Determine the unit fraction represented by one part of the bar. - Count the shaded parts rather than using a count supplied in the text. - The number of unit-fraction copies becomes the numerator.

1. The bar is divided into \(12\) equal parts, so each shaded part represents \(\frac{1}{12}\). 2. There are \(9\) shaded parts, giving \(9\) copies of \(\frac{1}{12}\). 3. Thus the sum is \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{9}{12}\).

\(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{9}{12}\)

- Identify the unit fraction used in both sums. - Count how many copies appear in each sum. - When the unit fractions are the same size, compare the numbers of copies.

1. The first sum has \(4\) copies of \(\frac{1}{6}\), so it equals \(\frac{4}{6}\). 2. The second sum has \(5\) copies of \(\frac{1}{6}\), so it equals \(\frac{5}{6}\). 3. The unit fractions are the same size, and \(5\) copies are more than \(4\) copies, so the second sum is greater.

The second sum is greater: \(\frac{5}{6}>\frac{4}{6}\).

- Count the equal parts in the whole to identify one unit fraction. - Count how many of those equal parts are shaded. - Write one copy of the unit fraction for each shaded part.

1. The bar is divided into \(5\) equal parts, so one part is \(\frac{1}{5}\). 2. Three parts are shaded, so the unit-fraction sum is \(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\). 3. Three copies of \(\frac{1}{5}\) make \(\frac{3}{5}\).

\(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}\)

- The denominator tells the size of the unit fraction. - The numerator tells how many copies are needed.

1. The denominator \(5\) tells us the unit fraction is \(\frac{1}{5}\). 2. The numerator \(4\) tells us to use four copies of that unit fraction. 3. Therefore, \(\frac{4}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\).

\(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\)

- Identify the size of one equal part in the model. - Count every shaded part across the wholes. - Regroup enough unit fractions to make one whole.

1. The model shows \(7\) shaded sixth-sized parts. 2. Seven copies of \(\frac{1}{6}\) make \(\frac{7}{6}\). 3. Six sixths make one whole, leaving one sixth, so \(\frac{7}{6}=1\frac{1}{6}\).

a) \(7\) copies b) \(\frac{7}{6}=1\frac{1}{6}\)

- Start with a unit fraction that is less than \(\frac{3}{4}\). - Subtract that unit fraction from \(\frac{3}{4}\). - Check whether the remainder is another unit fraction.

1. Choose a unit fraction less than \(\frac{3}{4}\), such as \(\frac{1}{2}\). 2. Find the difference: \(\frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}\). 3. The remainder is a different unit fraction, so \(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\).

\(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\)

- A unit fraction has numerator \(1\). - Ask what one of the ten equal parts is called as a fraction. - Relate the number of equal parts being counted to the numerator.

1. One tenth is the unit fraction \(\frac{1}{10}\). 2. Three tenths are three copies of that unit fraction: \(\frac{3}{10}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\). 3. The sum has \(3\) addends. That count matches the numerator \(3\), which tells how many tenths are present.

a) \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\) b) There are \(3\) addends, matching the numerator \(3\).

- Use the numerator to write \(\frac{2}{3}\) as repeated thirds. - Replace one third with the equivalent sum given in the problem. - Check that the three denominators in part 2) are different.

1. Two equal unit fractions give \(\frac{2}{3}=\frac{1}{3}+\frac{1}{3}\). 2. Replace one \(\frac{1}{3}\) with \(\frac{1}{4}+\frac{1}{12}\): \(\frac{2}{3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\).

1) \(\frac{2}{3}=\frac{1}{3}+\frac{1}{3}\) 2) \(\frac{2}{3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\)

- A unit fraction has a numerator of \(1\). - Find a common denominator before adding fractions with unlike denominators. - Consider unit fractions whose denominators are factors of \(6\).

1. Look for two unit fractions that can be rewritten with denominator \(6\). 2. Since \(\frac{1}{2}=\frac{3}{6}\) and \(\frac{1}{3}=\frac{2}{6}\), \(\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\).

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

- Read the total shaded amount from the bar. - Compare the number of shaded twelfths with the two unit twelfths already written. - Represent each remaining shaded part with one unit-fraction addend.

1. Read the bar: \(5\) of its \(12\) equal parts are shaded, so the target amount is \(\frac{5}{12}\). 2. The two written terms account for \(\frac{2}{12}\), leaving \(\frac{3}{12}\). 3. Write that missing amount as \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\).

\(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\)

- First determine what Leo's five identical addends total. - Use the bar to identify the size of one equal shaded part. - The number of shaded parts tells how many copies of that unit fraction are needed.

1. Five copies of \(\frac{1}{5}\) make \(\frac{5}{5}=1\), so Leo's sum represents a whole. 2. The displayed bar is divided into \(8\) equal parts, with \(5\) shaded. Each shaded part is therefore \(\frac{1}{8}\). 3. The correct sum is \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{5}{8}\).

Leo's sum equals \(1\), not the shaded amount. A correct sum is \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\).

- Identify the unit fraction represented by one equal part of a bar. - Count the shaded equal parts across both wholes. - Use one unit-fraction addend for each shaded part.

1. Each whole is divided into thirds, so each shaded part represents \(\frac{1}{3}\). 2. The model shows \(5\) shaded thirds in all, representing \(1\frac{2}{3}=\frac{5}{3}\). 3. Therefore, the unit-fraction sum is \(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\), which has \(5\) terms.

a) \(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\) b) \(5\) terms

- Read the starting number of shaded tenths from the bar. - Remove the stated number of shaded parts. - Use one \(\frac{1}{10}\) addend for each part that remains.

1. The bar is divided into tenths and begins with \(9\) shaded parts. Removing \(2\) leaves \(7\) shaded tenths. 2. The remaining unit-fraction sum is \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\). 3. Seven copies of \(\frac{1}{10}\) equal \(\frac{7}{10}\).

a) \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\) b) \(\frac{7}{10}\)

- Think of the numerator \(5\) as a count of five unit sixths. - Keep track separately of the copies that remain and the copies that were removed. - For the final equation, use sixths so the two groups visibly rebuild the original five copies.

1. Removing two of the five unit sixths leaves three copies: \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\). 2. The two removed copies total \(\frac{2}{6}=\frac{1}{3}\). 3. Using denominator \(6\), the parts recombine as \(\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\).

a) \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\) b) \(\frac{2}{6}=\frac{1}{3}\) c) \(\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\)

- Check the numerator of each addend against the definition of a unit fraction. - Ask how many copies of \(\frac{1}{6}\) make \(\frac{2}{6}\). - Verify that the rewritten right side still equals one half.

1. A unit fraction has numerator \(1\), so \(\frac{2}{6}\) is not a unit fraction. 2. The fraction \(\frac{2}{6}\) is two copies of \(\frac{1}{6}\). 3. Therefore, \(\frac{1}{2}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\), which is a valid unit-fraction sum because \(\frac{3}{6}=\frac{1}{2}\).

a) \(\frac{2}{6}\) is not a unit fraction because its numerator is not \(1\). b) \(\frac{1}{2}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\)

- Count the shaded equal parts in model a) and identify the unit fraction for one part. - Do the same for model b). - Compare the two numbers of unit-fraction addends.

1. Model a) shows \(3\) shaded halves, so the amount is \(\frac{3}{2}=1\frac{1}{2}\) and the unit-half sum is \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\). 2. Model b) shows the same amount as \(6\) shaded fourths, so the unit-fourth sum is \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\). 3. The fourths representation uses \(6-3=3\) more terms.

a) \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) b) \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\) c) The fourths representation uses \(3\) more terms.

- Read the shaded fraction from the bar first. - Rename that fraction in twelfths to find how many unit twelfths it contains. - Compare that count with Riley's eight copies.

1. The bar shows \(\frac{3}{4}\) shaded. 2. In twelfths, \(\frac{3}{4}=\frac{9}{12}\), so the bar represents nine copies of \(\frac{1}{12}\). 3. Riley used eight copies, so one more \(\frac{1}{12}\) is needed.

Riley is not correct. Add one copy of \(\frac{1}{12}\); the correct sum has nine copies of \(\frac{1}{12}\).

- Use the model to identify the unit fraction and count its shaded copies. - Regroup enough copies to make one whole. - Simplify the fraction represented by the copies left over.

1. Each shaded part is \(\frac{1}{6}\), and the model contains \(8\) shaded parts, so the total is \(\frac{8}{6}\). 2. Six sixths make one whole, leaving \(\frac{2}{6}\). 3. Since \(\frac{2}{6}=\frac{1}{3}\), the total is \(1\frac{1}{3}\). Maya is correct.

Yes. \(\frac{8}{6}=1\frac{2}{6}=1\frac{1}{3}\).

- Divide the number of identical terms equally among the three groups. - For part a), write one unit fraction for each term in a group. - For part b), combine and simplify the four twelfths.

1. Divide the twelve terms equally among three groups: \(12\div3=4\) terms per group. 2. Each group is \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{4}{12}\). 3. Since \(\frac{4}{12}=\frac{1}{3}\), each group equals \(\frac{1}{3}\).

a) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\) b) \(\frac{1}{3}\)

- Count the small sections in each color group; every small section has the same value. - For the shaded total, combine the counts before writing the numerator. - For the unshaded amount, count the white sections and use the same unit fraction.

1. The blue group has \(3\) unit twelfths, the orange group has \(2\), and the green group has \(4\). 2. Altogether there are \(3+2+4=9\) shaded unit twelfths. The expanded sum is \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\). 3. Three sections are unshaded, so the unshaded amount is \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\).

a) Blue: \(3\); orange: \(2\); green: \(4\) b) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\) c) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

- Ask what the denominator tells you about the size of each equal part. - Joining several fourth-sized parts does not cut those parts into a new size. - Count the number of unit fourths to determine the numerator of the sum.

1. Each addend is one fourth, so all three pieces have the same size. Joining them changes only how many fourths there are, not the size of each part. 2. There are three copies of \(\frac{1}{4}\), so the sum is \(\frac{3}{4}\), not \(\frac{3}{12}\).

a) The denominator stays \(4\) because every addend is still a fourth-sized part. b) \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

- First check whether every numerator is \(1\). - If every term is a unit fraction, check whether all the denominators are the same. - Only when the same unit fraction repeats do you need to combine the count into one numerator.

1. A uses three copies of the same unit fraction \(\frac{1}{8}\), so it is in the first category and totals \(\frac{3}{8}\). 2. B contains only unit fractions, but it uses both \(\frac{1}{4}\) and \(\frac{1}{8}\), so it is in the second category. 3. C contains \(\frac{2}{5}\), whose numerator is not \(1\), so it is in the third category. 4. D uses two copies of the same unit fraction \(\frac{1}{3}\), so it is in the first category and totals \(\frac{2}{3}\). 5. E contains only unit fractions, but it uses \(\frac{1}{2}\) and \(\frac{1}{4}\), so it is in the second category.

A: repeated copies of the same unit fraction; \(\frac{3}{8}\) B: all terms are unit fractions, but they are not all the same unit fraction C: at least one term is not a unit fraction D: repeated copies of the same unit fraction; \(\frac{2}{3}\) E: all terms are unit fractions, but they are not all the same unit fraction

- Treat each table count as a number of copies of the same unit fraction. - Add the counts to determine the numerator of the total in twelfths. - For part c), compare the group counts because every term has the same size.

1. Group A contributes three \(\frac{1}{12}\) terms, group B contributes one, and group C contributes five. The complete expanded sum is \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\). 2. The total is \(\frac{9}{12}=\frac{3}{4}\). 3. Group C has \(5\) copies, so it contributes \(\frac{5}{12}\), which is greater than group A's \(\frac{3}{12}\) and group B's \(\frac{1}{12}\).

a) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\) b) \(\frac{9}{12}=\frac{3}{4}\) c) Group C, \(\frac{5}{12}\)

- Six identical unit fractions will create an equivalent fraction with numerator \(6\). - Think about how the numerator \(3\) can become \(6\) without changing the fraction's value. - Apply the same scale factor to the denominator.

1. Six identical unit fractions combine to a fraction whose numerator is \(6\). 2. Find an equivalent fraction for \(\frac{3}{4}\) with numerator \(6\): multiply the numerator and denominator by \(2\). 3. \(\frac{3}{4}=\frac{6}{8}\), so the six identical terms are each \(\frac{1}{8}\).

\(\frac{1}{8}\)

- Think of one whole measure as eight equal eighth-note units. - Rewrite the half note in eighths. - How many eighths are missing from one whole measure? - Which combinations of quarter notes and eighth notes make that missing fraction?

1. The notes already use \(\frac{1}{2}+\frac{1}{8}=\frac{4}{8}+\frac{1}{8}=\frac{5}{8}\) of the measure. 2. For a), the unfilled part is \(1-\frac{5}{8}=\frac{3}{8}\). 3. For b), one quarter note and one eighth note fill \(\frac{2}{8}+\frac{1}{8}=\frac{3}{8}\). 4. Another way is three eighth notes: \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\).

a) \(\frac{3}{8}\) b) One quarter note and one eighth note; or three eighth notes.

- Start with a unit fraction smaller than \(\frac{2}{5}\), then find what remains. - For part 1), ask whether the remainder after your first choice is also a unit fraction. - For part 2), try decomposing a remaining fraction into two distinct unit fractions, and check that no denominator repeats.

1. For two unit fractions, subtract \(\frac{1}{3}\): \(\frac{2}{5}-\frac{1}{3}=\frac{6}{15}-\frac{5}{15}=\frac{1}{15}\). Therefore, \(\frac{2}{5}=\frac{1}{3}+\frac{1}{15}\). 2. For three unit fractions, begin with \(\frac{1}{4}\): \(\frac{2}{5}-\frac{1}{4}=\frac{8}{20}-\frac{5}{20}=\frac{3}{20}\). Since \(\frac{3}{20}=\frac{1}{10}+\frac{1}{20}\), one representation is \(\frac{2}{5}=\frac{1}{4}+\frac{1}{10}+\frac{1}{20}\).

1) \(\frac{2}{5}=\frac{1}{3}+\frac{1}{15}\) 2) \(\frac{2}{5}=\frac{1}{4}+\frac{1}{10}+\frac{1}{20}\)

- For part a), use the numerator to determine the number of twelfths. - For part b), look for unit fractions that can be rewritten with denominator \(12\). - Add using a common denominator to check your answer.

1. For a), \(\frac{5}{12}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\). 2. For b), one choice is \(\frac{1}{4}+\frac{1}{6}\). Using a common denominator, \(\frac{1}{4}+\frac{1}{6}=\frac{3}{12}+\frac{2}{12}=\frac{5}{12}\). 3. Another valid choice is \(\frac{1}{3}+\frac{1}{12}=\frac{4}{12}+\frac{1}{12}=\frac{5}{12}\).

a) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\) b) For example, \(\frac{1}{4}+\frac{1}{6}\) or \(\frac{1}{3}+\frac{1}{12}\).

- Rewrite unit fractions with denominator \(12\). - For part b), try decomposing one unit fraction from part a) into two smaller unit fractions. - Check each sum using a common denominator.

1. For a), \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\). Another valid answer is \(\frac{1}{2}+\frac{1}{12}\). 2. For b), \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}=\frac{4}{12}+\frac{2}{12}+\frac{1}{12}=\frac{7}{12}\).

a) For example, \(\frac{1}{3}+\frac{1}{4}\) or \(\frac{1}{2}+\frac{1}{12}\). b) For example, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}\).

- Read the shaded fraction from the bar. - Look for an equivalent fraction whose numerator matches the number of identical addends. - The denominator of that equivalent fraction tells the size of one unit fraction.

1. The bar shows \(3\) of \(4\) equal parts shaded, so the shaded amount is \(\frac{3}{4}\). 2. To express that amount with \(9\) equal unit-fraction terms, rename \(\frac{3}{4}\) with numerator \(9\): \(\frac{3}{4}=\frac{9}{12}\). 3. Therefore, the nine identical addends are copies of \(\frac{1}{12}\).

\(\frac{1}{12}\)

- Read the shaded fraction from the bar and consider its simplest form. - In a repeated unit-fraction sum, the numerator counts the number of identical terms. - Ask what happens to that numerator when you make an equivalent fraction with more equal parts.

1. The bar shows \(\frac{3}{4}\). 2. In simplest form, \(\frac{3}{4}\) means \(3\) copies of the unit fraction \(\frac{1}{4}\). 3. Any equivalent fraction with a different denominator would multiply the numerator \(3\) by a whole number and therefore use more than \(3\) identical unit-fraction copies. 4. The smallest possible number of terms is \(3\): \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\).

\(3\) terms: \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)

- Read each colored segment as a fraction of the twelve equal parts. - Choose one colored unit fraction and imagine dividing that amount into two equal pieces. - Check that each new expression has four unit-fraction terms and still makes one whole.

1. The blue part is \(\frac{6}{12}=\frac{1}{2}\). Splitting it into two equal parts gives \(\frac{1}{4}+\frac{1}{4}\), so \(1=\frac{1}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{6}\). 2. The orange part is \(\frac{4}{12}=\frac{1}{3}\). Splitting it into two equal parts gives \(\frac{1}{6}+\frac{1}{6}\), so \(1=\frac{1}{2}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\). 3. The green part is \(\frac{2}{12}=\frac{1}{6}\). Splitting it into two equal parts gives \(\frac{1}{12}+\frac{1}{12}\), so \(1=\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{12}\).

Split \(\frac{1}{2}\): \(\frac{1}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{6}\) Split \(\frac{1}{3}\): \(\frac{1}{2}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\) Split \(\frac{1}{6}\): \(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{12}\)

- A unit fraction has numerator \(1\). - Simplify each candidate only as far as needed to decide whether its numerator can become \(1\). - Look for a common factor shared by the numerator and \(12\).

1. A unit fraction has numerator \(1\), so each fraction must simplify to a fraction with numerator \(1\). 2. \(\frac{1}{12}\) is already a unit fraction. Also, \(\frac{2}{12}=\frac{1}{6}\), \(\frac{3}{12}=\frac{1}{4}\), \(\frac{4}{12}=\frac{1}{3}\), and \(\frac{6}{12}=\frac{1}{2}\). 3. The remaining fractions from \(\frac{5}{12}\) through \(\frac{11}{12}\) do not simplify to a numerator of \(1\).

\(\frac{1}{12}=\frac{1}{12}\), \(\frac{2}{12}=\frac{1}{6}\), \(\frac{3}{12}=\frac{1}{4}\), \(\frac{4}{12}=\frac{1}{3}\), \(\frac{6}{12}=\frac{1}{2}\)

- For part a), use the definition of a unit fraction, not only the denominator. - For part b), think about an equivalent unit fraction for \(\frac{2}{4}\). - Find a second decomposition by treating \(\frac{3}{4}\) as copies of one unit fraction.

1. A unit fraction must have numerator \(1\). Therefore, \(\frac{2}{4}\) is not a unit fraction, even though it is equivalent to \(\frac{1}{2}\). 2. One valid rewrite is \(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\). 3. Another valid rewrite is \(\frac{3}{4}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\).

a) \(\frac{2}{4}\) is not a unit fraction because its numerator is not \(1\). b) \(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\); \(\frac{3}{4}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)

- Use the twelve equal sections of the bar as the common unit. - Count how many twelfths belong to each color before combining them. - The total number of unit fractions is the sum of the three color counts.

1. The blue amount is \(\frac{6}{12}=\frac{1}{2}\), so it contributes six copies of \(\frac{1}{12}\). 2. The orange amount is \(\frac{4}{12}=\frac{1}{3}\), so it contributes four copies of \(\frac{1}{12}\). 3. The green amount is \(\frac{2}{12}=\frac{1}{6}\), so it contributes two copies of \(\frac{1}{12}\). 4. Altogether there are \(6+4+2=12\) copies of \(\frac{1}{12}\), which make \(1\).

a) Blue: \(6\) twelfths; orange: \(4\) twelfths; green: \(2\) twelfths b) Twelve copies of \(\frac{1}{12}\); total \(1\)

- For part a), replace a term only when the tree says that term splits. - Terms that do not split remain in the final leaf list. - For parts b) and c), count the final terms and verify them with twelfths.

1. One original half remains a leaf: \(\frac{1}{2}\). 2. The other half is replaced by \(\frac{1}{3}\) and \(\frac{1}{6}\), and the sixth is then replaced by two twelfths. 3. The final leaves are \(\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{1}{12}\), for four leaves. 4. Their total is \(\frac{6}{12}+\frac{4}{12}+\frac{1}{12}+\frac{1}{12}=1\).

a) \(\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{1}{12}\) b) \(4\) leaves c) \(\frac{6}{12}+\frac{4}{12}+\frac{1}{12}+\frac{1}{12}=1\)

- Count all equal parts in the bar to identify the unit fraction. - Count the shaded parts to identify how many copies of that unit fraction are used. - Generalize from what the two numbers in a fraction tell you about equal parts.

1. The bar has \(8\) equal parts, so one part is \(\frac{1}{8}\). Five parts are shaded, so the shaded fraction is \(\frac{5}{8}\). 2. Therefore, \(\frac{5}{8}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\). 3. In any fraction, the denominator tells the size of one equal part, so it determines the unit fraction. The numerator tells how many of those equal parts are counted.

The model shows \(\frac{5}{8}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\). In general, the denominator names the unit fraction and the numerator tells how many copies of that unit fraction are present.

- Verify each sum using a common denominator. - Compare the first unit fractions in the two methods. - Check the next larger unit fraction to determine whether it would exceed \(\frac{3}{10}\).

1. For Method A, \(\frac{1}{4}+\frac{1}{20}=\frac{5}{20}+\frac{1}{20}=\frac{6}{20}=\frac{3}{10}\). 2. For Method B, \(\frac{1}{5}+\frac{1}{10}=\frac{2}{10}+\frac{1}{10}=\frac{3}{10}\). 3. The greatest unit fraction not exceeding \(\frac{3}{10}\) is \(\frac{1}{4}\): \(\frac{1}{3}>\frac{3}{10}\), while \(\frac{1}{4}<\frac{3}{10}\). Therefore, Method A follows the greedy rule.

a) Both methods are correct. b) Method A follows the greedy rule because \(\frac{1}{4}\) is the greatest unit fraction less than \(\frac{3}{10}\).

- Estimate the reciprocal of \(\frac{3}{14}\) to find the first denominator. - Use a common denominator to subtract the fractions. - Check whether the remainder is a unit fraction.

1. Since \(14\div3\approx4.67\), the smallest integer denominator that gives a unit fraction less than \(\frac{3}{14}\) is \(5\). Thus, the first term is \(\frac{1}{5}\). 2. Subtract: \(\frac{3}{14}-\frac{1}{5}=\frac{15}{70}-\frac{14}{70}=\frac{1}{70}\). 3. Therefore, \(\frac{3}{14}=\frac{1}{5}+\frac{1}{70}\).

\(\frac{3}{14}=\frac{1}{5}+\frac{1}{70}\)

- Test unit fractions slightly less than \(\frac{1}{4}\). - After choosing one term, subtract it from \(\frac{1}{4}\) to find the other. - The two unit fractions in each sum must be different.

1. Try unit fractions just less than \(\frac{1}{4}\). 2. Starting with \(\frac{1}{5}\), the remainder is \(\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\). Thus, \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\). 3. Starting with \(\frac{1}{6}\), the remainder is \(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\). Thus, \(\frac{1}{4}=\frac{1}{6}+\frac{1}{12}\). 4. Verify: \(\frac{1}{5}+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{1}{4}\), and \(\frac{1}{6}+\frac{1}{12}=\frac{2}{12}+\frac{1}{12}=\frac{1}{4}\).

1) \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\) 2) \(\frac{1}{4}=\frac{1}{6}+\frac{1}{12}\)

- At each stage, choose the greatest unit fraction that does not exceed the current amount. - Use common denominators for each subtraction. - Stop when the remainder is a unit fraction.

1. The greatest unit fraction less than \(\frac{4}{5}\) is \(\frac{1}{2}\). 2. Subtract: \(\frac{4}{5}-\frac{1}{2}=\frac{8}{10}-\frac{5}{10}=\frac{3}{10}\). 3. The greatest unit fraction less than \(\frac{3}{10}\) is \(\frac{1}{4}\), because \(\frac{1}{3}>\frac{3}{10}\). 4. Subtract: \(\frac{3}{10}-\frac{1}{4}=\frac{6}{20}-\frac{5}{20}=\frac{1}{20}\). 5. Therefore, \(\frac{4}{5}=\frac{1}{2}+\frac{1}{4}+\frac{1}{20}\).

\(\frac{4}{5}=\frac{1}{2}+\frac{1}{4}+\frac{1}{20}\)

- Start with a large unit fraction that is less than \(\frac{3}{4}\). - Subtract it to find the remainder. - Split the remainder into two different unit fractions.

1. Begin with \(\frac{1}{2}\): \(\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\). 2. Decompose the remainder: \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\). 3. Therefore, \(\frac{3}{4}=\frac{1}{2}+\frac{1}{5}+\frac{1}{20}\). 4. Another valid answer is \(\frac{3}{4}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}\).

One possible answer is \(\frac{3}{4}=\frac{1}{2}+\frac{1}{5}+\frac{1}{20}\).

- If five identical unit fractions are combined, what number would count the copies in the numerator? - List a few numerators of fractions equivalent to \(\frac{2}{3}\). - Decide whether a numerator of \(5\) can occur in that pattern.

1. Five identical unit fractions would combine into one fraction whose numerator is \(5\), because there would be five equal unit-fraction copies. 2. Equivalent fractions for \(\frac{2}{3}\) are made by multiplying the numerator and denominator by the same whole number. Their numerators are \(2,4,6,8,\ldots\). 3. Since \(5\) is not one of those possible numerators, no equivalent fraction for \(\frac{2}{3}\) can represent exactly five identical unit-fraction copies.

No. \(\frac{2}{3}\) cannot be written as exactly five identical unit fractions.

- Read the three original unit fractions from the colored bar before making any split. - A split replaces one unit fraction by two equal smaller unit fractions without changing the total value. - Track both the number of splits and the number of terms after each split.

1. The bar represents \(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1\), which starts with three terms. 2. Split \(\frac{1}{2}\) into \(\frac{1}{4}+\frac{1}{4}\). There are now four quarter terms. 3. Split one \(\frac{1}{4}\) into \(\frac{1}{8}+\frac{1}{8}\). 4. Split a second \(\frac{1}{4}\) into \(\frac{1}{8}+\frac{1}{8}\), then split a third \(\frac{1}{4}\) the same way. After four splits total, one possible result is \(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=1\). 5. Each split replaces one term by two, so every split increases the term count by exactly \(1\). Starting with \(3\) terms, four splits always produce \(7\) terms.

a) One possible result is \(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=1\). b) Every four-split result has \(7\) terms.