QuanticaPrealgebraOpen in the course\(\tfrac{3}{4}\) and \(0.75\) are two names for one number. This lesson covers switching between them in both directions, and shows what decides whether a fraction produces a stopping decimal or one that runs forever.
Problem
A trail sign reads \(0.7\) km to the next lookout. The ranger's logbook only accepts fractions over a power of ten. Write \(0.7\) as a fraction.
Show a hint
- The first slot to the right of the decimal point has a special name. A single digit sitting there is not seven of something big, it is seven of something small. What is the name of that place, and what number goes underneath when you count that many of them?
- The last digit is \(7\) and it lives in the tenths place, so the denominator is \(10\). The digits after the point become the numerator, which here is just \(7\). Put the \(7\) on top and the \(10\) on the bottom.
Show the full solution
The \(7\) sits in the tenths place, so \(0.7\) is seven tenths, which is \(\boxed{7/10}\).
Every terminating decimal works this way. The last filled place names the denominator, and the digits after the point read straight across as the numerator.
Problem
A ribbon measures \(0.45\) m. Start with \(\frac{45}{100}\), find \(\gcd(45,100)\), and reduce. Write \(0.45\) as a fraction in simplest form.
Show a hint
- Start where the reading tells you to start. Two places past the point means hundredths, so \(0.45 = \frac{45}{100}\). Now ask yourself the Chapter 4 question. What is the largest number that divides both \(45\) and \(100\) at the same time?
- List what divides each one. The number \(45\) splits as \(3 \times 3 \times 5\), and \(100\) splits as \(2 \times 2 \times 5 \times 5\). The only factor they truly share is a single \(5\), so \(\gcd(45,100) = 5\). Divide the top and the bottom by \(5\).
Show the full solution
Two digits past the point means hundredths, so \(0.45 = \frac{45}{100}\). Since \(45 = 3 \times 3 \times 5\) and \(100 = 2 \times 2 \times 5 \times 5\), the only shared prime is a single \(5\), so \(\gcd(45,100) = 5\). $$\frac{45}{100} = \frac{45 \div 5}{100 \div 5} = \frac{9}{20}$$ \(\boxed{9/20}\)
The top \(9 = 3 \times 3\) and the bottom \(20 = 2 \times 2 \times 5\) share no prime, which is how you know it is all the way reduced.
Problem
Write \(2.5\) as a single improper fraction in simplest form. Read all digits as \(\tfrac{25}{10}\), then reduce.
Show a hint
- Read all the figures at once, ignoring the point for a moment. The digits are \(25\), and one digit sits past the decimal point, so the bottom is \(10^1 = 10\). That gives you \(\tfrac{25}{10}\) before you tidy it up.
- Now reduce \(\tfrac{25}{10}\). Both numbers share a factor of \(5\), so divide top and bottom by \(5\). You get \(\tfrac{25 \div 5}{10 \div 5}\), and since the top stays bigger than the bottom, leave it as an improper fraction rather than a mixed number.
Show the full solution
Read all the digits as one number, \(25\). One digit sits past the point, so the denominator is \(10^1 = 10\), giving \(2.5 = \tfrac{25}{10}\). Both share a factor of \(5\), so $$\frac{25}{10} = \frac{25 \div 5}{10 \div 5} = \frac{5}{2}.$$ Since \(5 > 2\) this is already improper, which is what was asked. \(\boxed{5/2}\)
The whole part out front changes nothing. Only the count of digits after the point sets the power of ten underneath.
Problem
Write \(-1.72\) as a single improper fraction in simplest form, keeping the sign. Convert the size \(1.72\) to \(\tfrac{172}{100}\), reduce using \(\gcd(172,100)=4\), then reattach the minus sign.
Show a hint
- The negative sign is just along for the ride. Convert the size 1.72 on its own. Two decimal places means hundredths, so the bottom of the fraction is 100.
- Reduce \(\tfrac{172}{100}\) by its greatest common divisor. Since \(\gcd(172,100)=4\), divide both the top and the bottom by 4, then write the minus sign in front of the result.
Show the full solution
Convert the size first. The last digit of \(1.72\) is in the hundredths place, so \(1.72 = \tfrac{172}{100}\). Since \(\gcd(172,100) = 4\), $$\frac{172}{100}=\frac{172\div 4}{100\div 4}=\frac{43}{25}.$$ Put the minus sign back and you have \(\boxed{-43/25}\).
It is fully reduced because \(43\) is prime. The sign takes no part in the reducing, so set it aside and reattach it at the end.
Problem
A depth sensor reports \(\frac{47}{100}\) m. The denominator is already \(100\). Drop the digits \(47\) into the tenths and hundredths places. Write \(\frac{47}{100}\) as a decimal.
Show a hint
- The bottom number tells you the smallest place value in play. A denominator of \(100\) means the last digit of the numerator should land in the hundredths place, two spots to the right of the decimal point.
- Take the numerator \(47\). The \(7\) goes in the hundredths place and the \(4\) goes in the tenths place. There is no whole part, so write a \(0\) in the ones place to hold it open, then the decimal point, then \(47\).
Show the full solution
A denominator of \(100\) means hundredths, so the numerator fills two places to the right of the point. The \(7\) drops into the hundredths place and the \(4\) sits in the tenths place. Since \(47 < 100\) there is no whole part, so a \(0\) holds the ones place. That gives \(\frac{47}{100} = \boxed{0.47}\).
This is the same place-value reading you used going the other way, just run backward. Two zeros in the denominator, two places after the point.
Problem
Write \(\tfrac{3}{5}\) as a decimal by multiplying numerator and denominator by \(2\) so the bottom becomes \(10\). Read off the decimal.
Show a hint
- You cannot just swap the \(5\) for a \(10\) on its own, because that would change the value. Whatever you do to the bottom you must also do to the top. What single number turns \(5\) into \(10\) when you multiply by it?
- Multiply both \(3\) and \(5\) by \(2\). That gives \(\tfrac{6}{10}\), which is the same value as \(\tfrac{3}{5}\) since you only multiplied by \(\tfrac{2}{2}\), which equals \(1\). Six tenths reads straight off the tenths place.
Show the full solution
Since \(5 \times 2 = 10\), multiply top and bottom by \(2\). $$\frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10}$$ Six tenths is exactly what the first place after the point holds, so the answer is \(\boxed{0.6}\).
Multiplying top and bottom by the same number is multiplying by \(\tfrac{2}{2} = 1\), so the value never moves. Only the name changes.
Problem
Write \(\tfrac{11}{20}\) as a decimal. The denominator \(20\) does not reach \(10\), so aim for the next power of ten it divides into evenly.
Show a hint
- You cannot reach 10, so aim higher. What is the next power of ten after 10, and does 20 fit into it a whole number of times?
- Since \(20 \times 5 = 100\), multiply both the top and the bottom by 5. That gives \(\tfrac{55}{100}\), and a denominator of 100 reads straight off as hundredths.
Show the full solution
\(20\) is bigger than \(10\), so aim for \(100\). Since \(100 \div 20 = 5\) with nothing left over, multiply top and bottom by \(5\). $$\frac{11}{20} = \frac{11 \times 5}{20 \times 5} = \frac{55}{100}$$ That is \(55\) hundredths, or \(\boxed{0.55}\).
It should land just past \(0.5\), since \(\tfrac{10}{20} = 0.5\) and you have one twentieth more than that.
Problem
A wrench is \(\tfrac{7}{8}\) inch wide. Since \(8=2^3\), supply three matching \(5\)s to reach \(2^3 \times 5^3=1{,}000\), then read off the decimal. Write \(\tfrac{7}{8}\) as a decimal.
Show a hint
- Decimals are just fractions whose denominators are powers of ten. So aim the bottom at a power of ten. You already have three 2s in the \(8\). How many 5s would you need to add so the 2s and 5s match up and form \(10\times 10\times 10\)?
- To turn \(8 = 2^3\) into \(1{,}000 = 2^3\times 5^3\) you are missing exactly three 5s, and \(5^3 = 125\). Multiply the top and bottom of \(\tfrac{7}{8}\) by \(125\), then read off the three decimal places.
Show the full solution
\(8 = 2^3\), so the bottom carries three \(2\)s and no \(5\)s. A power of ten pairs its \(2\)s and \(5\)s evenly, and \(1{,}000 = 2^3 \times 5^3\), so the denominator is short exactly three \(5\)s. Multiply top and bottom by \(5^3 = 125\). $$\frac{7}{8} = \frac{7\times 125}{8\times 125} = \frac{875}{1000}$$ A denominator of \(1{,}000\) means three decimal places, so \(\boxed{0.875}\).
You only ever have to top up whichever of \(2\) or \(5\) the denominator is missing.
Problem
Factor each denominator: \(\frac{7}{16}\), \(\frac{9}{40}\), \(\frac{5}{12}\), \(\frac{3}{25}\). Only \(2\)s and \(5\)s allow termination. Which fraction cannot terminate? Give its denominator.
Show a hint
- Factor each denominator into primes. \(16 = 2^4\), \(40 = 2^3 \times 5\), \(25 = 5^2\), and \(12 = 2^2 \times 3\). Now look for any prime that is not a 2 or a 5.
- A power of ten is \(2^n \times 5^n\), so it never contains a factor of 3. Three of the denominators use only 2s and 5s, so you can supply whatever is missing and climb to \(10\), \(100\), or \(1{,}000\). The denominator that hides a 3 can never be scaled into a power of ten, because no multiplier of 2s and 5s will ever cancel that 3.
Show the full solution
Factor the denominators. \(16 = 2^4\), \(40 = 2^3 \times 5\), \(25 = 5^2\), and \(12 = 2^2 \times 3\). A power of ten is \(2^n \times 5^n\), so the first three can be scaled up to one, but the \(3\) inside \(12\) cannot. So \(\dfrac{5}{12}\) is the fraction that never terminates, and its denominator is \(\boxed{12}\).
Multiplying by more \(2\)s and \(5\)s only adds factors. It can never remove a \(3\) that is already sitting on the bottom.
Problem
A board is \(5.75\) feet. The carpenter marks off eighths of a foot. Convert \(5.75\) to a fraction with denominator \(8\). How many eighths of a foot are in \(5.75\)?
Show a hint
- The decimal \(5.75\) and the fraction it equals are the same number written two ways. Turn \(5.75\) into a fraction first, then think about how to rewrite that fraction so its bottom number is \(8\). Once the bottom is \(8\), the top tells you how many eighths.
- Write \(5.75\) as \(\frac{575}{100}\), which reduces to \(\frac{23}{4}\). To rename \(\frac{23}{4}\) with a bottom of \(8\), multiply the top and bottom by \(2\). Read off the new top number.
Show the full solution
The \(75\) sits in the hundredths place, so \(5.75 = \frac{575}{100}\). Both share a factor of \(25\), and that reduces to \(\frac{23}{4}\). Since \(4 \times 2 = 8\), multiply top and bottom by \(2\). $$\frac{23}{4} = \frac{23 \times 2}{4 \times 2} = \frac{46}{8}$$ There are \(\boxed{46}\) eighths in \(5.75\).
Asking how many of a unit fit inside a number is just renaming the number with that unit on the bottom and reading the top.
Problem
Find the reciprocal of \(2.5\) and write it as a decimal. Convert \(2.5\) to a fraction, flip it, then convert back to a decimal.
Show a hint
- Start by writing it as a fraction. The digits of \(2.5\) are \(25\), and there is one place after the point, so \(2.5 = \frac{25}{10}\). Now reduce that to lowest terms before you flip, since a tidy fraction is much easier to turn over.
- In lowest terms \(2.5 = \frac{5}{2}\). Flipping a fraction means swapping the top and bottom, so the reciprocal of \(\frac{5}{2}\) is \(\frac{2}{5}\). Now you just need \(\frac{2}{5}\) as a decimal. Scale the bottom up to \(10\) by multiplying top and bottom by \(2\).
Show the full solution
\(2.5 = \frac{25}{10}\), which reduces to \(\frac{5}{2}\). Flipping gives the reciprocal \(\frac{2}{5}\). Multiply top and bottom by \(2\) to reach tenths, \(\frac{2}{5} = \frac{4}{10}\), and four tenths is \(\boxed{0.4}\).
Flipping is easy on a fraction and awkward on a decimal, so switch names first. Check it with \(2.5 \times 0.4 = 1\).
Problem
A calculator shows exactly \(0.84\). The divisor was less than \(30\) and the division was already in simplest form. Reduce \(\tfrac{84}{100}\) and read off the dividend.
Show a hint
- The screen shows \(0.84\), which is just another way of writing \(\tfrac{84}{100}\). Strip that fraction all the way down to lowest terms and study the denominator you land on. That denominator has to match the real divisor, since the student says nothing could be cancelled anymore.
- Once \(\tfrac{84}{100}\) is in lowest terms its denominator is the smallest whole number you could have divided by. The clue says the actual divisor is under \(30\), so see whether that lowest-terms denominator fits. If it does, it must BE the divisor, and the matching number on top is your dividend.
Show the full solution
\(0.84\) means eighty-four hundredths, so \(0.84 = \tfrac{84}{100}\). Their \(\gcd\) is \(4\), so $$\frac{84}{100} = \frac{84 \div 4}{100 \div 4} = \frac{21}{25}.$$ The division was already in simplest form, so the divisor had to be \(25\), \(50\), \(75\), and so on. Only \(25\) is under \(30\), which puts \(\boxed{21}\) on top.
The screen could show a clean ending decimal because \(25 = 5 \times 5\) is built from nothing but \(5\)s.
Practice these ideas
Practice
A sprint timer showed \(0.3\) of a second. Write \(0.3\) as a fraction over a power of ten.
Show the solution
The \(3\) sits one spot right of the point, in the tenths place, so \(0.3 = \frac{3}{10}\), which is \(\boxed{3/10}\).
It is already in lowest terms, since \(3\) is prime and \(10 = 2 \times 5\) shares nothing with it.
Practice
A bottle is labeled \(0.6\) full. Write \(0.6\) as a fraction in simplest form.
Show the solution
The \(6\) is in the tenths place, so \(0.6\) means six tenths, \(\frac{6}{10}\). Both are even and \(\gcd(6,10) = 2\), so $$\frac{6}{10}=\frac{6\div 2}{10\div 2}=\frac{3}{5}.$$ The bottle is \(\boxed{3/5}\) full.
Simplest form is the point where the top and bottom share no factor bigger than \(1\), and \(3\) and \(5\) do not.
Practice
\(0.85\) of students voted to keep the lunch menu. Write \(0.85\) as a fraction in lowest terms.
Show the solution
Two digits past the point means hundredths, so \(0.85 = \frac{85}{100}\). Since \(85 = 5 \times 17\) and \(100 = 5 \times 20\), the greatest common divisor is \(5\). $$\frac{85}{100} = \frac{85 \div 5}{100 \div 5} = \frac{17}{20}$$ The poll result is \(\boxed{17/20}\).
\(17\) is prime and does not divide \(20\), so nothing more comes out.
Practice
A drill bit is stamped \(0.375\) of an inch. Start from \(\tfrac{375}{1{,}000}\) and reduce. Write \(0.375\) in simplest form as a fraction.
Show the solution
The last digit lands in the thousandths place, so \(0.375 = \tfrac{375}{1{,}000}\). Since \(375 = 125 \times 3\) and \(1{,}000 = 125 \times 8\), the gcd is \(125\), and dividing both by it gives \(\boxed{3/8}\).
Check it by going back the other way, \(3 \div 8 = 0.375\).
Practice
Convert \(-2.6\) to a single improper fraction in simplest form, keeping the sign. Write your answer as one fraction.
Show the solution
Convert the size first. The \(6\) is in the tenths place, so \(2.6 = \frac{26}{10}\). Since \(\gcd(26,10) = 2\), that reduces to \(\frac{26 \div 2}{10 \div 2} = \frac{13}{5}\). Put the sign back for \(\boxed{-13/5}\).
It is improper and in lowest terms already, since \(13\) is prime and bigger than \(5\).
Practice
A rain gauge reads \(\tfrac{73}{100}\) of a liter. The denominator is already a power of ten. Write \(\tfrac{73}{100}\) as a decimal.
Show the solution
A denominator of \(100\) counts hundredths, so the numerator fills the first two places after the point. Drop \(73\) into the tenths and hundredths spots with a \(0\) in front, and the gauge reads \(\boxed{0.73}\) of a liter.
When the denominator is already a power of ten there is nothing to divide. You just read the numerator into its places.
Practice
A crew has paved \(\tfrac{4}{5}\) of a trail. Double the numerator and denominator to reach a denominator of \(10\). Write \(\tfrac{4}{5}\) as a decimal.
Show the solution
Since \(5 \times 2 = 10\), multiply top and bottom by the same \(2\), so \(\frac{4}{5} = \frac{8}{10}\). Eight tenths sits in the first place after the point, giving \(\boxed{0.8}\).
The decimal stops after a single digit because \(5\) divides evenly into \(10\).
Practice
A jar is \(\tfrac{13}{20}\) full. Multiply top and bottom by \(5\) to get a denominator of \(100\), then read the decimal. Write \(\tfrac{13}{20}\) as a decimal.
Show the solution
Since \(20 \times 5 = 100\), multiply top and bottom by \(5\). $$\frac{13}{20} = \frac{13 \times 5}{20 \times 5} = \frac{65}{100}$$ Sixty-five hundredths is \(0.65\), so the jar is \(\boxed{0.65}\) full.
It terminates because \(20 = 2^2 \times 5\) carries only \(2\)s and \(5\)s.
Practice
A gauge block is \(\tfrac{5}{8}\) inch thick. Since \(8=2^3\), supply \(5^3=125\) to reach \(1{,}000\). Write \(\tfrac{5}{8}\) as a decimal.
Show the solution
\(8 = 2^3\), so the bottom holds three \(2\)s and no \(5\)s. The smallest power of ten using three \(2\)s is \(10^3 = 2^3 \times 5^3 = 1{,}000\), so supply \(5^3 = 125\). $$\frac{5}{8} = \frac{5 \times 125}{8 \times 125} = \frac{625}{1{,}000}$$ Six hundred twenty-five thousandths is \(\boxed{0.625}\).
Check it another way. \(\tfrac{4}{8} = 0.5\), and one more eighth is \(0.125\), so \(0.5 + 0.125 = 0.625\).
Practice
Three fractions in lowest terms: \(\dfrac{7}{8}\), \(\dfrac{4}{15}\), \(\dfrac{9}{40}\). Factor each denominator. Which one has a prime other than \(2\) or \(5\)? Give that denominator.
Show the solution
Factor each denominator. \(8 = 2^3\) and \(40 = 2^3 \times 5\) use only \(2\)s and \(5\)s, so both terminate. But \(15 = 3 \times 5\) carries a \(3\), and the fraction is already in lowest terms, so nothing cancels it. The answer is \(\boxed{15}\).
A power of ten is \(2^n \times 5^n\), so a denominator holding any other prime can never be scaled into one.
Practice
The full-power brightness factor is \(1.25\). The low-power factor is its reciprocal. Convert \(1.25\) to a fraction, flip it, then write the reciprocal as a decimal.
Show the solution
The last digit of \(1.25\) is in the hundredths place, so \(1.25 = \frac{125}{100}\), which reduces to \(\frac{5}{4}\). Flipping gives \(\frac{4}{5}\), and multiplying top and bottom by \(2\) gives \(\frac{8}{10}\). The low-power factor is \(\boxed{0.8}\).
Check it with \(1.25 \times 0.8 = 1\), which is what a reciprocal pair has to do.
Practice
A glaze recipe is \(\tfrac{17}{50}\) cobalt. Since \(50 \times 2 = 100\), multiply top and bottom by \(2\). Write \(\tfrac{17}{50}\) as a decimal.
Show the solution
Since \(50 \times 2 = 100\), multiply top and bottom by \(2\). $$\frac{17}{50} = \frac{17 \times 2}{50 \times 2} = \frac{34}{100}$$ A denominator of \(100\) means two decimal places, so the glaze is \(\boxed{0.34}\) cobalt.
\(50 = 2 \times 5^2\) holds only \(2\)s and \(5\)s, so a clean decimal was guaranteed.
Practice
A shelf is \(3.25\) feet. Convert to a fraction, reduce to \(\tfrac{13}{4}\), then rename with denominator \(8\). How many eighths of a foot are in \(3.25\)?
Show the solution
Two digits sit after the point, so \(3.25 = \frac{325}{100}\). Both share a factor of \(25\), which reduces it to \(\frac{13}{4}\). Since \(4 \times 2 = 8\), multiply top and bottom by \(2\). $$\frac{13}{4} = \frac{13 \times 2}{4 \times 2} = \frac{26}{8}$$ So the shelf holds \(\boxed{26}\) eighths of a foot.
Renaming does not change the length, only the size of the piece you count with. Check it with \(26 \div 8 = 3.25\).
Practice
A calculator shows exactly \(0.65\). The divisor was less than \(25\) and the fraction was in lowest terms. Reduce \(\tfrac{65}{100}\) to find the divisor. What whole number was divided?
Show the solution
\(0.65\) means sixty-five hundredths, so start from \(\frac{65}{100}\). Since \(65 = 5 \times 13\) and \(100 = 5 \times 20\), that reduces to \(\frac{13}{20}\). The divisor is the bottom number, and \(20\) is indeed under \(25\), so the number divided was \(\boxed{13}\).
\(13\) and \(20\) share no factor bigger than \(1\), which confirms the lowest-terms clue.