Fractions are Hard!

2.5 The key fraction property

We have that \(\dfrac{a}{b}\) is an answer to a division problem:


 \(\dfrac{a}{b}\) represents the amount of pie an individual boy receives when \(a\) pies are distributed among \(b\) boys.


What happens if we double the number of pies and double the number of boys? Nothing! The amount of pie per boy is still the same:




For example, this picture shows that \(\dfrac{6}{3}\)  and \(\dfrac{12}{6}\) both give two pies to each boy:


And tripling the number of pies and tripling the number of boys also does not change the final amount of pie per boy, nor does quadrupling each number, or one-trillion-billion-tupling the numbers!


\(\dfrac{6}{3} =\dfrac{12}{6} = \dfrac{18}{9}= \dots = \) two pies per boy.


This leads us to want to believe


KEY FRACTION PROPERTY:    \(\dfrac{xa}{xb} = \dfrac{a}{b}\) 

(at least for positive whole numbers – but maybe for other types of numbers too!).


For example,


\(\dfrac{3}{5}\)   (sharing three pies among five boys)


yields the same result as


\(\dfrac{3 \times 2}{5 \times 2}= \dfrac{6}{10}\)  (sharing six pies among ten boys),


and as

\(\dfrac{3 \times 100}{5 \times 100}= \dfrac{300}{500}\) (sharing 300 pies among 500 boys).


Going backwards …


\(\dfrac{20}{32}\) (sharing 20 pies among 32 boys)


is the same problem as

\(\dfrac{5 \times 4}{8 \times 4}= \dfrac{5}{8}\) (sharing five pies among eight boys).



Comment: Most people say we have “cancelled” a common factor of 4 from the numerator and the denominator and, in doing so, we have “simplified” the expression \(\dfrac{20}{32}\). (Admittedly, \(\dfrac{5}{8}\) is easier to conceptualize than \(\dfrac{20}{32}\).)



As another example \(\dfrac{280}{350}\) can certainly be made to look more manageable by noticing that there is a common factor of 10 in both the numerator and the denominator: \(\dfrac{280}{320}=\dfrac{28\times 10}{32 \times 10}=\dfrac{28}{35}\).


We can go further as 28 and 35 are both multiples of 7: \(\dfrac{28}{35}=\dfrac{4 \times 7}{5 \times 7}=\dfrac{4}{5}\).


Thus, sharing 280 pies among 350 boys gives the same result as sharing just 4 pies among 5 boys: much easier to conceptualize!


As 4 and 5 share no common factors, this is as far as we can go with this example (while staying with whole numbers).


Thinking Question: Jenny says that \(\dfrac{4}{5}\) does “simplify” further is you are willing to move away from whole numbers. She writes:

\(\dfrac{4}{5}=\dfrac{2 \times 2}{2\frac{1}{2} \times 2} = \dfrac{2}{2\frac{1}{2}}\).


Is she right? Does sharing \(4\) pies among \(5\) boys yield the same result as sharing \(2\) pies among \(2\frac{1}{2}\) boys? What do you think? (And is her expression “simpler” than the original?)



We now have four beliefs about how fractions should work.

BELIEF 1: \(\dfrac{a}{a} = 1\)

BELIEF 2: \(\dfrac{a}{1}=a\)

BELIEF 3: \(x \times \dfrac{a}{b} = \dfrac{xa}{b}\)           (Basic Multiplication Belief)

BELIEF 4: \(\dfrac{a}{b}=\dfrac{xa}{xb}\)                           (Key Fraction Belief)

These beliefs feel so natural and right for positive counting numbers, so right in fact, that we feel they should hold for all types of numbers. All the usual rules of arithmetic, such as the commutative rule \(x \times y = y \times x\) and the distributive rule \(x\left(y+z\right) = xy + xz\), also feel so natural and right to us that we like to believe they hold for all types of numbers too.


So let’s play the game of accepting the above four beliefs as working for all numbers, as well as the usual rules of arithmetic as working for all types of numbers. The game now is to find the logical consequences as accepting these beliefs.


EXAMPLE: Show that \(\dfrac{7}{3} \times 3\) equals \(7\) as a logical consequence of our beliefs.



\(\dfrac{7}{3} \times 3 = 3 \times \dfrac{7}{3}\)     by the usual commutative rule for arithmetic

\(= \dfrac{3 \times 7}{3}\)    by Belief 3

\(= \dfrac{3 \times 7}{3 \times 1}\)    by usual beliefs about arithmetic

\(= \dfrac{7}{1}\)         by belief 4

\( = 7\)               by belief 2.


In general, we prove this way that

LOGICAL CONSEQUENCE 5: \(\dfrac{a}{b} \times b = a\) .


Exercise: Write out the details of the proof the more abstract statement \(\dfrac{a}{b} \times b = a\).


Challenge Question:  Show that Belief 1, \(\dfrac{a}{a}=1\), actually follows as a logical consequence of Beliefs 4 and 2. (This means we don’t actually have to list it as a separate Belief!)




 Here’s an example of putting the Key Fraction Property to use.


Question: Which is larger: \(\dfrac{5}{9}\)  or \(\dfrac{6}{11}\)?


DON’T CROSS MULTIPLY!  Cross multiplication is an artifice that takes us out of this fraction story.

Look for a common denominator instead and keep ownership of the fraction thinking we’re doing.

Let’s compare \(\dfrac{5 \times 11}{9 \times 11}= \dfrac{55}{99}\)  and  \(\dfrac{6 \times 9}{11 \times 9} = \dfrac{54}{99}\) to make it very clear what is going on.


Question: Arrange the fractions  \(\dfrac{5}{9}\)  and \(\dfrac{6}{11}\) and \(\dfrac{15}{28}\) from smallest to largest.


Question: Which is larger: \(\dfrac{1}{11}\)  or \(\dfrac{1}{12}\)? (Actually think your way through this!)


Question: Find ninety-nine fractions that lie between \(\dfrac{1}{11}\) and \(\dfrac{1}{12}\).


Actually, let me give away the answer to this one. Here are 99 fractions between \(\dfrac{1}{11}\)  and \(\dfrac{1}{12}\):


\(\dfrac{1}{11.01}\), \(\dfrac{1}{11.02}\), \(\dfrac{1}{11.03}\), …, \(\dfrac{1}{11.99}\).


If you don’t like decimals within fractions, then use the Key Fraction Property to fix things up.


\(\dfrac{1}{11.01} = \dfrac{1 \times 100}{11.01 \times 100}= \dfrac{100}{1101}\), etc.


COMMENT: I personally don’t like dividing decimals. To me, \( 2.45 \div 0.2\), for example, is


\(\dfrac{2.45}{0.2} = \dfrac{2.45 \times 100}{0.2 \times 100} = \dfrac{245}{20}\).


Now I can see this is \(12\frac{5}{20} = 12.25\).


That Key Fraction Belief is mighty handy!


Question: a) Show that \(\dfrac{266\cdots6}{66\cdots65}\) with the same numbers of 6s in the number in the numerator as the number in the denominator, is sure to equal \(\dfrac{2}{5}\).

b)   Show that \(\dfrac{166\cdots6}{66\cdots64}=\dfrac{1}{4}\) and \(\dfrac{499\cdots9}{99\cdots98}=\dfrac{4}{8}\) and \(\dfrac{199\cdots9}{99\cdots95}=\dfrac{1}{5}\).

(Credit to Swetha Dravida for discovering these examples as a high-school student.)


Optional Very Final Comment: Between any two fractions one can find a billion fractions. And between any of those two a trillion more. And so on. The fractions densely fill up the number line. (Yet they take up zero space on the number line! See  March 2014 Cool Math Essay .)

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