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Section 19.1 Multiply Straight Across

Many students have learned that to multiply fractions, you must "multiply straight across." Not as many students have stopped to ask the question of why that's the right way to multiply fractions. In this section, we're going to explore this idea more deeply to bring some intuition to that manipulation.

We will start by looking at a simple diagram to represent multiplication with integers. One concept we use for multiplication is by having multiple groups with the same number of elements.

The same idea applies to fractions using the wedges from looking at parts of a whole.

If we think about the structure of a fraction, the numerator represents the number of pieces that we have. And so it would make sense that if we have groups of wedges we would be multiplying the numerator by that quantity.

So we have a clear picture of what it looks like to have multiple groupings of fractional quantities. But what about a fractional grouping? We will once again work from diagrams.

It shouldn't be a surprise that we ended up with the same result since multiplication is commutative. We can even visually see that the two pictures of \(\frac{9}{4}\) have the same number of shaded wedges (just in a different arrangement). But this is an important step because it then gives us access to understand multiplying a fraction by a fraction.

The last set of images gives us the big insight that we need to understand fraction multiplication. We went from \(\frac{2}{3}\) of 3 wedges of size \(\frac{1}{4}\) to 6 wedges of size \(\frac{1}{12}\text{.}\) Let's take a look at each of these parts separately.

  • 6 wedges: We started with 3 wedges, and each of these turned into 2 (smaller) wedges. So we have that the 6 comes from 3 groups of 2 wedges.

  • Wedges of size \(\frac{1}{12}\text{:}\) The starting wedge size was \(\frac{1}{4}\text{,}\) which corresponds to cutting the whole into 4 equal pieces. Each of these pieces was cut 3 into pieces, which gives a total of 12 pieces for the whole. This is why that the wedges are size \(\frac{1}{12}\text{.}\)

This observation shows us that fraction multiplication is essentially a two-step process. We multiply the numerators together to get an accurate count of the number of wedges, and we multiply the denominators together to get the correct wedge size.

Activity 19.1.1. Fraction Multiplication Using a Diagram.

Now that we have the conceptual framework in place, we can understand the why the phrase "multiply straight across" for fraction multiplication actually works. For a fraction, the numerator represents the number of wedges and the denominator represents the size of the wedges. When we multiply the numerators, we're counting for the total number of wedges (a number of groups of a certain size). When we multiply the denominators, we're establishing the size of the wedges (based on parts of a part of a whole).

Try it!

Draw a diagram to represent the product \(\frac{1}{2} \cdot \frac{3}{4}\text{.}\)

Solution.

The practice of multiplying fractions looks nothing like the concept. This is not an uncommon situation in life. Most adults know how to drive a car, but very few of them can actually explain the processes involved between pressing the gas pedal and the car moving. In this analogy, we've just finished describing how the engine works, and now we're going to just focus on driving the car.

The "rule" of multiplying straight across is expressed using algebraic symbols in the following manner.

The idea of multiplying straight across applies regardless of whether we're working with numerical fractions or fractions involving variables.

Activity 19.1.2. Fraction Multiplication.

Try it!

Calculate \(\frac{2}{5} \cdot \frac{6}{7}\) and \(\frac{3x}{4} \cdot \frac{7x}{2}\text{.}\)

Solution.

Activity 19.1.3. Reducing Before Multiplying Fractions.

In some situations, you will find that multiplying straight across will result in very large numbers.

\begin{equation*} \frac{32}{21} \cdot \frac{35}{24} = \frac{1120}{504} \end{equation*}

It turns out that by using a little foresight, large fractions can sometimes be avoided by reducing before multiplying. The idea is the same as reducing a fraction. We are going to look for factors that appear in the product for both the numerator and the denominator. This requires you to identify common factors between the terms in the numerator and the terms in the denominator.

\begin{equation*} \begin{aligned} \frac{32}{21} \cdot \frac{35}{24} \amp = \frac{4 \cdot 8}{3 \cdot 7} \cdot \frac{5 \cdot 7}{3 \cdot 8} \amp \eqnspacer \amp \text{Identify common factors} \\ \amp = \frac{4 \cdot \cancel{8}}{3 \cdot \cancel{7}} \cdot \frac{5 \cdot \cancel{7}}{3 \cdot \cancel{8}} \amp \amp \text{Reduce} \\ \amp = \frac{4}{3} \cdot \frac{5}{3} \\ \amp = \frac{20}{9} \end{aligned} \end{equation*}

Try it!

Calculate \(\frac{12}{25} \cdot \frac{40}{9}\) by reducing before multiplying.

Solution.
\begin{equation*} \begin{aligned} \frac{12}{25} \cdot \frac{40}{9} \amp = \frac{4 \cdot 3}{5 \cdot 5} \cdot \frac{8 \cdot 5}{3 \cdot 3} \amp \eqnspacer \amp \text{Identify common factors} \\ \amp = \frac{4 \cdot \cancel{3}}{5 \cdot \cancel{5}} \cdot \frac{8 \cdot \cancel{5}}{3 \cdot \cancel{3}} \amp \amp \text{Reduce} \\ \amp = \frac{4}{5} \cdot \frac{8}{3} \\ \amp = \frac{32}{15} \end{aligned} \end{equation*}

Activity 19.1.4. Reducing Before Multiplying Fractions with Variables.

The exact same idea can be applied to fractions with variables.

\begin{equation*} \begin{aligned} \frac{4x}{9y^3} \cdot \frac{15y^2}{2x^4} \amp = \frac{2 \cdot 2x}{3y \cdot 3y^2} \cdot \frac{5 \cdot 3y^2}{x^3 \cdot 2x} \amp \eqnspacer \amp \text{Identify common factors} \\ \amp = \frac{2 \cdot \cancel{2x}}{3y \cdot \cancel{3y^2}} \cdot \frac{5 \cdot \cancel{3y^2}}{x^3 \cdot \cancel{2x}} \amp \amp \text{Reduce} \\ \amp = \frac{2}{3y} \cdot \frac{5}{x^3} \\ \amp = \frac{10}{3 x^3 y} \end{aligned} \end{equation*}

Try it!

Calculate \(\frac{6t}{5} \cdot \frac{8}{3t^3}\) by reducing before multiplying.

Solution.
\begin{equation*} \begin{aligned} \frac{6t}{5} \cdot \frac{8}{3t^3} \amp = \frac{2 \cdot 3t}{5} \cdot \frac{8}{t^2 \cdot 3t} \amp \eqnspacer \amp \text{Identify common factors} \\ \amp = \frac{2 \cdot \cancel{3t}}{5} \cdot \frac{8}{t^2 \cdot \cancel{3t}} \amp \amp \text{Reduce} \\ \amp = \frac{2}{5} \cdot \frac{8}{t^2} \\ \amp = \frac{16}{5t^2} \end{aligned} \end{equation*}