Before I begin with the main topic, let me review how fractions are multiplied together. If you remember, this is relatively easy as all you do is multiply the numerators together and the denominators together:

\[\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{6}{12}

\]

Now let’s look at this problem a different way. Now multiplication is commutative which means the order of the things multiplied together does not matter. So we could develop the problem thus:

\[\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}

\]

Now that is a long line of work but this shows two things. One, you can create the same equivalent fraction by multiplying the numerator and the denominator by the same number:

\[\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{6}{12}

\]

But you can also do the reverse – if there is a common factor in the numerator and the denominator, you can effectively cancel them, replacing them with an implied “1”:

\[\frac{6}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}

\]

So if you see this common factor, you can simply write:

\[\frac{6}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}{{4}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{4}

\]

But notice there is another common factor:

\[\frac{2}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{{1}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}{{2}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}

\]

So \[

\frac{2}{4}{,}\hspace{0.33em}\frac{6}{12}{,}\hspace{0.33em}{\mathrm{and}}\hspace{0.33em}\frac{1}{2}

\] are all equivalent fractions but \[\frac{1}{2}\] is the one with the smallest numbers. These “smallest numbers” are found using things called *prime numbers* and I will talk about these later.

With this knowledge, let’s go back to the original problem and get to the final answer of \[\frac{1}{2}\] in far fewer steps:

\[\frac{3}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}{\rlap{/}{2}\hspace{0.33em}\times\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}

\]

Remember that the cancelled factors are being replaced by “1”s. We can leave them out of the denominator since there is another number there but we must keep one of the “1”s in the numerator since there is no other number there.

We will use this skill of finding the factors of numbers in my next post. This will lead up to how to add fractions with different denominators together.