A LONG time ago, I wrote the post titled “Index Laws (Part I)“, where I developed a conceptual basis for the index law for multiplication, namely:
\(a^m \times a^n = a^{m+n}\tag{Index Law I}\)Why write on this topic? I wrote a description in the linked post above describing my motivation in-depth, however, put simply, I feel that often these index laws are taught as magical rules where students don’t truly understand WHY they work. This can often lead to students struggling with simplifying more complex algebraic expressions, as they struggle to see how the rote-learned index laws fit with other methods learnt through school.
In this post, we’ll explore the index law for division, which we’ll refer to as Index Law II:
$$a^m \div a^n = a^{m-n}\tag{Index Law II}$$
Developing a Conceputal Understanding
Different forms but the same law.
Firstly, it important to recall that in math, division is just fractions in disguise. We already know that:
$$a \div b = \frac{a}{b}$$
Because of this, we can always rewrite Index Law II to have the appearance of fractions:
$$\frac{a^m}{a^n} = a^{m-n}\tag{This is still Index Law II}$$
This can be a potential point of confusion for students! Just because the index law above looks different to the one written earlier, they are actually describing the exact same thing. It’s just that one has been written using a division symbol, whereas the other has been written with fractions (but the division and fraction notation is interchangeable here). Just to really drive this point home, both of the following are explaining the exact same thing:
$$a^m \div a^n = a^{m-n}$$
$$\frac{a^m}{a^n} = a^{m-n}$$
Back to basics – simplifying fractions.
Lets go back to when we first learnt fractions and division with a very simple example. Consider a question asking you to simplify:
$$\frac{44}{8}$$
(Note: We could also equivalently demonstrate this point by solving \(44 \div 8\), but the explanation is extremely similar, and I believe the fraction notation is easier to conceptualise.)
How would you go about this? You start by trying to take out the highest factor that is common to both the numerator and the denominator (i.e. you extract the common factor).
Looking at the example, we see that the number \(4\) goes into \(44\), \(11\)-times. Similarly, the number \(4\) goes into \(8\), \(2\)-times.
Hence, our fraction becomes:
$$\frac{44}{8} = \frac{4 \times 11}{4 \times 2}$$
The \(4\)‘s cancel out, leaving us with our simplified equation:
$$\frac{\mathbf{4} \times 11}{\mathbf{4} \times 2} = \frac{\mathbf{4}}{\mathbf{4}} \times \frac{11}{2} = \frac{11}{2}$$
Making sense of Index Law II.
The reason that we recapped the mechanics behind that simple example is that this is exactly how Index Law II operates. This is best seen through the lens of a very simple example, so let’s consider solving:
$$\frac{2^5}{2^2}$$
Now, you could use a calculator quite easily to solve this, but later we’ll be using pronumerals (i.e. \(x\)‘s) which will make things much harder (you won’t be able to use your trusty calculator). So what’s the answer to this question? Straight away, we should be able to use index law II to realise that the answer is \(2^3 = 8\), because:
$$\frac{2^5}{2^2} = 2^{5-2} = 2^3 = 8$$
But, understanding why this is the case is important, as otherwise it will be easy to get lost when we get to harder questions where there are heaps of different scary algebraic things taking place.
Let’s start by expanding the powers in our original fraction:
$$\frac{2^5}{2^2} = \frac{2 \times 2 \times 2 \times 2 \times 2}{2 \times 2}$$
Similar to the example we did earlier, if we want to simplify/solve the fraction, we want to take out the highest factor that is common to both the numerator and denominator. In this case, we see that both the numerator and denominator have the factor \(2^2 = 2 \times 2\) in common, as seen below:
$$\frac{2^5}{2^2} = \frac{\mathbf{2 \times 2} \times 2 \times 2 \times 2}{\mathbf{2 \times 2}}$$
We can also see clearly that this is the highest power that does into the numerator and denominator, as after all, we’ve run out of things in the denominator. With this highest factor isolated, we cancel it from the fraction:
$$\frac{2^5}{2^2} = \frac{\mathbf{2 \times 2} \times 2 \times 2 \times 2}{\mathbf{2 \times 2}} = \frac{\mathbf{2 \times 2}}{\mathbf{2 \times 2}} \times \frac{2 \times 2 \times 2}{1} = \frac{2 \times 2 \times 2}{1}$$
In the third part of the equation above, students may be confused as to why we write \(\frac{2 \times 2}{2 \times 2} \times \frac{2 \times 2 \times 2}{1}\), rather than \(\frac{2 \times 2}{2 \times 2} \times \frac{2 \times 2 \times 2}{0}\). Firstly, you can never divide by \(0\), so we would be breaking math straight away. Secondly, this is to preserve the original fraction that we had, and comes back to how you can split fractions using multiplication. The denominator in the original question is \(2^2 = 2 \times 2\). We need this denominator to stay consistent throughout, and so if we were to change it to a \(0\), we now have \(2 \times 2 \times 0 = 0\), which is obviously different to the denominator of our original fraction.
Anyway, back to solving the equation. By realising that dividing any value by \(1\) is equivalent to the original value, we have that:
$$\frac{2 \times 2 \times 2}{1} = 2 \times 2 \times 2 = 2^3$$
With this, we’ve shown conceptually how Index Law II actually works using a simple example. Now, you could have just used the simple Index Law II formula presented earlier, but now you know exactly how and why it works.
Another example.
To take this a step further, let’s consider another example, but this time they question will feature pronumerals only (i.e. \(x\)‘s). Consider you were asked to simplify:
$$x^7 \div x^3$$
Well, let’s start by rewriting this as a fraction to make it easier to visualise:
$$\frac{x^7}{x^3}$$
Now, rather than expanding the powers (which will take a lot longer now that the numerator is \(x^7\)), let’s think about how many \(x\)‘s we have on the numerator and the denominator.
In the numerator, we have \(x^7\), meaning that if we were to expand this, we would have \(7\) \(x\)‘s being multiplied together. However, in the denominator we have \(x^3\), meaning that we have only \(3\) \(x\)‘s being multiplied together.
This means that all-together, we can cancel \(3\) \(x\)‘s from the numerator and denominator, which means we are extracting out a factor of \(x^3\):
$$\frac{x^7}{x^3} = \frac{\mathbf{x^3}}{\mathbf{x^3}} \times \frac{x^4}{1} = x^4$$
Our simplified result is \(x^4\). Now we could also use the Index Law II formula here directly, which we’ll show below to verify our answer:
$$x^7 \div x^3 = x^{7-3} = x^4$$
Harder Questions
Question 1.
Simplify the following expression:
$$\frac{a^4b^2}{ab^2}$$
Now, in this question, I feel that using the shorthand Index Law II formula can be more complicated! It is much easier to solve this by considering our conceptual understanding, and how we are looking for the highest factor of the numerator and denominator.
We start by looking at the power of the pronumeral \(a\) in the numerator and denominator. In the numerator we see that the power of \(a\) is \(4\) (\(a^4\)). In the denominator, we see that we just have an \(a\). This just means that we have \(a\) to the power of \(1\) (\(a^1\)). Overall, this means that the highest factor of \(a\) that can be extracted is simply \(a^1 = a\).
Similarly, now we will look at the power of \(b\) in the numerator and denominator. We see that in the numerator and the denominator, we have \(b^2\). Hence, the highest factor of \(b\) is \(b^2\).
Hence, overall, the highest factor of the numerator and denominator is \(ab^2\) (Note, this could also be written as \(b^2a\) as order does not matter for multiplication). Therefore, our expression simplifies to:
$$\frac{a^4b^2}{ab^2} = \frac{\mathbf{ab^2}}{\mathbf{ab^2}} \times \frac{a^3}{1} = a^3$$
We cannot simplify this any further, so we’re done 🙂
Question 2.
Simplify the following expression:
$$\frac{y^2}{y^3}$$
When considering the highest common factor of the numerator and denominator, we see that it is \(y^2\). Cancelling out \(y^2\) gives us:
$$\frac{y^2}{y^3} = \frac{\mathbf{y^2}}{\mathbf{y^2}} \times \frac{1}{y} = \frac{1}{y}$$
Hence, our final simplified answer is \(\frac{1}{y}\).
Now, something interesting happen if instead we were to use the Index Law II formula taught earlier. Let’s try it here:
$$\frac{y^2}{y^3} = y^{2-3} = y^{-1}$$
What?! How can we get \(\frac{1}{y}\) when doing it one way, and then \(y^{-1}\) when doing it another way? Well, it actually turns out that these solutions are different ways of writing the same thing, and we’ll learn more about that next time.
