Inverse Operations

I said a little about operations that are inverses to each other when I talked about square roots. Operations (or functions on numbers – I may use operations and functions interchangeably) are inverses of each other if those operations, when performed sequentially on a number, result in the original number.

For example, consider the operations of adding, then subtracting 4 to/from a number. Let’s use 5:

5 + 4 = 9, 9 – 4 = 5, the original number. More directly,

5 + 4 – 4 = 5 + 0 = 5

So no surprise, addition and subtraction are inverse operations. It doesn’t matter what the starting number is or the number you are adding or subtracting. In general, if you start with a number x, + y and – y are inverse operations:

x + yy = x

The order of these operations does not matter – I can first subtract 4 then add 4. Multiplication and division are also inverse operations. For example,

5 × 2 = 10, 10 ÷ 2 = 5 or 5 × 2 ÷ 2 = 5 × 1 = 5

Other less familiar inverse operations are squaring a number and taking a number’s square root. For example, I can square 4 then take it square root or take its square root the square the result and I will get 4 back in either direction:

\[
\begin{array}{l}
{\sqrt{{4}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{16}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\
{{\left({\sqrt{4}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{4}}
\end{array}
\]

In general,

\[
\begin{array}{l}
{\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}}\\
{{\left({\sqrt{x}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{x}}
\end{array}
\]

There are other ones like this as well: cube and cube roots, raise to the fourth power and fourth roots, etc.

Another less well known one, and one that confuses many of my students, is exponential and logarithm operations.

When x is the base and a number, say 2, is in the exponent, as in x², this is a power operation. However, when x is the exponent, and a number called e is in the base, this is called an exponential operation, .

I won’t spend much time on e, but it is an irrational number like 𝜋 and there are several reasons why it is the ‘base of choice’. So if the operation is to take a number and make it the power of e, you will generate a number. Many calculators have an ‘.’ key where you can do this. So for example, e² = 7.389 … . What inverse operation can I perform on this number to get the ‘2’ back? This operation is called the log operation.

Now logs can use any base but again, the base e is most commonly used in math. The notation for taking the log of a number is logₑx which is often shortened to ln x. The shorthand ‘ln’ stands for ‘natural log’ as e is a natural base to use in math (topic for another post). If you have an ‘ ‘ key on your calculator, you probably also have an ‘logₑx‘ or ‘ln x‘ key as well.

So what does ‘ln x‘ mean? This operation is asking the mathematical question: “What number do I need to raise e to, so I get x?”. Can you see how this is the opposite of ‘eˣ ‘ where I actually raise e to the x power? So, ln() is asking the maths question: “What power do I need to raise e to, so that I get ?”. Isn’t it obvious that x is the number I have to raise e to, to get ? So and ln x are inverse operations. If you take the number we generated before, 7.389 … , and hit the ln x key on your calculator, you will get the original ‘2’ back.

There are many other inverse functions. In trigonometry, the sin x key in your calculator will give the sine of the angle x. Sometimes 𝜃 is used instead of x. The inverse sine, sometimes called the arcsine, is the inverse of this. The notation you may see on your calculator for this operation is ‘ASIN x‘ or ‘SIN ̅¹x‘. SIN ̅¹x is asking the maths question: ” What angle has x as its sine?”. So SIN ̅¹[SIN(x)] is x.

There are inverse functions associated with the other trig functions as well. You will usually see inverse function keys as the same key on a calculator, you just need to hit another key first to activate the inverse definition for the key.

Square roots

Now eventually, I would like to be able to solve equations like \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}
\]. But you see that if I am to get x by itself, I need something to get rid of the exponent “2”. This is where square roots come in. But let’s first talk about inverse operations.

You may have already noticed that addition and subtraction are inverse operations because if you add a number to something and then subtract it or vice versa, you are left with the original something:

x + 3 – 3 = x,       x – 3 + 3 = x

The same can be said of multiplication and division:

\[
{3}{x}\hspace{0.33em}\div\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}x}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

The square root of a number is the inverse of squaring a number. Where squaring a number means “what do I get when I multiply a number by itself”, the square root of a number means “what number multiplied by itself is equal to the original number”. The math symbol which means “take the square root of this number” is called a radical and looks like\[
\sqrt{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}
\]. You will see later that there are other ways to indicate a square root. Now a few examples:

\[
\begin{array}{l}
{\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}}\\
{\sqrt{4}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\
{\sqrt{100}\hspace{0.33em}{=}\hspace{0.33em}{10}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{10}^{2}\hspace{0.33em}{=}\hspace{0.33em}{100}}
\end{array}
\]

 

So let’s illustrate the “inverseness” of this:

\[
\begin{array}{l}
{\sqrt{{5}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{5}}\\
{\sqrt{{2}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{2}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}}\\
{\sqrt{{10}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{10}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{10}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{10}}
\end{array}
\]

 

Are you starting to see the potential of this to get rid of the “2” in \[
{x}^{2}
\]?

Now I’ve only been half truthful so far. For the square root examples I’ve shown you, I only showed you one of the possible solutions. Is 5 the only square root of 25? Think about it. Is there not another number that when multiplied by itself is 25? What about -5?  When you take the square root of a number, you actually are introducing two solutions:

\[
\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}\pm\hspace{0.33em}{5}
\]

The symbol ± means “plus or minus” and indicates that the two solutions are +5 and -5.

Let’s use this new tool to solve an equation on my next post.