4.9 The modulus function

When graphing a value inside a modulus function, it is often helpful to first sketch the function without the modulus. For example, when graphing $y=\left\lvert-\left(x-5\right)^{2}+1\right\rvert$, the graph of $y=2x-8$ would look like this:

As the modulus function must never be unhappy (i.e. take a negative value), we need to turn that frown upside down!¹⁵¹⁵ 15 Ok, I’ll stop, I promise. Using the quadratic formula to find the roots, we get that they are at $x=4$ and $x=6$. Therefore, everything to the left of $4$ and to the right of $5$ needs to be reflected it the $x$-axis.

One way to define the modulus function is, well, as a function. As the function always needs to be positive, we can write that

What this means is that in the case where $x$ (the input to the modulus function) is bigger than zero, we just return $x$. If $x$ is smaller than zero, we multiply it by $-1$ in order to make it positive.

To write the modulus function as an algebraic expression, we’re interested in functions which are always positive. The one which (probably?) springs to mind is squaring. Using this, we can define the modulus function as ¹⁶¹⁶ 16 Note that this is just the square root of the magnitude. It is also important to never give into the temptation to say ”oh this is just the square root of a square number, it’s just the original number.” This is wrong. In the expression $\sqrt{x^{2}}$ when we square $x$, both $x$ and $-x$ are mapped to the same value, so stating that the square root of the square is just the original value is not true for negative numbers.

This is really helpful for solving some equations.

Squaring both sides gives the following result

which can then be expanded to give

and then, being careful not to subtract anything incorrectly (which I always do) the equation can be reduced to this quadratic

We can then sketch this to work out when it would be greater than zero (or zero would be less than the curve).

From the graph, the inequality is true whenever $x$ is less than the first time it is zero, and whenever the inequality is greater than the second time it is zero. To find when $3x^{2}-20x+32=0$, we can use the quadratic formula, and thus obtain an equation for the values of $x$.

Subtracting a positive number from a positive number (note that $\sqrt{20^{2}-4(3)(32)}$ is a positive number) gives a smaller value than adding something to a positive number. Therefore, we can deduce (with aid from a calculator) that the smaller value of $x$ is $\frac{3}{8}$ and the larger value of $x$ is $4$, leaving us with two regions in which the inequality is satisfied.

This is just a quadratic ¹⁹¹⁹ 19 Review the section on hidden quadratics if you’re not sure what this is. TODO: write this section, and we can substitute $z=\sqrt{(x-1)^{2}}$ to get that

We can sketch this graph to work out where the inequality is true ²⁰²⁰ 20 The other way to do this is to think about where one of the brackets is positive, and the other negative, but in my experience it’s a more error-prone method.

From the graph, the inequality is true whenever $2<z<8$ (i.e. $z$ is between the roots).

Thus we have $2<\sqrt{(x-1)^{2}}<8$. This means that both of these inequalities are true:

If we square both sides, we get that

Here’s a time to be careful; $(x-1)^{2}$ is always positive. When we take the square root, however, we can have either the positive or negative square root. Note that the negative square root is just $-1$ times $\sqrt{\text{whatever}}$, and thus we need to flip the inequality.

Overall then we have these four inequalities.

Which can all be arranged to obtain that the following four inequalities must all be true for the inequality to hold

Examining these, it is clear that there are two regions in which this is true: $-7<x<-1$ and $3<x<9$. This is the complete set of regions which satisfy the inequality. We can write it using set theory notation as

where $\cup$ denotes the "union" operator (which means that overall we have the set of objects in either $\left\{-7<x<-1\right\}$, $\left\{3<x<9\right\}$, or in both of them - which in this case is nothing, as the two ranges don’t overlap).

Instead, we can think about the definition of the modulus function. For each side of the equation we have

Thinking about the cases, the graphs can intersect either

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  Where the argument to the modulus function is less than zero, and thus has been multiplied by negative one, for both functions.

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  Where the argument of one of the functions is less than zero, and the other is greater than zero.

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  Where the argument of both functions is greater than zero.

In the first case, the values of $t$ where the two curves intersect would be when

this is just the same as

Which is the same equation as for case three.

The only other case is when

Note that it doesn’t matter which one is less than zero (and has thus been multiplied by $-1$), as we can just multiply both sides by $-1$ to get from one to the other.

Thus solving Equation 4.101, we have

And solving Equation 4.102 we have

This is not the most exciting proof, but it helps to illustrate a general principle when proving things about the modulus function: consider the positive and negative cases separately.

We would like to show that $a$ is less than or equal to its modulus. Here we can find two exhaustive cases by the law of the excluded middle (see Section 23.4 for details, but it just means that \sayeither something is true, or it is false), that is that either $a\geqq 0$ or $a<0$. If these cases suspiciously mirror the definition²¹²¹ 21 See Section 4.9.2 if you can’t remember the definition. of the modulus function, that’s because they do.

Why are these cases useful? Because they allow us to assume more about the structure of the modulus function. This comes at a cost - that we have lost generality (that is, we now know more about $a$ but only in specific cases, so we must proceed case-by-case to prove this in general).

We can begin with the first case - the one where $a\geqq 0$. First, we can consider what we know about $|a|$ in this case, which is that $|a|=a$. This is enough to prove the inequality as

We must then consider the other case, i.e. when $a<0$ in this case $|a|=-a$, that is $|a|>0$, therefore in this case $a<0<|a|$ which implies that $a<|a|$ and thus that $a\leqq|a|$