## Modulus Rules

We start our journey through the mathematical world of analysis by studying the modulus function, defined as follows:

$mod:{\mathbb R} \rightarrow {\mathbb R}, x \rightarrow |x|$.

$|x| = x \text{ if } x \geq 0$.

$|x| = -x \text{ if } x < 0$.

The modulus function is critical in real analysis, and can be thought of as giving the ‘distance’ between x and 0.  The graph of the modulus is given below:

You are probably already familiar with the modulus function, and as a result, many of the following results will probably come as little surprise to you.  However, the point of going through such results is not just to display facts to be learnt, but to demonstrate how mathematical knowledge builds upon itself in a rigorous and logical manner, with simple results being used to prove progressively less simple ones.  For if we cannot be sure of the foundations upon which mathematical analysis is built, then how can we be sure of the intricate and marvellous results that follow from them?

We start with a result so simple that you may consider it utterly trivial:

### $|x| \geq 0 \text{ for all real x, with } |x| = 0 \Leftrightarrow x = 0$.

You may be thinking that this result is surely completely obvious from the definition and the graph above, and indeed it is obvious!  But this is not an excuse for laziness.  Fortunately, since the result follows so easily from the definition, constructing a formal proof is extremely simple.

### Proof

Let x be an arbitrary real.

If x > 0 then |x| = x > 0.

If x < 0 then |x| = -x > 0.

If x = 0 then |x| = x = 0.

This exhausts all the possibilities.

$\Box$.

Now we move on to a result which, while not particularly inspiring or profound in itself, will nevertheless be extremely useful in proving far more interesting results.

### Theorem 1.2

$-|x| \leq x \leq |x| \text{ for all real x }$.

### Proof

Let x be an arbitrary real.  Then -|x| ≤ 0 ≤ |x| by Theorem 1.1

If x ≥ 0 then x = |x| $\implies$ -|x| ≤ 0 ≤ x = |x|

$\implies$ -|x| ≤ x ≤ |x|

If x < 0 then x = -(-x) = -|x| $\implies$ -|x| = x ≤ 0 ≤ x

$\implies$ -|x| ≤ x ≤ |x|

This exhausts all the possibilities.

$\Box$.

Having thus proved that any real number x is bounded between its modulus and negative modulus, we are now in a position to prove a result which makes the entirety of real analysis possible.

### Theorem 1.3

$\text{Let } \varepsilon > 0 \text{. Then, for any real x }$$|x| < \varepsilon \iff -\varepsilon < x < \varepsilon$.

Before you proceed to the proof, make sure that you first fully understand what the theorem is saying.  Only then can you appreciate the importance of what is being proved, which is that whenever a real number, x, is trapped between any other real number and its negative, we can express this double inequality as a single inequality involving the modulus (and vice versa).

Such inequalities are the foundation of real analysis, and are how the modulus function comes to be used in the subject – hence why it is extremely important that we rigorously understand how it works.

### Proof

Fix ε > 0 and let x be an arbitrary real such that |x| < ε

$\implies$ -|x| ≤ x ≤ |x| < ε                  by Theorem 1.2

$\implies$ -ε < -|x| ≤ x ≤ |x| < ε           since |x| < ε $\implies$ -|x| > -ε

$\implies$ – ε < x < ε

Now, to prove the other direction, let -ε < x < ε

$\implies$ ε > -x > -ε                     (multiplying everything through by -1)

$\implies$ -ε < ±x < ε

$\implies$ |x| < ε                                 since |x| = x or –x

$\Box$.

Now we move on to an incredibly useful result, which though thankfully easy (if a little tedious) to prove, is not as trivial as it may appear on first glance.  We will often need it, so it is important to know that it always holds!

### Theorem 1.4

$|xy| = |x|.|y| \text{ for all real x and y }$.

### Proof

Let x and y be arbitrary reals.

If x = 0 or y = 0, then |xy| = 0 = |x|.|y|

If x > 0 and y > 0, then xy > 0.  Thus, |xy| = xy = |x|.|y|

If x < 0 and y < 0, then xy > 0.  Thus |xy| = xy = (-|x|)(-|y|) = |x|.|y|

If  x and y have different signs then, without loss of generality, assume x > 0 and y < 0.  (If not, relabel x and y.)   Then xy < 0.  Thus |xy| = -(xy) = (x)(-y) = |x|.|y|

This exhausts all the possibilities

$\Box$.

With the above result proved, we are in a position to tackle some easy corollaries – easy results which follow almost instantly from a more complicated result.

### Corollary 1.5

$|-x| = |x| \text{ for all real x }$.

### Proof

Let x be an arbitrary real.  Then,

$|-x| = |(-1)x| = |-1|.|x| = |x|$.

$\Box$.

### Corollary 1.6

$|x|^2 = x^2$.

### Proof

Let x be arbitrary.

$|x|^2 = |x|.|x| = |x.x| = |x^2| = x^2 \text{, since } x^2 \geq 0$.

﻿                                                                                                                                                    $\Box$.

Now, at last, we are in a position to prove a theorem so fundamental and so vital for analysis, that it even has its own name!

### $|x+y| \leq |x| + |y| \text{ for all real x and y }$.

Given that the modulus can be thought of as giving the ‘distance’ of the number from the origin, the truth of this theorem is very easy to demonstrate geometrically, if nevertheless requiring a little thought to prove in practice.  It is easy to see that if x and y both have the same sign (ie both positive or both negative) then both sides will be equal.  If not however, then since both then move in ‘opposite’ directions on the number line, it is surely trivial to see that this must be less than the sum of the two moduli.

Such appeals to geometry however, while useful for seeing what is going on, do not constitute a proof.  Pictures can be misleading, as can intuition.  We require clear proof, based on valid mathematical logic.  If you haven’t already been doing so, it is strongly recommended that you try to prove this result yourself before reading below.  Indeed, it is recommended that you try this before reading any proof.  Not only will you develop your own skills of mathematical reasoning by engaging your brain and not simply being a passive reader, but there is no better way to remember a proof than to have first constructed it independently yourself.

### Proof

For contradiction, assume that the theorem is not true.  Then, there exist real numbers x and y such that |x + y| > |x| + |y|

$\implies$ |x + y|2 > (|x| + |y|)2 since the squaring of non-negative terms  preserves inequalities.

$\implies$ (x + y)2 > x2 + y2 + 2|x|.|y|             by corollary 1.6

$\implies$ x2 + y2 + 2xy > x2 + y2 + 2|x|.|y|

$\implies$ xy > |x|.|y|

$\implies$ xy > |xy|

But Theorem 1.2 says that xy ≤ |xy|, so we have a contradiction.  The theorem must be true, and the result that was previously derided as being uninspiring turns out to be vital in proving it!

$\Box$.

Notice that each step in the proof above is reversible, so if we wanted, we could start with the fact that xy ≤ |xy| for all real x and y and deduce the result from there.  Either way is perfectly valid.

Our work on the modulus function is now complete, and we have at our disposal all the weapons that we need to tackle the analytical world of limits, functions, sequences and much more…