Proving an inequality involving absolute value; how do I justify using a conjunction (and) instead of a disjunction (or)?

Question

I'm putting together the following the proof, and I have a question about one of the final steps.

Definition of absolute value:

$\forall x \in \mathbb{R}, (x \geq 0 \Rightarrow |x| = x) \wedge (x < 0 \Rightarrow |x| = -x)$

We want to prove:

$\forall a, r, x \in \mathbb{R}, |x - a| < r \Rightarrow (a - r < x < a + r)$

Let $a, r, x, \in \mathbb{R}$. We assume $|x - a| < r$. We want to prove that $a - r < x < a +r$.

To do so, we will divide our proof into two cases.

Case 1: $x - a \geq 0$.

Then, by the definition of absolute value, $|x - a| = x - a$. So, by our assumption, $x - a < r \Longleftrightarrow x < a + r$.

Case 2: $x - a < 0$.

Then, by the definition of absolute value, $|x - a| = -x + a$. So, by our assumption, $-x + a < r \Longleftrightarrow a - r < x$.

Now, if I were to conjoin my result from case 2 with the result from case 1, I would obtain the desired statement: $(a - r < x) \wedge (x < a + r) \Longleftrightarrow a - r < x < a + r$.

My question is: how is the conjunction justified? How do I know it's okay to conjoin (use an "and" statement), rather than disjoin (use an "or" statement)?

Is it always the case that when we divide proofs into cases, the cases form a conjunction?

Thank you in advance!

Answer 1

With the help of the comments here, and one of my friends, I've finally worked out a proof that works for me.

Definition of absolute value:

$\forall x \in \mathbb{R}, (x \geq 0 \Rightarrow |x| = x) \wedge (x < 0 \Rightarrow |x| = -x)$

We want to prove:

$\forall a, r, x \in \mathbb{R}, |x - a| < r \Rightarrow (a - r < x < a + r)$

Let $a, r, x, \in \mathbb{R}$. We assume $|x - a| < r$.

What is left to prove is $a - r < x < a +r$.

In order to do so, we will divide our proof into two cases.

---

Case 1: $x - a \geq 0$.

Then, by the definition of absolute value, $|x - a| = x - a$.

So, by our assumption, $x - a < r \Longleftrightarrow x < a + r$.

Since $0 \leq x - a < r$, we know that $r > 0$.

So, $0 \leq x- a \Longleftrightarrow a \leq x \Longleftrightarrow a -r < x$.

By the preceding, we can conclude $ a - r < x < a + r$.

---

Case 2: $x - a < 0$.

Then, by the definition of absolute value, $|x - a| = -x + a$.

So, by our assumption, $-x + a < r$. And if we add $x - r $ to both sides, we obtain $ a - r < x$.

Since $a - r < x$, we can add $-a$ to both sides to obtain $-r < x - a$, and since we know $x - a < 0$, we obtain $-r < x - a$ < 0. Adding $r$ to all sides produces $0 < x - a + r < r$, so we can conclude $r > 0$.

We know $x - a < 0$, and if we add $a$ to both sides, we obtain $x < a$. We showed $r > 0$, so we can add $r$ to the right side without affecting the inequality, to obtain $x < a + r$.

Since we've shown $a - r < x$ and $x < a + r$, we can conclude $a - r < x < a + r$.

$\blacksquare$

Answer 2

Use the obvious fact: |z| < r implies -r < |z|.

Case 1. -r < x - a < r.
Desired conclusion follows in one step.

Case 2. -r < a - x < r.
Desired conclusion follows in two steps.