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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!

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    $\begingroup$ No, when you divide a proof into cases, the cases do not form a conjunction. Here's an easy counterexample: Consider any real number $x$. Case 1: $x\ge 0$. Case 2: $x<0$. Therefore, $x\ge0\wedge x<0$? $\endgroup$ – Rahul Apr 20 at 14:41
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    $\begingroup$ Instead, what you have to do is prove the same thing in both cases. Then you know the thing is true no matter what case you're in. In this problem, you have to prove that $a-x<x<a+r$ in both cases. $\endgroup$ – Rahul Apr 20 at 14:42
  • $\begingroup$ @Rahul What you said makes perfect sense. My new question is, can I derive the conclusion from the second implication from the definition as a vacuous truth? $\endgroup$ – Calculemus Apr 20 at 16:15
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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.

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  • $\begingroup$ How does one prove the obvious fact though? First thing that came to mind was multiplying both sides of the antecedent by $-1$, but that gives $-r < -|z|$, which isn't the consequent. $\endgroup$ – Calculemus Apr 21 at 17:58
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    $\begingroup$ @Calculemus. r is positive. $\endgroup$ – William Elliot Apr 22 at 3:30
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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 $a - r < x < a + r$.

$\blacksquare$

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