- #1

SithsNGiggles

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## Homework Statement

The following statement holds for all [itex]a,b,c,d \in \mathbb{Z}:[/itex]

[itex]a | 1\mbox{ iff } a=\pm 1[/itex]

Write a proof for this part of Theorem 2.3.6 (which just lists other properties I've written proofs for).

## Homework Equations

## The Attempt at a Solution

First I wrote a proof for the left-to-right conditional:

Assume [itex]a|1[/itex]. Then, by definition, [itex]\exists k\in\mathbb{Z}[/itex] such that [itex]ak = 1[/itex].

Suppose [itex]k>1[/itex]. Then [itex]ak = 1 \Rightarrow a = \frac{1}{k} \not\in\mathbb{Z}[/itex].

Suppose [itex]k=1[/itex]. Then [itex]ak = 1 \Rightarrow a = 1[/itex].

Suppose [itex]-1 < k < 1[/itex]. Then [itex]k = 0[/itex], but [itex]0a \not=1[/itex] for any [itex]a[/itex].

Suppose [itex]k=-1[/itex]. Then [itex]ak = 1 \Rightarrow a = -1[/itex].

Suppose [itex]k<-1[/itex]. Then [itex]ak = 1 \Rightarrow a = -\frac{1}{k} \not\in\mathbb{Z}[/itex].

The first, third, and fifth scenarios all lead to contradictions. In the remaining cases, [itex]a=\pm1[/itex].

(I had a feeling my approach here was a bit tedious, but if it's acceptable, I'm fine with it.)

Right-to-left:

Assume [itex]a=\pm1[/itex]. To show that [itex]a|1[/itex], suppose that [itex]ak=1[/itex] for some [itex]k[/itex]. If [itex]k \in\mathbb{Z}[/itex], then both [itex]k|1[/itex] and [itex]a|1[/itex], the latter of which I want to show.

Case 1: Let [itex]a=1[/itex]. Then we have [itex](1)k = 1 \Rightarrow k = 1 \in\mathbb{Z}[/itex].

Case 2: Let [itex]a=-1[/itex]. Then we have [itex](-1)k = 1 \Rightarrow k=-1 \in\mathbb{Z}[/itex].

Since both cases imply that [itex]k \in\mathbb{Z}[/itex], I conclude that [itex]a|1[/itex].