- #1

- 17

- 1

## Homework Statement

Show that for every α ∈ ℂ with α ≠ 0, there exists a unique β ∈ ℂ such that αβ = 1

## Homework Equations

Definition[/B]: ## \mathbb {F^n} ##

## \mathbb {F^n} ## is the set of all lists of length n of elements of ## \mathbb {F} ## :

## \mathbb {F} ## = {## (x_1,...,x_n) : x_j ∈ \mathbb {F} for j = 1,...,n ##}

**Definition**:

*addition in*## \mathbb {F^n} ##

## (x_1,...,x_n) + (y_1,...,y_n) = (x_1 + y_1,..., x_n + y_n) ##

**Definition**:

*scalar multiplication in*## \mathbb {F^n} ##

##λ(x_1,...,x_n) = (λx_1,...,λx_n) ##

##λ ∈ \mathbb {F}, (x_1,...,x_n) ∈ \mathbb {F^n}##

And the 7 other properties of fields: https://en.wikipedia.org/wiki/Field_(mathematics)#Classic_definition

## The Attempt at a Solution

## α ∈ ℂ → α = a + ui## ##a, u ∈ ℝ ##

##β ∈ ℂ → β = b + vi## ##b, v ∈ ℝ ##

##∃γ∈ℂ## such that ##γ = \frac{1}{α} = \frac {1}{a+ui} = c + di## ## c,d ∈ ℝ ##

Proof:

**Existence***: $$ \frac {1}{a+ui}\frac {(a-ui)}{(a-ui)} = \frac {(a-ui)}{(a^2 + u^2)} = \frac {(a)}{(a^2 + u^2)} +\frac {(-u)}{(a^2 + u^2)}i $$*

By multiplicative identity

By multiplicative identity

*By definition of real numbers:*

$$ = s + ti$$ $$s, t ∈ ℝ $$

*By definition of complex numbers:*

$$ = ψ ∈ ℂ $$

Let ## b = \frac {a}{a^2 + u^2}, v = \frac {-u}{a^2 + u^2}##, then

By substitution: $$ β = b + vi = \frac {a}{a^2 + u^2} + \frac {(-u)}{a^2 + u^2}i = \frac {1}{a+ui}$$

$$ → αβ = (a+ui)\frac {1}{(a+ui)} = 1$$

**Uniqueness**

Suppose ∃δ∈ℂ such that αδ = 1

$$ αβ = 1 $$ and $$ αδ = 1 $$

By transitivity of equality:

By transitivity of equality:

$$αβ = αδ $$

*By cancellation:*

$$ β = δ $$

My question is, have I correctly proved the uniqueness part? Was I also doing a bit too much with the existence portion of the proof? This very simple exercise comes from Axler's

*Linear Algebra Done Right*. I always felt a bit iffy proving uniqueness in linear algebra, and wish to brush up the subject again.