Understanding Existence and Uniqueness

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The discussion focuses on proving the existence and uniqueness of the multiplicative inverse for any non-zero complex number α. The proof demonstrates that for α = a + ui, the inverse can be expressed as β = (a - ui) / (a^2 + u^2), confirming that αβ = 1. Concerns are raised about the uniqueness proof, particularly regarding the use of cancellation laws and whether the proof sufficiently adheres to the field axioms. Suggestions are made to simplify the existence proof and clarify the well-defined nature of the inverse. Overall, the conversation emphasizes the importance of rigor in demonstrating both existence and uniqueness in linear algebra contexts.
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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

By multiplicative identity
: $$ \frac {1}{a+ui}\frac {(a-ui)}{(a-ui)} = \frac {(a-ui)}{(a^2 + u^2)} = \frac {(a)}{(a^2 + u^2)} +\frac {(-u)}{(a^2 + u^2)}i $$

By 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:

$$αβ = αδ $$
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.
 
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I'm not sure this is correct. If you assume the field axioms for ##\mathbb{C}##, then you have a unique inverse for any non-zero element, which is what you are trying to prove.

I suggest you can assume the field axioms for ##\mathbb{R}## and what you have proved already for ##\mathbb{C}## - commutivity, associativity, distributive law etc. But, I don't know exactly how your book approaches this.

For the uniqueness part, I'm not sure you can invoke a "cancellation" law. But, once you have proved existence, you might think about how you show uniqueness. Hint: similar to cancellation but using what you have shown already in this proof.

Finally, you took a lot of algebra to show that ##\frac{a - bi}{a^2 + b^2} (a+bi) = 1##. You could simply have shown that directly and thereby proved existence. Note that you need to say something about why ##\frac{a - bi}{a^2 + b^2}## is well-defined. Hint: why does this not apply to ##0##?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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