Question: Prove that if $t \in T$ and $q \in Q$, but $q \neq 0$ then $qt \in T$.
This is Exercise 2.7.13(a) from Mark E. Watkins, Jeffrey L. Meyer: Passage to Abstract Mathematics.
I'm currently practicing on proof writing and I want to know if I did this correctly. If it's right then I'm wondering why some proofs with easy to understand definitions or one line definitions are easy to endure while other proofs that require over three definitions are so hard that it requires deep thinking skills just to understand and write a correct proof.
Attempt: We are going to disprove this statement.
Definition 2.7.8 states that a number $s$ is an algebraic number when there exists some $ p \in Z[x]$ such that $p(s) = 0$. Let us denote the set
$A = [x \in C: $ x is algebraic]
By Proposition 2.7.9 all rational numbers are algebraic.
Moreover, the set $Q$ of rational numbers is $Q = [ \frac{a}{b} a,b \in Z $ and $b \neq 0]$
A number that is not algebraic is called transcendental. Thus, the set $T$ of transcendental numbers satisfies:
$T = [x \in C : x \notin A]$
For $t \in T$, $t$ is transcendental, so it's not algebraic.
. However, if $q \in Q$ it's rational, then by proposition 2.7.9, $q$ must be algebraic. Therefore, $q$ isn't transcendental. As a result, $t \in T$, but $q \notin T$, so $qt \notin T$. [What I'm trying to say is that since $t$ is transcendental and $q$ is rational, q doesn't belong in T because q is rational and algebraic. We can have $t \in T$, but $q \notin T$ due to to proposition 2.7.9 and definition 2.7.8]
Since transcendental means not-algebraic, the proof proceeds most naturally along the following lines. I'm leaving out the actual argument; this is just the structure. Note how this is much less verbose than the work in the question.
Proposition. Let $t \in {\mathbb C}$ be transcendental and $q \in {\mathbb Q}$, $q \neq 0$. Then $qt$ is also transcendental.
Proof. Suppose that $qt$ is not transcendental, i.e., algebraic. Then [insert argument here that leads to a contradiction, for instance by somehow concluding that $t$ must be algebraic as well.] Contradiction. Hence $qt$ is transcendental.