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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]

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  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Apr 27 '14 at 9:27
  • $\begingroup$ In this particular case, I think that a title like: "Is product of a transcendental number and a non-zero rational number again transcendental?" (or something similar) would help the reader to get the basic idea what the question is about without the need to read the whole question and looking for the place where the notation for $T$ is introduced. $\endgroup$ – Martin Sleziak Apr 27 '14 at 9:29
  • $\begingroup$ huh? but I've been leaving questions like this for months... I never had problems before. $\endgroup$ – usukidoll Apr 27 '14 at 9:30
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    $\begingroup$ I got lost at "We are going to disprove this statement"; the claim is actually true. Now I think that's what you're trying prove, but the formulation is awkward. Furthermore, your proof is very verbose - too verbose, I would say - and does not address the real point. (The critical step is "$t \in T$, but $q \not\in T$, so $qt \not\in T", but you give no argument for that.) $\endgroup$ – Magdiragdag Apr 27 '14 at 10:05
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    $\begingroup$ No, that's not a proof either. You should not assign a value to $q$ and $t$; you should give an argument that works for all $q$ and $t$. Giving specific $q$ and $t$ works if you want to give a counterexample. $\endgroup$ – Magdiragdag Apr 27 '14 at 10:10
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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.

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  • $\begingroup$ don't you mean $t \in T$... $t \in C$ means that t is complex. $\endgroup$ – usukidoll Apr 27 '14 at 10:22
  • $\begingroup$ ok so we're going the contrapositive route? that would be that If $qt \notin T$ then $t \in T$ and $q \in Q$, but $ q \neq 0$ $\endgroup$ – usukidoll Apr 27 '14 at 10:24
  • $\begingroup$ ah... the negation of transcendental. $T =[x \notin C: x \in A]$, so t is algebraic and so is q. $\endgroup$ – usukidoll Apr 27 '14 at 10:26
  • $\begingroup$ @usukidoll That's still not the argument: you still have to argue why $t$ is algebraic and at some point you will have to use what it means to be algebraic. $\endgroup$ – Magdiragdag Apr 27 '14 at 10:27
  • $\begingroup$ I'm not done yet.. taking it in baby steps :D ok and then since $q \neq 0$, then we have $ \frac{1}{q} \in Q$. so, $t = \frac{1}{q}(qt) \in A$ Therefore, $ t \in A$, and since $t$ is now algebraic, it's clear that $t$ can't be transcendental at all , so $ t \notin T$ $\endgroup$ – usukidoll Apr 27 '14 at 10:30

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