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Determine if $\mathbb{Z}[1/2] = \{\frac{a}{2^n} \mid a\in \mathbb{Z}, n \in \mathbb{N}\}$ is a subring of $\mathbb{Q}$.

  1. We will check that $(\mathbb{Z}[1/2], +)$ is a subgroup of $(\mathbb{Q},+)$. Let $x,y \in \mathbb{Z}[1/2]$. Now $x+y = \frac{a}{2^n}+\frac{b}{2^n} = \frac{a+b}{2^n} \in \mathbb{Z}[1/2]$. Also when $a = 0$ we have that $\frac{a}{2^n} = 0$ so the identity of $(\mathbb{Q},+)$ is in $(\mathbb{Z}[1/2], +)$. The inverses are also in $(\mathbb{Z}[1/2], +)$ since $-x+x= -\frac{a}{2^n}+\frac{a}{2^n} = 0$.

  2. For $x,y \in \mathbb{Z}[1/2]$ we have that $xy=\frac{ab}{2^{n+1}} \in \mathbb{Z}[1/2]$ so it's closed under multiplication.

  3. The identity w.r.t to $\cdot$ is in $\mathbb{Z}[1/2]$ since for $a=2$ and $n=1$ we have$\frac{a}{2^n} = 1$.

Is the proof correct or do I have holes somewhere?

Jen
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  • Your proof is correct, and indeed that is a subring of $\Bbb Q$ – Crostul Nov 15 '21 at 20:25
  • More conceptually, everything follows from the fact that the denominator set ${2^n}$ is a multiplicative submonoid, i.e. it is closed under multiplication (and contains $1$) - see the linked dupe. – Bill Dubuque Nov 15 '21 at 21:28
  • @Crostul it is not correct, but it is a sub ring. – Thomas Andrews Nov 15 '21 at 21:40
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    Please, when you ask for solution-verification, be sure to identify precisely where you have doubts and why. This site is not mean to be a general proof checking machine. – Bill Dubuque Nov 15 '21 at 21:45
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    If you were implicitly assuming that any two such fractions can be written with a common denominator then you should explicitly justify that in order to make it clear what you are doing (else the argument could instead be interpreted as a logical error - as Thomas explained). – Bill Dubuque Nov 15 '21 at 22:24

1 Answers1

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  1. You haven’t proven that the set is closed under addition and multiplication properly. You’ve allowed the numerators to differ, but not the denominators. You really need to show that $$\frac a{2^n}+\frac b{2^m}\\\frac a{2^n}\cdot\frac b{2^m}$$ are in your set when $a,b\in\mathbb Z,$ and $m,n\in\mathbb N.$ For multiplication, this is easy, but for addition you will need a little care. Specifically, if you want to avoid cases, $$\frac{a}{2^n}+\frac{b}{2^m}=\frac{2^ma+2^nb}{2^{m+n}}$$
  2. In the multiplication case, $2^n\cdot 2^n=2^{2n}\neq 2^{n+1}$ for $n\neq 1.$ That’s really a side note, since you don’t want to assume $m=n,$ anyway.
  3. Really a side-note, but if $0\in\mathbb N,$ we’d normally write the multiplicative identity as $\frac1{2^0}.$ $\frac2{2^1}$ works, too, of course.
  4. For the additive inverse, $-\frac a{2^n}$ is not technically in the right form. You want to note that $-x=-\frac a{2^n}=\frac{-a}{2^n}$ is in your set, since if $a$ is an integer, $-a$ is an integer.

Your approach for closure under addition and multiplication would work if you proved a lemma.

If $x,y\in \mathbb Z[1/2],$ then there exists $a,b\in\mathbb Z$ and $n\in\mathbb N$ such that $x=\frac{a}{2^n},y=\frac b{2^n}.$

Essentially, any pair of elements can be written with a common denominator also a power of $2.$

You’d still have to fix issues 2. and 4. above.

Given a non-empty set $S\subseteq \mathbb Z\setminus\{0\},$ we can define $\mathbb Z[S^{-1}]=\left\{\frac aq\mid a\in \mathbb Z,q\in S\right\}.$ This is a ring iff:

Condition: For each $m,n\in S,$ there is a $p\in S$ such that $mn\mid p.$

Basically, that condition ensures addition and product are closed in $\mathbb Z[S^{-1}].$

In your case, $S=\{2^n\mid n\in\mathbb N\},$ it is easy to prove the condition.

Thomas Andrews
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    Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 15 '21 at 21:32
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    This question and answer is about a correction to the specifics of the OP’s attempted answer. @BillDubuque – Thomas Andrews Nov 15 '21 at 21:34
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    But the correction could have been handled in a short comment after the dupe closure. I didn't do that since it is already in this asnwer. But that is a better way to handle dupes with minor errors, and it makes them easier to clean up. – Bill Dubuque Nov 15 '21 at 21:41
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    The very first comment tells the OP that the proof is correct, when there are three glaring errors. Comments are terrible because (1) they don’t allow formatting - no paragraph breaks. (2) No voting - that first comment can sit there forever being wrong right at the top. (Paragraph break) I understand the desire for no duplicates on this site, but I think we can take it too far. This was a question that showed some work and was distinct from other questions, even if the underlying question has been answered here before. – Thomas Andrews Nov 15 '21 at 21:52
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    In any event, what you and I consider “minor errors” can be pretty major to a student learning. – Thomas Andrews Nov 15 '21 at 21:53
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    This type of error is very common and there are better (and much simpler!) questions to use as canonical examples if that is what you wish to emphasize. Generally questions based on minor oversights do nothing but clutter search results on a topic. – Bill Dubuque Nov 15 '21 at 21:55
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    Sure, if the goal is only to maintain the site’s purity as a source of answers to common questions. But the added goal of helping people? I don’t see the harm in writing an answer instead of five comments, if the question is to be closed anyway. If there is some difference, maybe SE needs an architecture change. I often write comment-answers if I think they would help, but I also write answer-answers when I think that the user has shown effort and a longer, less disjoint form might help. – Thomas Andrews Nov 15 '21 at 22:03
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    If you ever tried to search for answers (or dupes) you would soon see how extremely difficult it is to locate good matches. This problem is greatly exacerbated if we allow questions based on trivial arithmetical or logical errors to survive. Pointing out the oversight in a comment still helps the user, and makes it easier to delete the question (if need be), and doesn't encourage of flood of further (dupe) answers. etc. – Bill Dubuque Nov 15 '21 at 22:13
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    That’s an architecture problem. There is no reason the site’s search function should give preference to closed questions, as I know it does. I’d rather be helpful and let the designers of the site fix their flaws. – Thomas Andrews Nov 15 '21 at 22:17
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    Once, reading an old answer, I noted a flaw in the answer. The answerer said he wouldn’t fix it because the site would bump it to the top. That’s taking the architecture’s flaws and making it worse. – Thomas Andrews Nov 15 '21 at 22:20
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    There are many "architectural" problems with the platform (that will likely never be fixed). Throwing in the towel is not the best way to handle them. And, again, a comment pointing out the simple oversight is still helpful. – Bill Dubuque Nov 15 '21 at 22:32