Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $x = [(x_i)]$ is a real number (equivalence class of Cauchy sequences of rational numbers). a) How would you define the additive inverse of $x$? b) Prove that your answer to part a is well-defined.

I know that when adding the opposite to $x$, the sum should equal zero. Is it possible to have a negative Cauchy sequence? Would the additive inverse of $x$ be $-x$?

As for proving the answer is well-defined, I believe I need to show: If $[(x_i)]\sim[(x_i)]'$, then $[(-x_i)]\sim[(-x_i)]'$ (The i's are supposed to be sub i's.)

Suggestions would be appreciated!

share|cite|improve this question
I've edited your question to include a link relevant to this construction and to get the subscripts. You should check your post - I am not sure that you wanted to write [(xi)]' there and in the end of the post probably one of $-x_i$ is supposed to be $-x_i'$. – Martin Sleziak Nov 8 '11 at 19:05
For you first question: Just notice that if $z_n=-x_n$ then $|z_n-z_m|=|x_n-x_m|$ for each $n$, $m$. Then you can see directly from the definition that the sequence $(z_n)=(-x_n)$ is Cauchy.\\ Your second question is very similar: If you know that $\lim_{n\to\infty} |x_n-y_n|=0$ what can you say about sequences $(-x_n)$ and $(-y_n)$? – Martin Sleziak Nov 8 '11 at 19:14
I would tag this as analysis more than number theory... – Patrick Da Silva Nov 8 '11 at 19:25

You have the right idea, but your notation is badly confused. First, the relation $\sim$ is a relation between Cauchy sequences, not between equivalence classes of them. You can write $(x_i)\sim(y_i)$, for instance, or $[(x_i)]=[(y_i)]$, which means exactly the same thing, but you can’t write $[(x_i)]\sim[(y_i)]$: it’s meaningless.

Next, as Martin suggested, you don’t want $[(x_i)]'$: that suggests that the Cauchy sequence $(x_i)$ is generating two equivalence classes, $[(x_i)]$ and $[(x_i)]'$. It would be simpler to avoid the primes altogether and just use a different letter from $x$.

For (a) you’re basically right: you want to define $-[(x_i)]$ to be $[(-x_i)]$. You’re also on the right track for (b): to show that this is well-defined, you must show that you get the same result no matter which member of the equivalence class $[(x_i)]$ you use to define $-[(x_i)]$. In other words, you must show that if $(x_i)\sim(y_i)$, then $(-x_i)\sim(-y_i)$.

To do this, translate the hypothesis, that $(x_i)\sim(y_i)$, and the desired conclusion, that $(-x_i)\sim(-y_i)$, into statements involving more basic notions by using the definition of $\sim$: you want to show that $$\lim_{n\to\infty}|x_n-y_n|=0 \implies \lim_{n\to\infty}|(-x_n)-(-y_n)|=0\;.$$ This should be pretty straightforward.

share|cite|improve this answer
I understand what you're doing for the most part. I'm just not sure why you used the limits. Did you do this because every Cauchy sequence of real numbers converses to a real number? – Lily Nov 10 '11 at 21:25

You've got the right approach. You need to prove that if a Cauchy sequence $y$ is equivalent to the Cauchy sequence $-x$, then the Cauchy sequence $y+x$ is equivalent to the additive identity.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.