# Additive Inverse of Cauchy Sequence

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!

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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.

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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.

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