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So I need to prove, from the definition of reals as Cauchy sequences of rationals, that $i$ is not a real number. The guidance given is to assume that $a\sim b$ are equivalent Cauchy sequences of rationals such that $\displaystyle \lim_{k\rightarrow \infty} a_{k}b_{k} = -1$. Apparently I'm to do this by contradiction.

What I've done so far is (in outline) to suppose these things, and let $P=\max{\{M,N\}}$ by the larger of the ranks of the two sequences such that both are constrained by any arbitrary $\varepsilon>0$ for all $p>P$. For abbreviation let $a_{p}=a, b_{p}=b$.

I've done a lot of algebraic manipulation on $|ab+1|$ and $|a-b|$ using the fact that both are positive and less than $\varepsilon>0$. I've also been trying to do this with the constraint that $\varepsilon <1$ since I thought maybe I could show that by combining them in the right ways I could get something bigger than 1. But so far no luck.

The algebra I've tried:

$|ab+1|+|a-b|$

$|ab+1|^2 + |a-b|$

$2|a-b|^2 + |ab+1|$

And many, many other permutations on these. But I haven't found any way to squeeze a contradiction out of any of them. Any guidance?

Thank you!

User12345
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Addem
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    What is your definition for $i$? If it is that $i^2 = -1$, this is an immediate consequence because any real number $x$ must satisfy $x^2\ge 0$. – Cameron Williams Apr 17 '14 at 20:46

2 Answers2

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HINT: If $a_k\sim b_k$ then without loss of generality you can assume that $a_k=b_k$. Therefore, $a_kb_k$ is now $(a_k)^2$. Use the fact that $x^2\geq 0$ for rational numbers (or prove that fact first), and conclude that a limit of squares cannot be a negative number.

Asaf Karagila
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This is a bit of a trick, but: $$a_kb_k = \tfrac12\big(a_k^2 + b_k^2 - (a_k-b_k)^2\big) \ge -\tfrac12(a_k-b_k)^2 \to 0$$ (Maybe it's worth comparing to the proof that multiplication is well-defined on reals — I suspect it must use some trick like this.)