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Prove that $\imath$ (defined by $\imath^2=-1$) does not have a position on the Real number line. That is, show that there does not exist two real numbers $a$ and $b$ such that $a<\imath<b$.

(I'm assuming I need to use some sort of definition of $<$ on $\mathbb{R}$?)

The reason I'm asking is that I want to see why there can't be a "hole" on the Real number line into which $\imath$ could fit.


Can someone offer a Cauchy sequence proof? I.e. assume to the contrary there exists a Cauchy sequence $\left(x_n\right)$ of real numbers that converges to $\imath$ and derive a contradiction.?

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Hint: show that for any real number, $a^2>0$. – Alex Becker Jan 6 '13 at 7:25
@AlexBecker: Great Hint. – Babak S. Jan 6 '13 at 7:27
If the problem is to show that there is no real number whose square is $-1$, that would make sense, and the fact that squares of real numbers are always nonnegative implies this. But I do not understand the formulation of the problem in terms of showing that $\imath$ is not "between" two real numbers. How would you even define "$\imath<b$" to check that it is false? I guess the statement is trying to set up a proof by contradiction. Suppose that $\imath$ is real, take $2$ real numbers $a$ and $b$ such that $a<\imath<b$, and find out what goes wrong? – Jonas Meyer Jan 6 '13 at 7:29
@Jonas:I sort of agree. But you could start by saying something like "Suppose that $i \in \mathbb{R}.$ Then $i \neq 0$ as $i^{2} \neq 0,$ so either $i < 0$ or $i >0.$" But, like you, I don't really see the point of the "between two real numbers" formulation. Ah, I see you amended your comment. – Geoff Robinson Jan 6 '13 at 7:36
Alex: $0^{2} = 0 \not{>} 0$ (sorry for being pedantic). – user123454321 Jan 6 '13 at 8:26
up vote 4 down vote accepted

The question is best answered by the fact that for any real number $a$, we have $a^{2} \geq 0$.

To answer your second question, suppose there are such $a, b \in \mathbb{R}$ with $a < i < b$. We have $ai < -1 < bi$ if $i > 0$ and $bi < -1 < ai$ if $i < 0$. Thus either case we have $-a < -i < -b$, which yields $b < i < a < i < b < i < a < \cdots$.

Unless we use weird rules for binary relations for symbol $<$ the last statement should be derivable using the transitivity of $<$ and it is a contradicting the trichotomy of $<$. I liked using Walter Rudin's undergraduate text to avoid these kinds of confusions, so I recommend the same reading!

(In the proof, I used trichotomy and transitivity of $<$ and the fact that real number system forms an ordered field. It is perfectly fine for you to ask why these are the facts. Studying those will make your mathematics even stronger, but don't try to absorb it in couple of days. Best of luck in your first analysis course!)

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You can stop at $-a < -\imath < -b$, because that contradicts our assumed statement that $a < b$. – chharvey Jan 6 '13 at 23:25
I know. But I wanted a prettier contradiction. – user123454321 Jan 8 '13 at 9:40
I found a simpler method. Assume $a < b$ and multiply throughout by $i$. If $i > 0$, then $ai < bi$, and do it again and get $-a < -b$, a contradiction. If $i < 0$ though, you get $ai > bi$, and again, $-a < -b$, the same result. – chharvey Mar 1 at 14:20

Recall that $x^2>0$ for all $x\neq 0$ in the real numbers. Clearly $i\neq 0$. Therefore $i^2>0$, but $i^2=-1<0$.


The Cauchy sequence argument will fail, but for other reasons. First note that $i\notin\mathbb R$ implies that if a sequences of real numbers has a limit in $\mathbb R$ it has to be a real number as well.

Now we can do one of the following things:

  1. Note that $\mathbb R$ is a complete metric space, so every Cauchy sequence of real numbers converges to a real number. In particular this means that it has a limit. So there is no such sequence which may converge to a number outside of $\mathbb R$.

  2. We can ask in $\mathbb C$ whether or not there is a Cauchy sequence of real numbers which converges to $i$. If there is then $i\in\mathbb R$. But we know that if $x\in\mathbb R$ then $|i-x|\geq 1$, so there is such Cauchy sequence either.

Because the real numbers are complete you cannot add $i$ into them, there are no "holes". But even if there were, for example in $\mathbb Q$ there are plenty of holes to add new numbers. You still can't add $i$ to the rational numbers and retain an ordered field. So by virtue of $\mathbb R$ being ordered, $i\notin\mathbb R$.

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