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By definition, $i^2=-1$, right?

But one can then clearly deduce that $(-i)^2=-1$. The only difference I see is that one is $-1$ times the second.

So what allows us to differentiate between $i$ and $-i$? Can they be used synonymous? That is, does nothing happen if all of a sudden we were to switch $i$ and $-i$ in math, so long as we are consistent?

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    $\begingroup$ we can hold the same argument for 2 and -2 right...? $\endgroup$
    – user394255
    Commented Jan 2, 2017 at 13:55
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    $\begingroup$ I think the point here is that there is no way to distinguish in a purely algebraic way the roots of an irreducible polynomial. Here $i$ and $-i$ are the two roots of the polynomial $x^2 + 1$ which is irreducible over the reals. Complex conjugation (an automorphism of the complex field that fixes each real number) exchanges the two, so that they cannot be distinguished algebraically. $\endgroup$
    – Andreas Caranti
    Commented Jan 2, 2017 at 13:59
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    $\begingroup$ @A.Molendijk No we cannot, for example $1$ and $-1$ can be differentiated intrinsically by the fact that $1$ solves $x^2=x$ while $-1$ does not. I believe the OP is asking whether a similar intrinsic characterization of $i$ vs $-i$ exists -- and the answer is no (which makes the question interesting). $\endgroup$
    – Did
    Commented Jan 2, 2017 at 14:00
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    $\begingroup$ @barakmanos ?? $(-i)=(-1)\times i$ and $i=(-1)\times(-i)$. $\endgroup$
    – Did
    Commented Jan 2, 2017 at 14:04
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    $\begingroup$ If anyone were to think that this question is trivial, it's worth noting that pretty much this single idea can be considered the basis of Galois theory - we can swap $i$ and $-i$ and all of algebra and arithmetic stays the same, so as soon as we ask "what other permutations of a field can do that?" we enter the automorphism theory. $\endgroup$
    – Wojowu
    Commented Jan 2, 2017 at 14:16

1 Answer 1

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There is some interesting mathematics bound up in your question. There is an automorphism of $\mathbb C$ as a field extension of $\mathbb R$ which is defined by sending $i$ to $-i$. Provided you are consistent (including very great care with signs) everything algebraic works nicely. This reflects the fact that in building $\mathbb C$ from $\mathbb R$ we make an arbitrary choice of a square root of $-1$ to call $i$.

For amusement there is an automorphism of $\mathbb Q(\sqrt 2)$ as an extension of $\mathbb Q$ which sends $\sqrt 2$ to $-\sqrt 2$ - for much the same reason. However, interesting things happen to the metric (distance between points) - so though the algebraic properties are retained, it is not true that "everything" remains the same.

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  • $\begingroup$ I think the same doesn't happen with square roots because we defined them so that the square root of a perfect square is a positive integer. $\endgroup$
    – J. C.
    Commented Jan 2, 2017 at 14:07
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    $\begingroup$ Indeed, with automorphisms we often have the metric "messing up". This is not the case with complex automorphism, however. And complex conjugation is very special with this regard - if we consider any two embedding of a number field in $\mathbb C$, then they will give rise to very different metrics, unless they are equal or complex conjugates. $\endgroup$
    – Wojowu
    Commented Jan 2, 2017 at 14:10

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