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There is 2 definition about imaginary numbers:-

  1. Imaginary unit, $i=\sqrt{-1}$
  2. The square of Imaginary unit, $i^2=-1$

But the later is used mostly, as known to me, because of many reasons. And my question is from the later(2nd) definition. Now consider this:- $$ i^2=-1$$ $$=> (1) i=\sqrt{-1} or (2) i=-\sqrt{-1}$$ Now, $$ i=-\sqrt{-1}$$ $$=> i=-i \text{ [from (1)]}$$

But actually $i$ cannot be equal to $-i$. So, where did I actually gone; where is my fault.

Curious learner
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    What are you doing in your second step, it's not clear. – AkatsukiMaliki Oct 17 '18 at 05:28
  • 1 --> 2, but 2 -/-> 1. That is, it's not the plus of minus square root. When you "multiply by 'step 2' i", you aren't actually doing that. You assumed the plus or minus square root of i^2 is +/- i. – Christopher Marley Oct 17 '18 at 05:31
  • No, my question is a different question. I'm asking about why not $i=-i$; where is the contradiction? I know where is the fault in the question that you are saying, my question can be duplicate of. – Curious learner Oct 17 '18 at 05:33
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    The equation $x^2=-1$ has two roots, $\pm i$, just as $x^2=9$ has $\pm3$. – Berci Oct 17 '18 at 05:34
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    $1^2=1$. Does that make $-1=1$? By the way, I have a personal vendetta against your definition 1 because it hints that square roots and complex numbers work together. They don't. Also, formally, definition 2 is the only one we use (and the one which is easier to teach, yet still 1 is so commonly used in introduction classes). – Arthur Oct 17 '18 at 05:34
  • @Berci So that I asked the question. – Curious learner Oct 17 '18 at 05:37
  • @Arthur My question is using the 2nd definition not using the first one. Although $i=-i$ cannot be proved using the 1st definition. – Curious learner Oct 17 '18 at 05:41
  • What do you mean by $\sqrt{-1}$? It is not clear which complex number the expression $\sqrt{-1}$ refers to, since there are two complex numbers whose square is $-1$. – littleO Oct 17 '18 at 05:45
  • @littleO I think that's what he's asking: Are there really two of them? Are they actually different? (And of course they are, I know that.) – Arthur Oct 17 '18 at 05:51

1 Answers1

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You can replace $i$ with $-i$ and get exactly the same algebraic structure and since we have no reason to prefer one over the other we choose the one with less symbols.

CyclotomicField
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  • Do you mean, in any way, $i=-i$. And that we can use either of the symbol anywhere in the calculation and that will not cause any problems or unmathematical things??? – Curious learner Oct 17 '18 at 05:52
  • It means $\mathbb{R}[i]$ is isomorphic to $\mathbb{R}[-i]$ – Kemono Chen Oct 17 '18 at 08:22
  • @Curiouslearner Yes, if you replace them then everything still works exactly as you would expect. I bet with a little effort you can even prove this on your own. – CyclotomicField Oct 17 '18 at 15:14