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For $ x^2+1=0 $ , can we say that: $ x=+i $ and $ x=-i $

Enock Kabibi
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  • Yes, and this question has been asked for many times – MafPrivate Sep 24 '19 at 14:44
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    What is your question? – It'sNotALie. Sep 24 '19 at 14:45
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    For $x^2+1=0$ we can say that $x=+i$ is a solution and we can say that $x=-i$ is a solution. Alternatively, given that $x^2+1=0$ we can say that $x=i$ or $x=-i$. What we cannot say is given that $x^2+1=0$ that $x=i$ and $x=-i$ as that would imply that a single value of $x$ is simultaneously equal to two different nonequal values. – JMoravitz Sep 24 '19 at 14:47
  • Well, no, to be strictly accurate. Maybe you meant $x=i$ OR $-i$. But don't worry too much no-one can tell them apart! After all how do you know which is which? – almagest Sep 24 '19 at 14:54
  • @almagest: couldn't you distinguish by saying that multiplication by $i$ corresponds to $90^o$ counterclockwise rotation in the complex plane, whereas multiplication by $-i$ corresponds to $90^o$ clockwise rotation? – J. W. Tanner Sep 24 '19 at 15:01
  • x=i is the only solution stated – Enock Kabibi Sep 24 '19 at 15:03
  • @J.W.Tanner that begs the question as to how we set up the "complex plane" in the first place, why we chose up and down/clockwise counterclockwise. We can't avoid that it was arbitrary. – fleablood Sep 24 '19 at 15:03
  • @J.W. Tanner, while counterclockwise is well defined (positivity is well defined and 1 is also well defined hence the real coordinate too, hence the purely complex one), complex multiplication is defined in terms of $i$ so we go circular, in other words, $i$ is the one with argument $\frac{\pi}{2}$ but we have an isomorphism over $\mathbb R$ that sends it to $-i$ etc – Conrad Sep 24 '19 at 15:04
  • @It'sNotALie. Why only +i is the only solution given but not - i – Enock Kabibi Sep 24 '19 at 15:06
  • @EnockKabibi "x=i is the only solution stated". Stated where? If it stated that $x^2 +1 = 0$ had only one solution or that $x=i$ or $x = -i$ were two solutions, then it is just plain wrong. ... And it's important to realize that the statement i)$x^2 - a =0$ and ii) $x =\sqrt a$ are NOT equivalent.... and that no matter how you define what $\sqrt{a}$ means, it conventionally means that $\sqrt{a}$ is ONE of two possible solutions to $x^2 -a = 0$. – fleablood Sep 24 '19 at 15:08
  • "Why only +i is the only solution given but not - i " Because your text is effing WRONG. – fleablood Sep 24 '19 at 15:09
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    "Why only +i is the only solution given but not - i". Given by whom under what circumstance? If someone casually said "the solution to $x^2 + 1=0$ is $i$" they may have been too casual. If you paid $200 for a text that says "$x^2+1=0$ has $i$ as it's only solution" then you were badly cheated. – fleablood Sep 24 '19 at 15:17
  • I've never seen that anywhere but I hope I should just read more of this. Thanks for your kind thoughts – Enock Kabibi Sep 24 '19 at 15:19
  • Bottom line: $i$ is a complex number. $i^2=-1$. All complex numbers are of the form $a + bi$ where $a,b\in \mathbb R$. From that, basic algebra will yield that the equation $x^2 + 1 =0$ will have two solutions. One is $i$ and the other is $-i$. – fleablood Sep 24 '19 at 15:21
  • People need to be careful. If you are trying to define what $i$ is by saying it is a solution to $x^2 + 1=0$, you should be careful to say that $i$ is A solution. – fleablood Sep 24 '19 at 15:23

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