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$$Lemma\text{:}\qquad \forall n\in \mathbb{R}, \ n^2 = (-n)^2 \ \because 1 = (-1)^2$$ $$\text{e.g.}\qquad \forall (x, y)\in \mathbb{R}, \ (x - y)^2 = \big(-(y - x)\big)^2 : (x - y)\in \mathbb{R}\ \land \ (x - y) = -(y - x)$$. $$\begin{align} \text{Pursuant to the Lemma, }\ \sqrt{-1}^2 &= (-\sqrt{-1})^2 \\ &= \big((-1)\sqrt{-1}\big)^2 \\ &= (\sqrt{-1}^2\sqrt{-1})^2 \\ &= (\sqrt{-1}^3)^2 \\ &= \sqrt{-1}^{3\times 2} \\ &= \sqrt{-1}^6 \\ \end{align}$$. $$\therefore \sqrt{-1}^2 = \sqrt{-1}^6$$ I know that with the rules of $\sqrt{-1}$, this is allowed. But doesn't this break the rules of the Lemma since $\mathbb{R} \subseteq \mathbb{C}$ and $\sqrt{-1}\in \mathbb{C}?$ The only difference this would be proposing in the Lemma would be that it is false for all $n\in \mathbb{R}$.

$(Lemma = \text{a subsidiary proposition; a helping theorem})$

My Attempt: $$\begin{align} \sqrt{-1}^2 &= (-\sqrt{-1})^2 \\ &= \big((-1)\sqrt{-1}\big)^2 \\ &= (-1)^2\sqrt{-1}^2 \\ &= 1\times \sqrt{-1}^2 \\ &= \sqrt{-1}^2 \\ &= -1 \\ \end{align}$$ This way, it seems like the Lemma must be true, but in the same way, we can write: $$\begin{align} \sqrt{-1} &= (\sqrt{-1}^2)^2\sqrt{-1}^2 \\ &= \sqrt{-1}^4\sqrt{-1}^2 \\ &= \sqrt{-1}^{4 + 2} \\ &= \sqrt{-1}^6 \\ \end{align}$$

Could you please explain to me what I am doing incorrectly (if I am, or if not, what I am interpreting incorrectly)? I only know the "algebra" part of imaginary numbers and not the "geometric" part. If you are using different techniques to explain the reasoning behind this, please be clear and show me step by step.

Thank you in advance.

Mr Pie
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    No, it's only 90% true. – mtheorylord Sep 08 '17 at 02:29
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    What's wrong with saying that $i^2 = i^6$? – Michael L. Sep 08 '17 at 02:29
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    @MichaelLee Nothing is wrong with the fact that $i^2 = i^6$, but it's just that the Lemma doesn't seem to agree... – Mr Pie Sep 08 '17 at 02:30
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    $100$% true? Can the truth be measured? Have you got a truemeter around there? – ajotatxe Sep 08 '17 at 02:32
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    @ajotatxe hah I wish I actually did tbh – Mr Pie Sep 08 '17 at 02:33
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    It looks like you've got a subset containment backwards. $\mathbb{R} \subseteq \mathbb{C}$, but $\sqrt{-1} \notin \mathbb{R}$. The lemma doesn't apply to it. –  Sep 08 '17 at 02:34
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    @user296602 I thought that since real numbers are contained within complex numbers, every number is a real number and every number is a complex number, including a complex number itself. Thank you for that. – Mr Pie Sep 08 '17 at 02:36
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    The set of males is contained in the set of humans. Does that mean every human is male? I still don't understand why you think $i^2=i^6$ disagrees with the lemma. Can you please tell us why you think that? – anon Sep 08 '17 at 02:37
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    @anon thank you for the comparison. I understand better now :) – Mr Pie Sep 08 '17 at 02:38
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    The "pursuant to the lemma" doesn't follow from the lemma since $i \notin \mathbb{R}$. That doesn't mean the pursuant is false. It just doesn't follow from the lemma. – FullofDill Sep 08 '17 at 02:53

2 Answers2

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Substitute $a=n$ and $b=-n$ in the well-known identity $a^2-b^2=(a+b)\cdot(a-b)$ to get $n^2-(-n)^2 = (n+(-n))\cdot(n-(-n)) = 0\cdot 2n = 0$.

kimchi lover
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$$(-n)^2 = (-n).(-n) = -1.n.-1.n = (-1).(-1).n.n = ((-1).(-1)).(n.n) = 1.n^2 = n^2$$

Mark Joshi
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  • If $n = \sqrt{-1}$ then we have our desired result $-1$, so the Lemma must be true and $\sqrt{-1}\notin \mathbb{R}$. I believe I was assuming that $\sqrt{-1}\in \mathbb{R}$ – Mr Pie Sep 08 '17 at 02:41
  • $$(+1) \ \ \ \checkmark$$ – Mr Pie Sep 08 '17 at 03:52
  • By the way, I also created another proof: Consider the equation $(a+b)(a-b)=a^2-b^2$. If this is equal to $0$, then $a^2=b^2$. Also, $(a+b)(a-b)=0$ so $a=\pm b$. Therefore, $b^2=(-b)^2$. $;\bigcirc$ ......... Oh wait crap, the answer below has already done that. – Mr Pie Jun 15 '18 at 01:23