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I'm a simple man living my life and enjoying mathematics now and then. Today during lunch my friend asked me about complex numbers and $i$. I told him what I knew and we went back to work.

After work I decided to read up on complex numbers and I somehow ended up with this equation:

$$ 1 = \sqrt 1 = \sqrt{(-1)(-1)} = \sqrt{(-1)} \ \sqrt{(-1)} = i \cdot i = i² = -1 $$

Somehow I got that $1 = -1.$ I can't see a contradiction. Did I just break math? What happened? Where is my mistake?

Jeff Strom
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bodacydo
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  • I think this question has been asked here a few times before. However, please note that in MathJax one can write $\sqrt{(-1)(-1)}$. It's coded as \sqrt{(-1)(-1)}. ${}\qquad{}$ – Michael Hardy Sep 09 '15 at 05:03

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The identity $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is only true when $a \ge 0$ and $b \ge 0$. Here you are saying $a=-1$ and $b=-1$ which violates the condition for the identity you used: $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}$

ChaoSXDemon
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  • As was similarly noted in avid19's link, the identity can be true for other values of $a$ and $b$. It just isn't guaranteed to always be true – Paul Sinclair Sep 09 '15 at 05:09
  • ooooooooooooooooh.................... – bodacydo Sep 09 '15 at 05:12
  • Additionally, if we use complex analysis, we will get the true answer. The truth lies in the fact that in the complex plane, we have a discontinuity every $2\pi$. We also know that $-1=e^{i\pi}$. Try to prove what you just did with complex numbers and you will get $1\ne-1$ – ChaoSXDemon Jan 23 '16 at 18:50
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When you use $\sqrt{a}$ where $a$ complex number or real number is not a unique number, i.e., $\sqrt{1}$ is not a unique number, $\sqrt{1}= 1, -1$. So $\sqrt{1}=1$ is not true. Thus you begin like this: $$1=\sqrt{(1)(1)}\ \ \ \ \text{or}\ \ \ \ -1=\sqrt{(-1)(-1)}$$