4? Positive and negative 4?
I just got into an argument with a buddy about this. He argues if it's not an i, it's not included as a imaginary number, but only the real positive number.
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1$2+2=4$, I see no argument whatsoever. Btw, what has this to do with linear algebra? – AdLibitum Aug 21 '15 at 14:18
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Is the question (1) whether it's true that $2+2=\sqrt{16}$, or (2) what all of the square roots of $16$ are, both real and complex? – Fabrosi Aug 21 '15 at 14:19
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1possible duplicate of How $\sqrt{x}$ is an function? – N. F. Taussig Aug 21 '15 at 14:31
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Both numbers when squared make $16$... $$\left(4\right)^2 = 16,\ \ (-4)^2 = (-1)^2(4)^2 = 1\cdot (4^2) = 16$$
By convention, if $x>0$ we typically allow $\sqrt x$ to represent the positive of the two solutions, and denote $\pm\sqrt{x}$ to consider both solutions.
If you are concerned about imaginary numbers, these only appear for square roots of negative numbers. I.E. $\sqrt {-16} = \pm 4i$. This is true for the following reason: $$\left(-4i\right)^2 = (-1)^2i^24^2 = (1)(-1)(16) = -16 = (i^2)(16) = (4i)^2$$
jameselmore
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The square root of 16 can either be 4 or -4. However, we usually take the positive square root by default, so it's just 4. Imaginary numbers come into play only when talking about 'square roots' of negative numbers. $2+2=4$ is, I think, accepted in general, but the square root issue is a separate question.
shardulc
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