5 ms·

Hah! So if, as per the article, "imaginary numbers are square roots of negative numbers" then I am happy to report that: 1 = sqrt(1) = sqrt((-1) * (-1)) = sqrt

• JadeNB 4y ago
  Of course it's a joke, but the problem with that argument isn't that imaginary numbers don't exist, but that the rule sqrt(a * b) = sqrt(a) * sqrt(b) doesn't hold for them. Or, rather, it does hold, but (as we should for real numbers, too!) we must regard sqrt(a) as being not a single number but a set of square roots of a (something we avoid doing in the real case by preferring non-negative numbers—a convenience not available for the complex numbers). With this understanding, the rule sqrt(a * b) = sqrt(a) * sqrt(b) holds, and your instantiation of it (with a = b = −1) shows that 1 and −1 are both equally valid square roots of 1, which is certainly true.
  • prof-dr-ir 4y ago
    Excellent. Indeed, as you say already for the real numbers one could have chosen sqrt(9) = -3: after all, (-3)^2 = 9 just as much as 3^2 = 9. Taking into account this inherent ambiguity in the definition of the square root function becomes crucial when extending it to the negative numbers, as my example shows. So, long story short: although not everyone agrees, personally I prefer to define the imaginary numbers by means of an entity i that obeys i^2 = -1, and avoid talking about "square roots of negative numbers" as the (otherwise fine) article does.
• trinovantes 4y ago
  Only if you ignore the implicit ± infront of sqrt