Imaginary Numbers

Excerpts

• Imaginary Numbers

  Hence, if our symbols are to mean the ordinary positive or negative numbers, there is no solution to $x^{2} = -2$, and the equation is in fact nonsense.

• Imaginary Numbers

  The equation $x^{2} + 1 = 3$ becomes $x^{2} = 2$, and this has two solutions, either $x = +\sqrt{2}$, or $x = -\sqrt{2}$.

• Imaginary Numbers

  Nothing can be proved by a succession of blots, except the existence of a bad pen or a careless writer.

• Imaginary Numbers

  All these requisites are satisfied by taking $(x, y) + (x', y')$ to mean the ordered couple $(x + x', y + y')$.

• Imaginary Numbers

  It is no paradox to say that in our most theoretical moods we may be nearest to our most practical applications.

Equations

• Imaginary Numbers

  (x, y) - (u, v) = (x, y) + (-u, -v)

  Subtracting an ordered couple is the same as adding the couple with both numbers negated.

• Imaginary Numbers

  x = ±\sqrt{(b - a)}

  The two solutions of x² + a = b are ±√(b − a), and they exist only when b is not less than a.

• Imaginary Numbers

  \sqrt{(-1)} \sqrt{c^{2}} = c\sqrt{(-1)}

  For a positive c, the square root of −c² equals c times √(−1), so the imaginary unit √(−1) carries the root of any negative number.

• Imaginary Numbers

  (x, y) + (x', y') = (x + x', y + y')

  By definition, the sum of two ordered couples is the ordered couple whose first number is the sum of the first numbers and whose second number is the sum of the second numbers.

• Imaginary Numbers

  (x, y) = (c, d) - (a, b)

  The unknown ordered couple (x, y) satisfying (x, y) + (a, b) = (c, d) is unique and equals (c, d) minus (a, b).

• Imaginary Numbers

  (x, y) - (u, v) = (x - u, y - v)

  Subtraction of ordered couples is defined componentwise: the first numbers are subtracted and the second numbers are subtracted.

• Imaginary Numbers

  (x, y) - (x, y) = (0, 0)

  Any ordered couple minus itself gives the couple (0, 0), which is therefore the zero ordered couple.

• Imaginary Numbers

  (x, y) + (0, 0) = (x, y)

  Adding the zero ordered couple (0, 0) to any ordered couple leaves it unchanged.

Problems

No exercises in this chapter.