1

Is $a+bi=c+di\iff a=c, b=d$, where $a,b\in\mathbb{R}$ something that requires proof? My instinct is telling me that proof is required to demonstrate that playing around with real numbers (using field operations) cannot make them 'escape' into the realm of non-real numbers. Is there a term for this, perhaps involving metric spaces?

Trogdor
  • 10,331
  • Correct me if I'm wrong someone, but doesn't this simply go along with the fact that $\mathbb{C}$ is an ordered field? – user28375028 Sep 08 '14 at 04:12
  • How do you define/construct the complex numbers? – Michael Albanese Sep 08 '14 at 04:12
  • 1
    If you consider $a + bi$ to mean the pair $(a,b)$, then it's obvious. But when you're first developing the theory of complex numbers, you'll want to check at some point that $a + bi = (a,0) + (b,0)(0,1)$ really is $(a,b)$, so you can switch to the notation $a + bi$ and it won't be incorrect. – Dave Sep 08 '14 at 04:13
  • @AlexMiller, $\mathbb{C}$ is not an ordered field. – jxnh Sep 08 '14 at 16:23

1 Answers1

1

As Dave keenly pointed out, one way of framing this question is whether the notations $(a, b)$ and $a + bi$ are compatible, or more precisely, whether the space of pairs $(a, b)$ and the space of complex numbers $a + bi$, each endowed respectively with their usual operations, are isomorphic (say, as rings) via the identification $\Phi: (a, b) \mapsto a + bi$, which is obviously a bijection.

In particular, we must check that:

  • $\Phi((a, b) + (a', b')) = \Phi((a, b)) + \Phi((a', b'))$ and
  • $\Phi((a, b) \cdot (a', b')) = \Phi((a, b)) \Phi((a', b'))$.

But the first identity is almost immediate, and the multiplication rule $$(a, b) \cdot (a', b') := (aa' - bb', ab' + a'b)$$ is chosen precisely so that the second identity holds too. (If this isn't clear, evaluate both sides of the identity to check for yourself.)

Travis Willse
  • 99,363
  • What is an example of another way to define $\mathbb{C}$? – Trogdor Sep 08 '14 at 04:20
  • 1
    The question is whether $a + bi$ means $(a,b)$ or $(a,0) + (b,0)(0,1)$. – Dave Sep 08 '14 at 04:21
  • @Trogdor (1) the set of direct similarity transformations of the Euclidean plane that fix the origin. Multiplication is composition, and addition is defined as for linear mappings. (2) The quotient ring of $\mathbf{R}[X]$ by the ideal generated by $X^2 + 1$. – Dave Sep 08 '14 at 04:25
  • @Trogdor: One way is as the quotient ring $\mathbb{R}[x] / (x^2 + 1)$. Here, $x$ plays the role of $i$: by forming the quotient of the polynomial ring by $x^2 + 1$, we declare that (in the quotient) $x$ is a quantity such that $x^2 = -1$. This is ain some sense a more natural construction, as the multiplication rule for complex numbers is now just a consequence of the usual rule for polynomial multiplication in $\mathbb{R}[x]$, rather than a declaration of some pair of bilinear forms that might, to the new student, look unmotivated. – Travis Willse Sep 08 '14 at 04:29
  • @Dave: Ah, I see your point---the notation is chosen so exactly that works, but one might want proof that the two notations (together with their respective usual operations) define the same algebraic structure via the identification $(a, b) \leftrightarrow a + bi$, which includes that $(a, b) = (a, 0) + (b, 0)(0, 1)$. – Travis Willse Sep 08 '14 at 04:35
  • @Dave: Usually one does not consider the zero transformation to be a similarity transformation, exactly so that the set of similarity transformations is a group. – Travis Willse Sep 08 '14 at 04:38
  • @Travis Right, that's what I meant. You're right about similarity transformations - I should have added in zero. – Dave Sep 08 '14 at 04:39
  • @Trogdor This is essentially the same as Dave's suggestion (1), but you can also view complex numbers as the set of matrices $\left(\begin{array}{c}a & -b\b & a\end{array}\right)$, $a, b \in \mathbb{R}$ where complex addition and multiplication are now just matrix addition and multiplication. – Travis Willse Sep 08 '14 at 04:47