3 ms·
I agree with your sequence until the last two. The second to last, that should be Cauchy sequences. And for the last, demonstrating that the algebraic closure
by btilly 8y ago
I agree with your sequence until the last two.
The second to last, that should be Cauchy sequences.
And for the last, demonstrating that the algebraic closure of the reals is the complex numbers from first principles is much harder than describing complex numbers as pairs of reals with a multiplication rule that (a,b) * (c,d) = (ac - bd, ad + bc) and then much later proving that it is algebraically closed through complex analysis.
For those who don't know the proof, the idea is this. Liouville's theorem says that if a function is differentiable everywhere, and it is bounded, then it must be constant. Now suppose that p(z) is a polynomial. Consider the function 1/p(z). You can show that as z approaches infinity, it approaches zero. It is not constant. Therefore it must not be differentiable or not bounded. It doesn't take too much work from there to prove that it blows up somewhere, and the spot that it blows up is a point where p(z) is 0.
Apply unique factorization for polynomials (see http://sites.millersville.edu/bikenaga/abstract-algebra-2/polyufd/polyufd.html http://sites.millersville.edu/bikenaga/abstract-algebra-2/po... for that proof) and you quickly get the fact that the complex numbers are algebraically closed.
- est31 8y agoA much easier to understand definition for the set of reals is probably to use infinite decimal expansions that don't end in 9999 etc. Equivalence classes of cauchy sequences is tougher to explain I'd say.
- btilly 8y agoIn that case, good luck coming up with a good definition around multiplication where 3 * 0.33333... works out right. And then proving arithmetic properties like the associative law. And then proving that when you do the reals in decimal, you get the same system as the reals in binary. It sounds harder, but is actually easier to go through the Cauchy sequence definition and then point out that the decimal representation naturally gives rise to a Cauchy sequence. So, for example, 3.1415926535... gives you (3, 31/10, 314/100, 3141/1000, ...). And as Cauchy sequences, of course, (1, 1, 1, 1,...) is easily proved to be the same as (9/10, 99/100, 999/1000, ...).
- est31 8y agoOh right, 3 * 0.33333 is an issue, yeah. Still, you need to define that equivalence relation on the Cauchy sequences and then explain what a set modulo a relation means.
- btilly 8y agoThis is all standard mathematics. The equivalence relationship is that the sequence (x_1, x_2, x_3, ...) is equivalent to (y_1, y_2, y_2, ...) if and only if the limit as n goes to infinity of x_n - y_n = 0. Formally, the real number represented by (x_1, x_2, x_3, ...) is the set of all Cauchy sequences which are equivalent to that one. Since "equivalent to" is transitive, any Cauchy sequence in that set will define the same set. Addition and multiplication are defined elementwise. Proving that they are well-defined is relatively straightforward. Their algebraic properties follow for free. Any rational number q can be mapped to the Cauchy sequence (q, q, q, ...) which leads to a unique real number that we somewhat sloppily call q again. I've left some details out, but this construction is well-understood, and is how we define the completion of a metric space.