Sergei Yakovenko's blog: on Math and Teaching

Tuesday, November 7, 2017

Lecture 2, November 7, 2017


The basic set theory allows us to construct a set \mathbb N=\{|,||,|||,||||,|||||,\dots\} with a function “next”, denoted by \mathrm{Succ}:\mathbb N\to\mathbb N\smallsetminus\{|\}, which is bijective. This set describes the process of counting objects and is the most basic structure. Starting from a distinguished element denoted by 1, we construct an infinite number of elements 2=\mathrm{Succ}(1),\ 3=\mathrm{Succ}(2),\ 4=\mathrm{Succ}(3) etc. There are two axioms guaranteeing that the set \mathbb N indeed coincides with what we call the set of natural numbers:

  1. \forall x\in\mathbb N\ \mathrm{Succ}(x)\ne 1
  2. Any element x\in\mathbb N is obtained by the iteration of \mathrm{Succ}: x=(\mathrm{Succ}\circ\cdots\circ\mathrm{Succ})(1).

Using this function and its partial inverse  one can define on \mathbb N the order and the operations of addition (as repeated addition of 1 which is just evaluation of \mathrm{Succ}) and multiplication (repeated addition).

However, not all equations of the form x+a=b or x\cdot a=b are solvable. One can enlarge \mathbb N by adding solutions of all such equations, obtaining the set of integer numbers \mathbb Z which is a commutative group with respect to the operation of addition, and finally the set of rational numbers \mathbb Q in which division is available by any nonzero number.

Division by zero is impossible: if we add “solution of the equation 0\cdot x=1” as a new imaginary element, then we will not be able to do some arithmetic operations on it. Still, if we are ready to pay this price, then the rational numbers can be extended by a new element so that, say, the function f(x)=1/x would be everywhere defined and continuous.

Details are available in the lecture notes here.


Sunday, November 13, 2011

Lecture 3, Nov 10, 2011 (Thu)

Number systems

Leopold Kronecker (1823-1891) famously quipped, “God made the natural numbers; all else is the work of man”. So we start working.

  1. Construct non-positive integers by adjoining “formal solutions” to the equations x+n=m for n\ge m
  2. Embed \mathbb N into \mathbb Z, identifying the above solution with the difference m-n for m>n.
  3. Define arithmetic operations on these “new numbers” via manipulations with the corresponding equations.
  4. Prove that with the “new numbers” the addition operation is always invertible, and x+n=m is always solvable with any n,m\in\mathbb Z.
  5. The construction can be essentially reproduced to define fractions as “formal solutions” to the equations of the form qx=p with p,q\in\mathbb Z. To avoid an obvious non-uniqueness, consider only case where p,q do not vanish simultaneously.
  6. Derive the formulas for addition/subtraction and multiplication/division. Note that these formulas sometimes give the forbidden combination 0\cdot x=0.
  7. Two ways to solve the problem:
    • keep the addition/subtraction always defined, but exclude the root of 0\cdot x=1, or
    • keep the “ideal element” and have a nice picture and lots of simplification in geometry, but live with arithmetic prohibitions.
  8. The ring \mathbb Q of rationals and the “circle” of the “rational projective line” \mathbb Q P^1:

Download the pdf file for the complete exposition.

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