Professor Flashman will discuss some examples of models from geometry, algebra, and set theory that provide proofs for both the possible and the impossible.
" The supreme concept with which it is customary to begin a transcendental philosophy is the division into the possible and the impossible.Early examples of the possible and the impossible from Greek mathematics.
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4. The object of a concept which contradicts itself is nothing, because the concept is nothing, is the impossible, e.g. , a twosided rectilinear figure."
Given a segment AB of one unit length, it is possible to construct a segment AC of two units length.
Proof: Draw a circle with center B and radius AB. Extend the radius AB to meet the circle at C Then the diameter AC has measure two units. EOP
Given a square ABCD of one unit area, it is possible to construct a square of two units area.
Proof: Construct the square on the diagonal AC of the given square. Then the square on AC has area of two square units. EOP
Given a square, it is not possible to find a segment for a common measuring unit between the side and the diagonal of that square.
Proof: Apply the Euclidean algorithm in an attempt to find a common measuring unit between the diagonal and a side of a square.This will fail. For suppose there is a common measuring unit for the two sides..
Now count the number of 2's on each side of the equation.
P^{2} and Q^{2 }must each
have an even number of 2's as factors.
[ P^{2 } has twice as
many 2's as P and similarly Q^{2 } has twice as many 2's as
Q ]
But then on the left side of the equation there are an even number of 2's as factors while on the right side there are an odd number of 2's. [One more than the even number of 2's that appear as factors of Q^{2}.]
This is impossible because there are (must be) the same number of 2's on both sides of the equation P^{2}=Q^{2} *2.
Thus
there is no segment that can be a common measuring
unit. EOP
In the context of real numbers, the statement(1) It is possible to solve the equation x^{2} = 2 with real numbers.
In the context of rational numbers, the statement"There is a number x with x^{2} = 2" is true! (2) It is not possible to solve the equation x^{2} = 2with rational numbers.
Issue: How can the same statement can be true in one context and false in another?
Response:There is a fundamental difference between the two contexts for the statements.
Furthermore, if we isolate all the statements that are true about both contexts, then the given statement cannot be a logical consequence of any collection of those statements.
[Why? Because the logical consequences of true statements about a context are also true about the context.]
Definition: A set of statements Tis consistent if for any statement A, it is not possible that both A and the negation of A are in T.
Remark: If a set of statements S has a model, then the set of all logical consequences T of the statements in the set (which includes the original set of statements) are consistent.
Algebraic or analytic: Contexts: Integers vs. Rational
Numbers vs. Real Numbers vs .....
[Take abstract algebra Math 343... Take real analysis Math 415] 
Geometric: Contexts: Points and/or lines in Euclidean
plane vs. projective plane vs. hyperbolic plane vs...
[Take geometry Math 371.] 
There is a solution to a linear equation.
3x=24; 3x=13 
Partition segments: 2AC= AB; KAC=AB
Proportions of line. AB:DE::AC:DX (Intersections of parallel lines/ similiar triangles) 
There is a solution to a quadratic equation. x^{2 }= 2.  bisect angle, double a square
Intersection of line with circle, circle with circle. [Rational vs. real geometry] 
There is a solution to a cubic equation. x^{3
}=
2.
There is a solution to a polynomial equation. 
Angle trisection is possible/impossible!
The duplication of a cube is possible/impossible. 
There exist algebraic irrational and transcendental numbers.  There are "nonconstructible" lengths. 
There is a common solution to a pair of linear equations.

A pair of lines has a common point.

There is a positive number h>0, so that: if a>0, then h < a.  If AB has length h and CD has length a and then there is some counting number N where Nh>a. 
There is a nonzero number x with nx = 0  There is a way to move an oriented object to change its orientation. 
An infinite subset of the natural numbers, integers, rational or algebraic numbers can be "counted"  The set of segments "constructible" from a given length can be counted. 
[a,b], [c,d]; [a,b], [a,b); [a,b], and the circle Set theory vs. "bicontinuous" 1:1 functions  A line segment, A Ray, A Circle. Set Theoretic vs. "bicontinuous" 1:1 geometric functions 
Any subset of the real numbers can be counted [can be measured.]  Any set of points on a line segment can be counted. [can be measured.] 
Issue (revisited): How can the same statement, A, can
be true in one context and false in another?
For the first model consider all statements that are true in both contexts together with the statement A that is true in the first context. We'll call this set of statements S1. Then the context in which the statement A is true is a model for S1.
For the second model consider all statements that are true in both contexts together with the negative of the statement A. We'll call this set of statements S2. Then the context in which the statement A is false is a model for S2.
But the existence of models for both S1and S2 means that it is not possible to prove either the statement A or its negation ~ A from the collection of statements true for both contexts.
Thus when a statement, A, and its negation, ~A, are true (possible) in different but related contexts it is not possible to prove the statement A from a set of statements true in both contexts.
In this sense we say that the statement A (and its negation ~A) is independent from any subset of the statements true for both contexts.
Definition: A set X is countable if there is a function f mapping the natural numbers onto X, so that for any x in X there is some natural number n where f (n) = x, i.e. X= { f (1), f (2), f (3), ... }
1/1  2/1  3/1  4/1  5/1  6/1  ... 
1/2  2/2  3/2  4/2  5/2  6/2  ... 
1/3  2/3  3/3  4/3  5/3  6/3  ... 
1/4  2/4  3/4  4/4  5/4  6/4  ... 
1/5  2/5  3/5  4/5  5/5  6/5  ... 
1/6  2/6  3/6  4/6  5/6  6/6  ... 
...  ...  ...  ...  ...  ...  ... 
0, 1/1, 1/2, 2/1,
3/1, 2/2, 1/3, 1/4, 2/3, 3/2,
4/1, ...
EOP
^{EOP}
Proof: More Godel counting. Each symbol is given a
number used for an exponent. Prime numbers are used as the base for each
symbol number, indicating the position of symbol in the statement of the
proposition. Each statement corresponds uniquely to the product of powers
of prime numbers.
EOP
Theorem: If a formal mathematical system is consistent, then there is a model for that system in which the underlying set of objects is countable.
Proof: Take Math 446. :)
Remarks:
There are formal mathematical systems for Set Theory which includes a formal theory for the real numbers.
Such a formal system can be demonstrated to be "relatively" consistent and is believed to be consistent.
Thus, by the Lowenheim  Skolem Theorem there is a model for the theory of real numbers in which the set of objects that correspond to real numbers in this model is a countable set and yet the theory says that the set of real numbers is not countable. How is this possible????
Fact 1. An algebraic number is the solution to a polynomial equation with integer coefficients.
Fact 2. A polynomial equation of degree N can have at most N distinct real number solutions.
Fact 3. There are a finite number of integer polynomial
equations of degree N where the sum of the absolute values of the coefficients
is less than a given number K.
Proof Procedure: (modified
slightly from Cantor's)
First list algebraic number roots from integer polynomials
of degree 1 with sum of the absolute values of the coefficients equal to
1. (This is just 1x=0 and 1x=0.)
Then use degree 1, sum 2;
(This is 2x=0, 2x=0, 1+1x=0, 1+1x=0, 11x=0, 11x=0.)
then degree 2, sum 1;
then degree 3, sum 1;
then degree 2, sum 2;
then degree 1, sum 3;. ...
EOP
Consider the set S of real numbers determined by having
the 2k th decimal be a 5 for every k.
For example: .358595953525450505050505... is in S.
A real number b in S is determined uniquely by
a sequence of integers, with each integer between 0 and 9, where the kth
integer of the sequence corresponds the 2k+1 th decimal place of b.
Suppose there is a function, f, from N to S.
Then let b=.b1 5 b3 5 b5 5 b7 5 b9 5 ... where we choose
b2k+1 so that it is not equal to the 2k+1st digit of f(k).
Thus b is not equal to f(k) for any k and f is not "onto."
Hence S is not a countable set. ( And so neither is [0,1].)
EOP
Suppose f: R > P(R).
Let B= {x such that x is not an element of f(x)}.
Suppose B = f(b) for some b.
If b is in B then b is not in f(b)=B, which is a contradiction.
So b is not in B, but then b is not in f(b), so b is
in B!
Thus B is not f(b) for any b, and f is not onto.
EOP