Infinity and Beyond Lecture 1 Infinity in Real
Infinity and Beyond
Lecture 1 Infinity in Real Arithmetic
What is infinity ( )? n It’s a “number” that is larger than all natural n n n numbers, i. e. for all n {0, 1, 2, 3, …}, n < . Questions: 1) Are there more than one infinity? 2) Can we perform algebraic operations with ? 3) Is + 1 = ? Is + = ? Is = ? 4) How about , / , 0 ? 5) Can we write 1/ = 0? How about 1/0? Answer: This depends on our definitions.
Is (or ) a real number? n We defined: For all n {0, 1, 2, 3, …}, n < . n Likewise: For all n {0, 1, 2, 3, …}, n > . n Question: Is (or ) a real number? n Answer: This depends on our definition of real numbers. n Usually we define real numbers in a way that excludes both and . n Question: What are real numbers?
The Set of Real Numbers R: n Definition 1: This is the set of all points on n n the real line. This definition is intuitive and visual, but faces the following problems: What does a line mean? Is a line physical? In this case are there infinitely many real numbers? Does the line have a beginning and an end? I. e. do and belong to it?
The Set of Real Numbers R: n Definition 2: This is the set of all decimal n n n infinite sequences of digits (including a decimal point). E. g. = 3. 14159265358979323846… Also 1/3 = 0. 3333333333… Caution: 1 = 1. 000000… = 0. 9999999… This set does not contain a sequence like … 9999…. (infinite in both directions). Thus, both , R.
The Set of Real Numbers R: n Definition 3: Instead of defining the set R, we n n define the structure (R, +, , <) by the following axioms: 1) (R, +, ) is a field, i. e. + and satisfy the usual properties, e. g. x (y + z) = x y + x z. 2) (R, <) is a linear order, i. e. for any x and y, either x < y or x = y or x > y, and the relation < is transitive, i. e. for all x, y, and z; x < y < z x < z. 3) < is congruent with respect to + and , i. e. for all x, y, and z; x < y x + z < y + z. Also x < y and z > 0 xz < yz. 4) Every nonempty subset of R that is bounded above, has a least upper bound.
Do and belong to R? n n n n Theorem: No. There is no real number that is larger than all natural numbers. Proof: If there were such a number (called ), then the set N of all natural numbers is bounded above by . Thus, using Axiom 4, we can get a least upper bound . From 1 < 0, it follows that 1 < . Since is a least upper bound of N, it follows that 1 is NOT an upper bound of N. Thus, there is an n N, such that 1 n. It follows that n + 1 < n + 2, contradicting the fact that is an upper bound of N.
The extended real numbers R* n In real arithmetic, we can choose to extend the set of real numbers R with the two new elements and , i. e. we define: R* = R { , }. n Note: We have only one and one . n (R*, <) is an extension of (R, <) by defining: n For all real numbers r, < r < , and < . n Note: (R*, <) is a linear order as before, i. e. for any x and y, either x < y or x = y or x > y, and the relation < is transitive, i. e. for all x, y, and z; x < y < z x < z.
Algebraic Properties of and n We extend the algebraic operations by the following definitions: For all r R, p R+, n R , nr+ = +r= + = n r + ( ) = ( ) + r = ( ) + ( ) = n p = p = n ( ) = ( ) n = = ( ) = n n = n = p ( ) = ( ) p = ( ) = ( ) = n 1/ = 1/( ) = 0 n Note: Since we can define x y = x + ( 1) y, and also x/y = x (1/y), subtraction and division involving and can be defined.
Example: Show that, for all r R, p R+, n R , n r = ( ) = n r = r ( ) = n r/ = r/( ) = 0 n /p = ( )/n = n /n = ( )/p =
Undefined Quantities involving and n The following quantities are left undefined: n + ( ), ( ) + , n and consequently , ( ), n 0 , 0, 0 ( ), ( ) 0, n and consequently / , /( ), ( )/( ) n Also, 1/0 is still not defined (informally, we can not choose between + and ), n and consequently /0, ( )/0 n Question: Why don’t we just define them as we like?
Answer: n If we tried defining the previous undefined quantities, we will ruin the following theorem. n Theorem: Whenever defined, arithmetic expressions involving and obey Axioms 1 and 3 of Slide 8. n E. g. , for all x, y, z R*=R { , }, the equation x + (y + z) = (x + y) + z holds, if both sides are defined. n Also, x < y and z > 0 xz < yz, etc. . n Proof: An exhaustive check of all properties.
Defining the Undefined n What goes wrong if we defined + ( ) = 0, say? n Answer: The required properties will not hold anymore. If they did, we can derive the contradiction 0 = 1 as follows: n 0 = + ( ) = (1 + ) + ( ) = 1 + ( )) = 1 + 0 = 1. n Exercise: Show that we can not define 0 , without ruining Axiom 1. Hint: Use 0 = 1 + ( 1) n What about 1/0?
Challenge! Give a definition of a set R** that contains all real numbers and (possibly many) infinities, with the operations of + and defined for ALL elements, and obeying all of Axioms 1, 2, 3.
Thank you for listening. Wafik
- Slides: 17