0≠1 as real numbers


Theorem.

The real numbers 0 and 1 are distinct.

There are four relatively common ways of constructing the real numbers. One can start with the natural numbersMathworldPlanetmath and augment it by adding solutions to particular classes of equations, ultimately considering either equivalence classesMathworldPlanetmathPlanetmath of Cauchy sequencesMathworldPlanetmathPlanetmath of rational numbers or Dedekind cuts of rational numbers. One can instead define the real numbers to be the unique (up to isomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath) ordered field with the least upper bound property. Finally, one can characterise the real numbers as equivalence classes of possibly infiniteMathworldPlanetmath strings over the alphabet  {0,1,2,3,4,5,6,7,8,9,.}  satisfying certain conditions. We offer a proof for each characterisation.

Cauchy sequences.

This construction proceeds by starting with a standard model of Peano arithmeticMathworldPlanetmathPlanetmath, the natural numbers ℕ, extending to ℤ by adding additive inverses, extending to ℚ by taking the field of fractions of ℤ, and finally defining ℝ to be the set of equivalence classes of Cauchy sequences in ℚ for an appropriately defined equivalence relation.

There is a natural embedding  i:ℕ→ℝ  defined by sending a given number x to the equivalence class of the constant sequenceMathworldPlanetmath (x,x,…).  Since i is injectivePlanetmathPlanetmath and 0 and 1 are elements of ℕ, to prove that  0≠1  in ℝ we need only show that  0≠1  in ℕ.

The name 1 is a label for the successorMathworldPlanetmathPlanetmath S⁢0 of 0 in ℕ. One of the axioms of Peano arithmetic states that 0 is not the successor of any number. Therefore  0≠S⁢0  in ℕ, and so  0≠1  in ℝ. ∎

Dedekind cuts.

This construction agrees with the previous one up to constructing the rationals ℚ. Then ℝ is defined to be the set of all Dedekind cuts on ℚ. Letting xℚ represent the name of an element of ℚ and xℝ represent the name of an element of ℝ, we define

0ℝ ={x∈ℚ|x<0ℚ}
1ℝ ={x∈ℚ|x<1ℚ}

The proof that 0ℚ≠1ℚ is similar to the previous proof. Observe that 0ℚ<1ℚ. Since no number is less than itself, it follows that 0ℚ∉0ℝ but 0ℚ∈1ℝ. Thus these Dedekind cuts are not equal. ∎

Ordered field with least upper bound property.

Here the fact that 0≠1 is a consequence of the field axiom requiring 0 and 1 to be distinct. ∎

Decimal strings.

If one defines

0 =(0,0,0,0,…)¯
1 =(1,0,0,0,…)¯

then since neither defining string ends with a tail of 9s and the strings differ in one position, their equivalence classes are distinct. ∎

Title 0≠1 as real numbers
Canonical name 0ne1AsRealNumbers
Date of creation 2013-03-22 15:23:15
Last modified on 2013-03-22 15:23:15
Owner mps (409)
Last modified by mps (409)
Numerical id 11
Author mps (409)
Entry type Theorem
Classification msc 54C30
Classification msc 26-00
Classification msc 12D99
Related topic DecimalExpansion