Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

Search

  • Filter Results
  • Location
  • Classification
    • Article type
    • Stage
    • Author
    • Embed Hypothes.is?
    • Cover Page
    • License
    • Show Page TOC
    • Transcluded
    • PrintOptions
    • OER program or Publisher
    • Autonumber Section Headings
    • License Version
    • Print CSS
    • Screen CSS
  • Include attachments
Searching in
About 1 results
  • https://math.libretexts.org/Courses/Coastline_College/Math_C285%3A_Linear_Algebra_and_Diffrential_Equations_(Tran)/10%3A_Some_Prerequisite_Topics/10.02%3A_Well_Ordering_and_Induction
    Let \(T\) consist of all integers larger than or equal to \(a\) which are not in \(S.\) The theorem will be proved if \(T=\emptyset .\) If \(T\neq \emptyset\) then by the well ordering principle, ther...Let \(T\) consist of all integers larger than or equal to \(a\) which are not in \(S.\) The theorem will be proved if \(T=\emptyset .\) If \(T\neq \emptyset\) then by the well ordering principle, there would have to exist a smallest element of \(T,\) denoted as \(b.\) It must be the case that \(b>a\) since by definition, \(a\notin T.\) Thus \(b\geq a+1\), and so \(b-1\geq a\) and \(b-1\notin S\) because if \(b-1\in\) \(S,\) then \(b-1+1=b\in S\) by the assumed property of \(S.\) Therefore, \(b-…

Support Center

How can we help?