Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Need help with analysis questions...

  • 27-11-2009 1:28pm
    #1
    Registered Users, Registered Users 2 Posts: 329 ✭✭


    1: Give a complete, formal proof of the following statement: If a E R, then there exists a negative
    integer n such that n < a.

    where R= real numbers and E = is an element

    I found most of these analysis questions extremely hard as I dont know where to start with them?? Can anyone give me a helping hand to make a start on this one...

    Cheers

    Nappy


Comments

  • Closed Accounts Posts: 6,151 ✭✭✭Thomas_S_Hunterson


    This page may help:
    http://en.wikipedia.org/wiki/Floor_and_ceiling_functions

    /edit: It's not identical, kind of related, but possibly confusing things.


  • Registered Users, Registered Users 2 Posts: 329 ✭✭Nappy


    Well as the real numbers is just the number line from negative infinity to positive infinity the statement seems obviously correct, But how to prove it?? All I can really take from above link is that n< floor a but dont know how to transform all this into a formal proof.

    Thanks for help


  • Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Presumably you were given a list of axioms for the real numbers in the context of this class. You need to use combinations of these axioms along with proof by contradiction to derive the result.
    If you post up the axioms I can help you more, of you're still stuck.


  • Registered Users, Registered Users 2 Posts: 2,149 ✭✭✭ZorbaTehZ


    If its [latex]n \in Q[/latex] then you can draw parallels from the theorem about integers that says given any [latex]x \in R[/latex] there is always some [latex]n \in N[/latex] such that [latex]n > x[/latex]. If you start assuming the converse is true, ie. [latex]n \geq a\ \forall_{n \in Q}[/latex] then you should be easily able to come up a contradiction - also you don't need that floor/ceiling stuff above at all to prove it.


Advertisement