Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Something that just popped into my mind...does there exist an  n\in \mathbb{R} \setminus \mathbb{Q} such that:
     (-1)^n \in \mathbb{R}?
    Could someone come up with a a proof that either confirms or refutes this?
    Offline

    0
    ReputationRep:
    so far as n is a whole number and not a fraction - you should always get real numbers
    Offline

    2
    ReputationRep:
    n = 1/3
    Offline

    18
    ReputationRep:
    (-1)^n = \exp ( n \log (-1) ) = \exp  ( n (i \pi + 2ki \pi)) = \exp(i \pi n(2k+1)) (k an integer)

    For this to be real, we need n(2k+1) = m (m an integer)

    Therefore n is rational. (Specifically a rational of the form \frac{m}{2k+1})
    Offline

    0
    ReputationRep:
    (Original post by SimonM)
    (-1)^n = \exp ( n \log (-1) ) = \exp  ( n (i \pi + 2ki \pi)) = \exp(i \pi n(2k+1)) (k an integer)

    For this to be real, we need n(2k+1) = m (m an integer)

    Therefore n is rational. (Specifically a rational of the form \frac{m}{2k+1})
    So basically as long as the denominator is odd it'll work. Nice proof by the way!
    • PS Helper
    Offline

    14
    PS Helper
    For \alpha \not \in \mathbb{Q} and z \in \mathbb{C} we define z^{\alpha} = \exp ( \alpha \log z ) = \exp (\alpha (\log |z| + i\arg z) ) = |z|^{\alpha} (\cos (\alpha \arg z) + i\sin (\alpha \arg z)), which real only when \alpha \arg z = k\pi for some k \in \mathbb{Z}. Can this happen when z=-1 and \alpha \in \mathbb{R} \backslash \mathbb{Q}? [Bear in mind that the interval on which you can choose arg z to lie is arbitrary.]

    (Original post by imzir)
    so far as n is a whole number and not a fraction - you should always get real numbers
    (Original post by Bobifier)
    n = 1/3
    The OP said that n \in \mathbb{R} \backslash \mathbb{Q}, so neither whole numbers nor 1/3 are suitable choices for n here.
    • Thread Starter
    Offline

    0
    ReputationRep:
    So in answer to my question, no. Thanks for clarifying that.
    Offline

    2
    ReputationRep:
    (Original post by nuodai)
    The OP said that n \in \mathbb{R} \backslash \mathbb{Q}, so neither whole numbers nor 1/3 are suitable choices for n here.
    I realised that as I posted, and at the time I'm sure I meant to delete it, but somehow it didn't quite seem to come to anything...
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 4, 2011

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.