Turn on thread page Beta

Clever way of showing which of these two fractions is greater? watch

Announcements
    • Thread Starter
    Offline

    1
    ReputationRep:
    Anyone got a nifty way to show which of the following is larger?

    1) 1/(1+x)

    2) (1/x) - 1
    Offline

    0
    ReputationRep:
    (Original post by Economister)
    Anyone got a nifty way to show which of the following is larger?

    1) 1/(1+x)

    2) (1/x) - 1
    For what x?
    Offline

    14
    ReputationRep:
    Suppose 1) = 2)

    1/(1+x) = (1/x) - 1

    1 = (x+1)/x - 1 - x
    1 = 1 + 1/x - 1 - (x^2)/x
    x = -x^2 + 1
    x^2 + x - 1 = 0

    the roots of that are

    -1.61803

    and

    0.61803

    If x > 0.61803 , 1) > 2)

    If -1.61803 < x < 0 then 1) > 2)
    If 0 < x < 0.61803 then 2) > 1)

    If x < -1.61803 then 1) > 2)


    That's the best that can be done. Neither is DEFINITELY larger than the other. But 1) is much more likely to be bigger as 2) is only bigger in a very small range.

    Also they are not comparable if x = 0 or x = -1 as one will be infinite.


    EDIT: I think I have corrected my inequalities now
    Offline

    12
    ReputationRep:
    (Original post by hassi94)
    Suppose 1) = 2)

    1/(1+x) = (1/x) - 1

    1 = (x+1)/x - 1 - x
    1 = 1 + 1/x - 1 - (x^2)/x
    x = -x^2 + 1
    x^2 + x - 1 = 0

    the roots of that are

    -1.61803

    and

    0.61803

    If x > 0.61803 , 1) > 2)

    If -1.61803 < x < 0.61803 then 2) > 1)

    If x < -1.61803 then 1) > 2)


    That's the best that can be done. Neither is DEFINITELY larger than the other. But 1) is much more likely to be bigger as 2) is only bigger in a very small range.

    Also they are not comparable if x = 0 or x = -1 as one will be infinite.
    A rather dangerous statement me-thinks ...... I could equally argue that either statement is true for an infinite "number" of values ....
    Offline

    14
    ReputationRep:
    (Original post by H.C. Chinaski)
    A rather dangerous statement me-thinks ...... I could equally argue that either statement is true for an infinite "number" of values ....
    Yeah you're right, but he's doing economics not maths so it would be applied more to real life and that's why I derived that statement.

    But yeah I agree probably not the best statement to make
    Offline

    15
    ReputationRep:
    (Original post by hassi94)
    If -1.61803 < x < 0.61803 then 2) > 1)
    Take x=-0.1:

    1/(1+(-0.1)) = 1.1111....

    1/(-0.1) - 1 = -11

    2) < 1)

    The mistake you made was you jumped from talking about equalities to inequalities without properly considering the signs of everything you multiply each side by, causing the inequality to flip round.

    The approach I would take would be to consider 1/x - 1 - 1/(1+x) and see when it's positive, negative or undefined.
    Spoiler:
    Show
    It helps to add fractions and factorise
    Offline

    10
    ReputationRep:
    1) 1/(1+x)

    2) (1/x) - 1
    Offline

    14
    ReputationRep:
    (Original post by ttoby)
    Take x=-0.1:

    1/(1+(-0.1)) = 1.1111....

    1/(-0.1) - 1 = -11

    2) < 1)

    The mistake you made was you jumped from talking about equalities to inequalities without properly considering the signs of everything you multiply each side by, causing the inequality to flip round.

    The approach I would take would be to consider 1/x - 1 - 1/(1+x) and see when it's positive, negative or undefined.
    Spoiler:
    Show
    It helps to add fractions and factorise
    Yeah sorry my bad just realised how wrong that was haha. Don't know why I didn't consider the fact I was multiplying by an unknown quantity and how that could affect the inequality.

    My bad! :P
    Offline

    11
    ReputationRep:
    (Original post by hassi94)
    Suppose 1) = 2)

    1/(1+x) = (1/x) - 1

    1 = (x+1)/x - 1 - x
    1 = 1 + 1/x - 1 - (x^2)/x
    x = -x^2 + 1
    x^2 + x - 1 = 0

    the roots of that are

    -1.61803

    and

    0.61803

    If x > 0.61803 , 1) > 2)

    If -1.61803 < x < 0.61803 then 2) > 1)

    If x < -1.61803 then 1) > 2)


    That's the best that can be done. Neither is DEFINITELY larger than the other. But 1) is much more likely to be bigger as 2) is only bigger in a very small range.

    Also they are not comparable if x = 0 or x = -1 as one will be infinite.
    I think you have the range of values incorrect possibly because you haven't use inequalities and somewhere you may have got a sign wrong - I haven't checked fully though. For this, I would use the sign test (but before you do this you will have to collect fractions and make a table of the critical points).

    Edit: LOL posted at the same time - oh well
    Offline

    14
    ReputationRep:
    (Original post by ttoby)
    Take x=-0.1:

    1/(1+(-0.1)) = 1.1111....

    1/(-0.1) - 1 = -11

    2) < 1)

    The mistake you made was you jumped from talking about equalities to inequalities without properly considering the signs of everything you multiply each side by, causing the inequality to flip round.

    The approach I would take would be to consider 1/x - 1 - 1/(1+x) and see when it's positive, negative or undefined.
    Spoiler:
    Show
    It helps to add fractions and factorise
    I've corrected it now I believe.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 9, 2011

University open days

  1. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  2. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
  3. Birmingham City University
    General Open Day 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.