Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    The following question has me quite puzzled.
    x^2>x for x>1

    I don't know how I can prove this to be right or wrong.
    Offline

    17
    ReputationRep:
    (Original post by ShredAlert)
    The following question has me quite puzzled.
    x^2>x for x>1

    I don't know how I can prove this to be right or wrong.
    You can safely multiply through inequalities by positive quantities (like x here, for example) and not have to worry about the sign changing direction.

    You need to start from what you know and show what you want; what is it that you know about x here?
    Offline

    11
    ReputationRep:
    (Original post by ShredAlert)
    The following question has me quite puzzled.
    x^2>x for x>1

    I don't know how I can prove this to be right or wrong.
    Perhaps:
    Spoiler:
    Show
    Consider x = (1 + d)
    Where d is very small.
    Expand the brackets and show that x2 > x
    Then consider x = (1 - d) and show that when x < 1 the above inequality does not hold.
    • Thread Starter
    Offline

    0
    ReputationRep:
    So this is what I got so far.
    x2>x
    x2-x>0
    x(x-1)>0

    kind of hit a brick wall after this.
    Offline

    16
    ReputationRep:
    (Original post by ShredAlert)
    So this is what I got so far.
    x2>x
    x2-x>0
    x(x-1)>0

    kind of hit a brick wall after this.
    If x>1 is x positive or negative
    If x>1 is (x-1) positive or negative
    Offline

    11
    ReputationRep:
    (Original post by ShredAlert)
    So this is what I got so far.
    x2>x
    x2-x>0
    x(x-1)>0

    kind of hit a brick wall after this.
    If x(x-1)>0 both factors in the product must be either greater than 0 or less than 0.
    (Negative x Negative), (Positive x Positive) both make positive.
    What does this tell you about x?
    Offline

    10
    ReputationRep:
    (Original post by ShredAlert)
    So this is what I got so far.
    x2>x
    x2-x>0
    x(x-1)>0

    kind of hit a brick wall after this.
    Since x is positive you can just divide by x.
    Offline

    17
    ReputationRep:
    (Original post by ShredAlert)
    So this is what I got so far.
    x2>x
    x2-x>0
    x(x-1)>0

    kind of hit a brick wall after this.
    Generally speaking, you shouldn't start from what you're trying to show - usually it's bad logic (although it's not so much a problem here given that the steps are reversible, but still).

    What you know is that x>1. Can you start from this and get x^2>x? (hint: it's literally a one step proof using the first line of my previous post).
    Offline

    17
    ReputationRep:
    (Original post by joostan)
    If x(x-1)>0 both factors in the product must be either greater than 0 or less than 0.
    (Negative x Negative), (Positive x Positive) both make positive.
    What does this tell you about x?
    (Original post by metaltron)
    Since x is positive you can just divide by x.
    Without further explanation/comments, all that you've done here is shown that x^2>x => x>1; which is not what the OP is trying to show.
    Offline

    11
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Without further explanation/comments, all that you've done here is shown that x^2>x => x>1; which is not what the OP is trying to show.
    Having worked through surely one could then work back and show that this is true such that x>1 <=> x2 > x (or x<0)and hence the question.
    Or have I misunderstood. . .
    Offline

    2
    ReputationRep:
    Wouldn't a sketch be of any help? :/
    Offline

    10
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Without further explanation/comments, all that you've done here is shown that x^2>x => x>1; which is not what the OP is trying to show.
    The further explanation was that x is positive, so you can have double implication arrows.
    Offline

    17
    ReputationRep:
    (Original post by joostan)
    Having worked through surely one could then work back and show that this is true such that x>1 <=> x2 > x
    Or have I misunderstood. . .
    That wasn't evident from your post.

    (Original post by metaltron)
    The further explanation was that x is positive, so you can have double implication arrows.
    Your double implications would require even further explanation - when showing that x^2 > x => x>1 (which is not necessarily true), you're not allowed to assume x>0.
    Offline

    10
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Of course, but that wasn't evident from either post.

    As for Metaltron's; your double implications would require even further explanation - when showing that x^2 > x => x>1, you're not allowed to assume x>0.
    well it says x>1 .
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Generally speaking, you shouldn't start from what you're trying to show - usually it's bad logic (although it's not so much a problem here given that the steps are reversible, but still).

    What you know is that x>1. Can you start from this and get x^2>x? (hint: it's literally a one step proof using the first line of my previous post).
    According to your suggestion I have done the following.
    If x>1
    multiplying both sides by x gives us x2>x
    therefore the statement is true?
    Please tell me if this is correct and if so sufficient in detail.
    Thanks for all the help.
    Offline

    17
    ReputationRep:
    (Original post by metaltron)
    well it says x>1 .
    Which is exactly where the flaw in your logic lies. You claim to be able to use double implications here but, given that the converse is false, this isn't possible. (i.e. x^2>x ≠> x>1) So far, you've claimed that x^2>x, divided by x (assuming that x>1) and shown that x>1.

    (Original post by ShredAlert)
    According to your suggestion I have done the following.
    If x>1
    multiplying both sides by x gives us x2>x
    therefore the statement is true?
    Please tell me if this is correct and if so sufficient in detail.
    Thanks for all the help.
    Yes.
    Offline

    10
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Which is exactly where the flaw in your logic lies. You claim to be able to use double implications here but, given that the converse is false, this isn't possible. (i.e. x^2>x ≠> x>1) So far, you've claimed that x^2>x, divided by x (assuming that x>1) and shown that x>1.


    Yes.
    I assumed x>0 not x>1. So the end argument says:

    Since x>1, x >0 so x^2>x. The other way round is if x is positive x^2 > x goes to x > 1 which is consistent with x being positive. Also, this is so messy that I have to make clear that I was just finishing off what the OP wrote.
    Offline

    0
    ReputationRep:
    It's straight forward.

    x²>x and we know that x>1.
    So you can divide both sides by x since x is not 0, and also the sign wont change since x is positive.

    we get
    x²/x>x/x which is equivalent to x>1: this is true. thus proven.


    If you want a more complicated solution just prove by induction.
    Offline

    17
    ReputationRep:
    (Original post by Giveme45)
    It's straight forward.

    x²>x and we know that x>1.
    So you can divide both sides by x since x is not 0, and also the sign wont change since x is positive.

    we get
    x²/x>x/x which is equivalent to x>1: this is true. thus proven.


    If you want a more complicated solution just prove by induction.
    You're starting with what you're trying to show, which you shouldn't do.

    x&gt;1 implies that x-1&gt;0 so (x-1)(x-1)&gt;0

    You can do this last step because one of the axioms is that if a&gt;0, b&gt;0 then ab &gt;0

    x^2 - 2x + 1 &gt; 0

    x^2 &gt; 2x -1

    Now using the fact that x&gt;1

    x^2 &gt; x + x -1 &gt; x

    There's probably a much quicker and neater way of doing it though.
    Offline

    17
    ReputationRep:
    (Original post by metaltron)
    I assumed x>0 not x>1.
    Why would you do that?

    Since x>1, x >0 so x^2>x.
    That's fine. (in fact, this finishes the entire proof as TenofThem alludes to earlier in the thread [and is not what you said in any of your above posts!])

    The other way round is if x is positive x^2 > x goes to x > 1 which is consistent with x being positive.
    Not fine - you're not allowed to assume x>0 here. You need to show that if x^2>x then x>1 here without any further assumptions about x (including x>0).

    [Unless you're doing something odd like showing that if x>0 then (x^2>x <=> x>1) - but I have no idea why you'd be doing that here given that it deviates from the question at hand before coming back to it]
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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