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

    13
    ReputationRep:
    Hi!

    As part of my homework, I have been set the following question:

    'You are given that y is not 0, and that x>y. Now suppose that \sqrt{x-y} = \sqrt{x} - \sqrt{y}.

    (a) Show that (\sqrt{x} - \sqrt{y})^2 = x - 2 \sqrt{x} \sqrt{y} +y.
    (b) Deduce that y(x-y) =0, and hence that either y=0 or x=y.
    (c) What can you deduce about \sqrt{x-y} and \sqrt{x} - \sqrt{y}?'

    I've done the first two parts, but I'm not sure what to write for part (c).

    While writing this, I've noticed that x=y is not possible, because earlier in the question it states that x>y. Hence I assume the answer must be to say that y must be 0, but I may be missing something? Is that really all I have to write?
    Offline

    15
    ReputationRep:
    (Original post by K-Man_PhysCheM)
    Hi!

    As part of my homework, I have been set the following question:
    'You are given that y is not 0, and that x>y. Now suppose that \sqrt{x-y} = \sqrt{x} - \sqrt{y}.

    (a) Show that (\sqrt{x} - \sqrt{y})^2 = x - 2 \sqrt{x} \sqrt{y} +y.
    (b) Deduce that y(x-y) =0, and hence that either y=0 or x=y.
    (c) What can you deduce about \sqrt{x-y} and \sqrt{x} - \sqrt{y}?'

    I've done the first two parts, but I'm not sure what to write for part (c).

    While writing this, I've noticed that x=y is not possible, because earlier in the question it states that x>y. Hence I assume the answer must be to say that y must be 0, but I may be missing something? Is that really all I have to write?
    You're also given that y is not 0.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by B_9710)
    You're also given that y is not 0.
    Oh yeah... OK, so I just say that the supposed \sqrt{x-y} = \sqrt{x}-\sqrt{y} is false/impossible? Thank you!
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by K-Man_PhysCheM)
    Oh yeah... OK, so I just say that the supposed \sqrt{x-y} = \sqrt{x}-\sqrt{y} is false/impossible? Thank you!
    Both solutions are not possible therefore \sqrt{x-y} \not= \sqrt{x} - \sqrt{y}
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by RDKGames)
    Both solutions are not possible therefore \sqrt{x-y} \not= \sqrt{x} - \sqrt{y}
    Ok, thank you!
 
 
 
  • 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
    Brussels sprouts
    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.