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

    5
    ReputationRep:
    Hi, I've done the proof for the question reading "prove that x element of R, is an invert know element IF N(x)=1. And just wanted to know if I had went wrong anywhere! Also I'm stuck on the last part, the deducing the invertible elements ? Be great if anyone could help thanks! Attachment 511063

    Name:  ImageUploadedByStudent Room1457381978.578258.jpg
Views: 73
Size:  123.1 KB


    Posted from TSR Mobile
    • Thread Starter
    Offline

    5
    ReputationRep:
    @Zacken


    Posted from TSR Mobile
    • Thread Starter
    Offline

    5
    ReputationRep:
    @Bath_Student say again?


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by maths10101)
    ...
    Heya! Your tag didn't work so I didn't get a notification and this is in the wrong forum (should be in maths, in the future stick to posting questions in the maths forum), sorry for not seeing it sooner. I'll have a look at your question now.
    Offline

    22
    ReputationRep:
    (Original post by maths10101)
    ...
    Okay, let's do the deduce bit. The only invertible elements of the ring are going to satisfy zw = 1 where w is the inverse to z. Then you know that |z|^2 |w|^2 = |zw|^2 = 1 from the fact that N(x) = 1, yeah?

    Now try writing z = a+ib and w = c+ id and see if you can take it from there. If not:

    more hints
    This is the same thing as saying (a^2 +b^2)(c^2 + d^2) = 1 since z = a+ ib and w = c+id.


    What does this imply about a^2 + b^2 and c^2 + d^2?

    More hint
    You know that (because ring), a, b, c, d \in \mathbb{Z}. So a^2 + b^2 and c^2 +d^2 are both non-negative integers which must both equal 1 since they need to multiply to 1 in the integers.


    You know that a^2 + b^2 = 1, so... what does that mean for a^2 and b^2?

    More hints
    So you need a^2 ,b^2 \leq 1 which means that you should be able to easily check your solutions are of the form (a,b) = (\pm 1, 0) or...? I'll let you fill in the blanks.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by Zacken)
    Okay, let's do the deduce bit. The only invertible elements of the ring are going to satisfy zw = 1 where w is the inverse to z. Then you know that |z|^2 |w|^2 = |zw|^2 = 1 from the fact that N(x) = 1, yeah?

    Now try writing z = a+ib and w = c+ id and see if you can take it from there. If not:

    more hints
    This is the same thing as saying (a^2 +b^2)(c^2 + d^2) = 1 since z = a+ ib and w = c+id.


    What does this imply about a^2 + b^2 and c^2 + d^2?

    More hint
    You know that (because ring), a, b, c, d \in \mathbb{Z}. So a^2 + b^2 and c^2 +d^2 are both non-negative integers which must both equal 1 since they need to multiply to 1 in the integers.


    You know that a^2 + b^2 = 1, so... what does that mean for a^2 and b^2?

    More hints
    So you need a^2 ,b^2 \leq 1 which means that you should be able to easily check your solutions are of the form (a,b) = (\pm 1, 0) or...? I'll let you fill in the blanks.
    A lot of thanks for this help mate!..I'll look into it in a second to ensure I am okay with it...however in terms of the picture I uploaded, is the method/what I've done, look okay to show N(X)=1?




    Posted from TSR Mobile

    @Zacken
    Offline

    22
    ReputationRep:
    (Original post by maths10101)
    A lot of thanks for this help mate!..I'll look into it in a second to ensure I am okay with it...however in terms of the picture I uploaded, is the method/what I've done, look okay to show N(X)=1?




    Posted from TSR Mobile

    @Zacken

    Your tagging still isn't working!

    I think your method is okay.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by Zacken)
    Okay, let's do the deduce bit. The only invertible elements of the ring are going to satisfy zw = 1 where w is the inverse to z. Then you know that |z|^2 |w|^2 = |zw|^2 = 1 from the fact that N(x) = 1, yeah?

    Now try writing z = a+ib and w = c+ id and see if you can take it from there. If not:

    more hints
    This is the same thing as saying (a^2 +b^2)(c^2 + d^2) = 1 since z = a+ ib and w = c+id.


    What does this imply about a^2 + b^2 and c^2 + d^2?

    More hint
    You know that (because ring), a, b, c, d \in \mathbb{Z}. So a^2 + b^2 and c^2 +d^2 are both non-negative integers which must both equal 1 since they need to multiply to 1 in the integers.


    You know that a^2 + b^2 = 1, so... what does that mean for a^2 and b^2?

    More hints
    So you need a^2 ,b^2 \leq 1 which means that you should be able to easily check your solutions are of the form (a,b) = (\pm 1, 0) or...? I'll let you fill in the blanks.
    Alrite, so the blanks would just be
    "of the form (+-1,0) or (0,+-i). correct?
    or is there more of this proof?
    Offline

    22
    ReputationRep:
    (Original post by maths10101)
    Alrite, so the blanks would just be
    "of the form (+-1,0) or (0,+-i). correct?
    or is there more of this proof?
    Nopes, that's pretty much it - methinks.
 
 
 
  • 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
    What's your favourite Christmas sweets?
    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.