Results 1 to 5 of 5

Math Help - Discrete Quadratic Fields and Integers (Easy, Yes, No Examples)

  1. #1
    Member
    Joined
    Jun 2010
    From
    United States
    Posts
    200

    Discrete Quadratic Fields and Integers (Easy, Yes, No Examples)

    Hello All,

    I had a few finite examples from my text I had some questions about. They are discussing Q[Sqrt(-5)] , and I know that it is i*Sqrt(5) = Sqrt(-5). Given this information, it gives some examples and it said I should be able to identify which examples are in Q[Sqrt(-5)]

    Example 1: 2+3*Sqrt(-5)

    Example 2: (3+8*Sqrt(-5))/2

    Example 3: (3+8*Sqrt(-5))/5

    Example 4: 3/5

    Example 5: i*Sqrt(-5)

    From observation I inferred that Example 5 was in Q[Sqrt(-5)], but then i noticed that it says i*Sqrt(-5) and not i*Sqrt(5). I don't understand how Example 4 could be part of a negative rooted Quadratic field either.

    Could someone explain these? If you could, please provide a small reason as to why each one is or isn't part of Q[Sqrt(-5)].

    Thank you, all the help is appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Samson View Post
    Hello All,

    I had a few finite examples from my text I had some questions about. They are discussing Q[Sqrt(-5)] , and I know that it is i*Sqrt(5) = Sqrt(-5). Given this information, it gives some examples and it said I should be able to identify which examples are in Q[Sqrt(-5)]

    Example 1: 2+3*Sqrt(-5)

    Example 2: (3+8*Sqrt(-5))/2

    Example 3: (3+8*Sqrt(-5))/5

    Example 4: 3/5

    Example 5: i*Sqrt(-5)

    From observation I inferred that Example 5 was in Q[Sqrt(-5)], but then i noticed that it says i*Sqrt(-5) and not i*Sqrt(5). I don't understand how Example 4 could be part of a negative rooted Quadratic field either.

    Could someone explain these? If you could, please provide a small reason as to why each one is or isn't part of Q[Sqrt(-5)].

    Thank you, all the help is appreciated!
    So a + bi\sqrt{5} \in \mathbb{Q}(\sqrt{-5}) if and only if a,b\in\mathbb{Q}. Example 1 already matches the form, so you just look at it and say "yes". For example 2 you can do simple manipulation to get it into the form (3/2)+(8/2)*Sqrt(-5) which matches the form a + bi\sqrt{5}. Likewise for example 3. Example 4, let b = 0, which is acceptable. Example 5 is slightly trickier, just expand to get i*sqrt(-5) = i*i*sqrt(5) = -sqrt(5) which you should be able to see is not in Q[sqrt(-5)].

    Edit: Well I interchanged a + bi\sqrt{5} with a + b\sqrt{-5} without meaning to, but it should still be clear enough.
    Last edited by undefined; July 22nd 2010 at 12:45 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2010
    From
    United States
    Posts
    200
    Quote Originally Posted by undefined View Post
    So a + bi\sqrt{5} \in \mathbb{Q}[-5] if and only if a,b\in\mathbb{Q}. Example 1 already matches the form, so you just look at it and say "yes". For example 2 you can do simple manipulation to get it into the form (3/2)+(8/2)*Sqrt(-5) which matches the form a + bi\sqrt{5}. Likewise for example 3. Example 4, let b = 0, which is acceptable. Example 5 is slightly trickier, just expand to get i*sqrt(-5) = i*i*sqrt(5) = -sqrt(5) which you should be able to see is not in Q[-5].
    Thank you! A quick clarification, so when we have (3/2)+(8/2)*Sqrt(-5), it doesn't matter that a and b are not integers does it? And how do we check if they exist in Q[-5] ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Samson View Post
    Thank you! A quick clarification, so when we have (3/2)+(8/2)*Sqrt(-5), it doesn't matter that a and b are not integers does it? And how do we check if they exist in Q[-5] ?
    Notice the "Q" in Q[sqrt(-5)]. So you just check the definition, a+b*Sqrt(-5) where a and b are rational. There is nothing further to check.
    Last edited by undefined; July 21st 2010 at 06:17 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2010
    From
    United States
    Posts
    200
    Quote Originally Posted by undefined View Post
    Notice the "Q" in Q[-5]. So you just check the definition, a+b*Sqrt(-5) where a and b are rational. There is nothing further to check.
    Okay! Thank you very much for your help! Much props to be awarded!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fields and gaussian integers
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 9th 2010, 06:44 PM
  2. Primes in Quadratic Fields
    Posted in the Number Theory Forum
    Replies: 31
    Last Post: September 9th 2010, 02:09 PM
  3. [SOLVED] Quadratic Fields and Norms
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: July 21st 2010, 02:56 AM
  4. Examples of Identity of Finite Sum of Integers
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: April 11th 2010, 04:59 AM
  5. easy discrete RV question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 2nd 2010, 04:06 PM

/mathhelpforum @mathhelpforum