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

Jun 2010
200
3
United States
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!
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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 \(\displaystyle a + bi\sqrt{5} \in \mathbb{Q}(\sqrt{-5})\) if and only if \(\displaystyle a,b\in\mathbb{Q}\). 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 \(\displaystyle a + bi\sqrt{5}\). 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 \(\displaystyle a + bi\sqrt{5}\) with \(\displaystyle a + b\sqrt{-5}\) without meaning to, but it should still be clear enough.
 
Last edited:
  • Like
Reactions: Samson
Jun 2010
200
3
United States
So \(\displaystyle a + bi\sqrt{5} \in \mathbb{Q}[-5]\) if and only if \(\displaystyle a,b\in\mathbb{Q}\). 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 \(\displaystyle a + bi\sqrt{5}\). 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] ?
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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:
  • Like
Reactions: Samson
Jun 2010
200
3
United States
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!