anonymous
  • anonymous
Ordered Fields S={a+b(sqrt2) | a,b in Q} I need to show that this set is an ordered field under addition and multiplication. The first property of addition states: A1: For all x,y in R, x + y in R and if x=w and y=z, then x + y = w + z. What do I substitute to show this. Is x = a, or is x = a+b(sqrt2)???
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
@zzr0ck3r Could you help maybe ?
anonymous
  • anonymous
A field (F, + ,×) together with a total order ≤ on F is an ordered field if the order satisfies the following properties: if a ≤ b then a + c ≤ b + c if 0 ≤ a and 0 ≤ b then 0 ≤ a×b is this right?
anonymous
  • anonymous
Those look like a few of the order axioms. But I believe what I need to show is that S is closed with respect to only addition and multiplication. But maybe that is saying that A1 and M1 doesnt apply. I'm really not sure the more I read the more confused I am getting.

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anonymous
  • anonymous
There are also 5 properties under addition, 5 properties under multiplication and one distributive. Do I need to show all 11, or just 1 of each.
anonymous
  • anonymous
You only need to show closure as the question indicates, since everything comes from \(R\) all the other properties fall from the tree. Yes, x needs to equal the form of the yet to be proven field. So, \(x=a_1+b_1\sqrt{2}\) and \(x=a_2+b_2\sqrt{2}\). Sum them and show they have the same form, and multiply them and show they have the same form, that is, the form \(a+b\sqrt{2}\).
anonymous
  • anonymous
There is a theorem that states you only need three thing to show it is field if the elements come from a field. Closure under both addition and multiplication, and the unit is in the set.
anonymous
  • anonymous
(a+b(sr2))+(c+d(sr2))=(a+c)+(b+d)(sr2) (a+b(sr2))(c+d(sr2))=(ac+2bd)+(ad+bc)(sr2) 1/(a+b(sr2))=a/(a^2-2b^2)+(-b/(a^2-2b^2))(sr2) So how does that look? Is that good enough to show what I want?
anonymous
  • anonymous
Yes, add to that: Since \(a+b \in R\) and \(b+d \in R\) by closure of \(R\), it follows \( (F,+,R) \) is closed. Since \(ac+2bd \in R\) and \(ad+bc \in R\) by closure of \(R\), it follows \((F,*,R) \) is closed. For division, which is really multiplication by an inverse, is the same with a caviat, as long as \(a^2-b^2 \neq0\).
anonymous
  • anonymous
does it matter that sqrt2 is irrational though. I know a and b are in Q so then they are in R. But since it is a+b(sqrt2), does it matter that it is irrational? Or are you only worried about what a and b are?
anonymous
  • anonymous
It does not matter. Only the form of the elements matters, along with the properties described above.(closure under +, closure under *, the unit)

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