Show that $F=\{a+b\sqrt3\;|\;a,b ∈Q\}$ is a field

Question

We just started Linear Algebra 2 and this is the question that was given to me:

Show that $F=\{a+b\sqrt3\;|\;a,b ∈Q\}$ is a field under operations ⊕ and ⊙, where ⊕ is given by $(a+b\sqrt3)\;⊕\;(c+d\sqrt3) = > (a+c)+(b+d)\sqrt3$ and ⊙ is given by $(a+b\sqrt3)\;⊙\;(c+d\sqrt3) = > (ac+3bd)+(ad+bc)\sqrt3$.

My answer started off something like this.

Addition and multiplication on $F$ will generate elements in $F$. Therefore $F$ is closed under addition and multiplication. Since the components are also elements of $Q$, associative, commutative and distributive properties will also hold.

And now I have to prove the existence of identity elements and existence of inverses.

Additive identity : $(a+b\sqrt3)\;⊕\;0 = (a+0)+(b+0)\sqrt3=a+b\sqrt3$

Multiplicative identity : $(a+b\sqrt3)\;⊙\;1 = (a∙1)+b\sqrt3 = a+b\sqrt3$ - For this part however, I'm not sure if I did this correctly. I know I should be following the directions that they gave to multiply, but the value is simply 1. What would I have to do in this scenerio?

Then I had to prove the inverses, and the additive inverse was easy enough, I proved it like so.

Additive inverse: $-(a+b\sqrt3) = (-a-b\sqrt3)$

Multiplicative inverse: ??

So I got lost at this point and I searched around, and found a similar post, but I have a few questions. (I'm going to change the posts numbers to relate with my question)

In the post, the additive inverse was written as $0+0\sqrt{3}$, but my initial thought was to simply write $0$. Why is the other way the right way to write it? In the end isn't $0+0\sqrt{3}$ equal to zero regardless?
Same goes for the multiplicative identity portion, it was written as $1+0\sqrt{3}$, but my thought was to simply write $1$. Why is $1+0\sqrt{3}$ the proper way to write it (I'm assuming)

And lastly, I'm wondering if anyone has a good place where I could read/watch a video on how to do these type of questions. Normally I refer to my textbook but for some reason this chapter, and this chapter alone, isn't in our textbook.

Thank you

My mistake haha the arrow came out of random, didn't mean for that to be there in the first place — Hello Mellow, Sep 13 '17 at 06:53
You could use commands to make the arrows, it just use $=$, since that is what you actually mean. — Arthur, Sep 13 '17 at 09:06
Last time I checked there were infinitely many rational numbers, so the resulting field is not finite. Nor is there a vector field (= a concept from multivariable calculus and its abstractions) in sight. So I removed tags [tag:finite-fields] and [tag:vector-fields]. Please be careful in picking tags. — Jyrki Lahtonen, Sep 13 '17 at 09:11

score 0 · Answer 1 · answered Sep 13 '17 at 06:52

0

Some details just needed to be filled as you pointed out the multiplicative inverse $(a+b\sqrt{3})^{-1} = \dfrac{1}{a+b\sqrt{3}}= \dfrac{a-b\sqrt{3}}{a^2-3b^2}= \dfrac{a}{a^2-3b^2} + \dfrac{-b}{a^2-3b^2}\sqrt{3}$. The rest of your work is fine.

answered Sep 13 '17 at 06:52

DeepSea

77,651

Oh I see, so it's okay to simply write out "The additive inverse is $0$" rather than $0+0\sqrt{3}$
And there was another question similar to this, but F was written as $[a,b]$. How would the multiplicative inverse of that be found? I know now you just take the conjugate for my previous question, but what about for $[a,b]$
– Hello Mellow Sep 13 '17 at 06:56
@HelloMellow No, you should write "The additive inverse is $0+0\sqrt3$". That's because the additive inverse is always called $0$, and the fact that you know what it should be called does not demonstrate that the element exists, or which element it really is. – Arthur Sep 13 '17 at 09:08

Show that $F=\{a+b\sqrt3\;|\;a,b ∈Q\}$ is a field

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