Let Q[sqrt(3)]= {a + b(sqrt3) : a, b is in the set of rational numbers}?

Question

Show: Q[sqrt(3)] is a subfield of R (the real numbers).
I know to show this, the following must be true:
1) a + b is an element of Q
2) (a)(b) is an element of Q
3) Of is an element of Q
4) a + x = 0f has a solution in Q
5) 1f is an element of Q
6) if a is an element of Q and a is not equal to 0f, then the solution of (a)(x) =1f is in Q

I'm not really sure how to show any of these, so any help would be greatly appreciated! Thanks!

Answer 1

I assume that in the questions themselves, you mean a and b to be elements in Q, as opposed to the a,b used to describe Q. That's what would make sense, anyway. I'll use x,y as generic elements in Q, and let x=a+b(sqrt3), and y=c+d(sqrt(3)), where a,b,c,d are rational.

1) x+y=(a+c)+(b+d)(sqrt(3)), and a+c, b+d are rational because rationals are closed under addition, so a+b is in Q
2) xy=(a+b(sqrt3))(c+d(sqrt3)) =
(ac+3bd) + (bc+ad)(sqrt3), and ac+3bd, bc+ad are rational, so ab is in Q. To confirm these are rational, you can always write, for instance, a=p/q and b,c,d in similar form, and do the arithmetic, using the fact that the integers are closed under addition and multiplication, to show you get an integer over an integer again.
3) By Of I assume you mean the zero element. 0 is in Q, because you can set a,b=0.
4) We show x+y has a solution in Q, where x,y in Q. Let x=a+b(sqrt3), y=c+d(sqrt3) as before. Let c=-a, and d=-b. Then if a,b are rational, so are c,d. So y is in Q, and x+y=0.
5) 1 is in Q: let x=a+b(sqrt3), a=1,b=0. Then x=1.
6) We show xy=1 has a solution, where x=a+b(sqrt3), y=c+d(sqrt3). Assume x is nonzero. Then 1/x = 1/(a+b(sqrt3)). Multiply top and bottom by (a-b(sqrt3)). Then
1/x = (a-b(sqrt3)) / (a^2 - 3b^2). Then let c=a/(a^2-3b^2), and d=-b/(a^2-3b^2). Then y=c+d(sqrt3) is in Q, and xy=1.

Answer 2

take x=a+b(sqrt3)
y=c+d(sqrt3)
x+y=a+b(sqrt3)+c+d(sqrt3)
=(a+c)+(b+d)sqrt3,
so it is an element of Q[sqrt3]

3) 0 = 0+0sqrt3, therefore is an element of Q[sqrt3]

4) suppose you have A=a+b sqrt3 and
x=y+w sqrt(3) (this is the variable)
such that
A+x=0
then
a+b sqrt3 + y+w sqrt(3)=0
so a+y=0, => y=-a
and b+w=0 => w=-b,
and there you have, the solution is unique.

5) 1 =1 +0sqrt3, so it is an element of Q[sqrt3]

hopefully you will get started with these ones

Answer 3

For answer (c) i'm guessing that's going to study "c)Q is closed for addition yet P isn't closed for addition." the respond is (a) because of the fact, as Puggy suggested Q isn't closed below addition, on the grounds that pi and -pi are in Q yet their sum is in P. besides the undeniable fact that, P is closed below addition, because of the fact for any 2 rational numbers a/b and c/d, their sum is (advert + bc)/bd, this is rational.