Hi,
I am trying to find a solution for the following problem:
F is given as follows:
F = {(a,b) where a,b are rational numbers}
Let's define:
(a, b) + (c, d) = (a + c, b + d)
(a, b) * (c, d) = (ac+2bd, ad + bc)
Prove that F is an algebraic field
I was able to prove that all the axioms hold here, till I needed to find the
zero, the one, and the inverse the "a in the power of -1". Please forgive my
English, I am not a native speaker.
For the Zero, I found it must be (0, 0)
For the "One" (assume (c,d) is the one) I get
ac + 2bd = a
ad + bc = b
Which can happen for arrbitrary (a,b) if c=1 and d=0.
That is the one is (1, 0)
Again, assume (c,d) is the inverse.
So I need to find c and d such that
ac+2bd = 1
ad+bc = 0
for any a,b
I can't find any because a and b are in both terms.
I suspect my choice of the one is not so successful or maybe this can't be
proved at all?
Thanks for you help
I am trying to find a solution for the following problem:
F is given as follows:
F = {(a,b) where a,b are rational numbers}
Let's define:
(a, b) + (c, d) = (a + c, b + d)
(a, b) * (c, d) = (ac+2bd, ad + bc)
Prove that F is an algebraic field
I was able to prove that all the axioms hold here, till I needed to find the
zero, the one, and the inverse the "a in the power of -1". Please forgive my
English, I am not a native speaker.
For the Zero, I found it must be (0, 0)
For the "One" (assume (c,d) is the one) I get
ac + 2bd = a
ad + bc = b
Which can happen for arrbitrary (a,b) if c=1 and d=0.
That is the one is (1, 0)
Again, assume (c,d) is the inverse.
So I need to find c and d such that
ac+2bd = 1
ad+bc = 0
for any a,b
I can't find any because a and b are in both terms.
I suspect my choice of the one is not so successful or maybe this can't be
proved at all?
Thanks for you help