Hi,
problem:
Let be the set of all real numbers of the form
, where are rational.
(a) Is a field?
(b) What if are required to be integers?
attempt:
(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form as long as are rational?
(b)
I am unable to find an axiom that "fails", but say I want to write the real number . I don't think I can do this if are integers..
Any comments are very welcome!
Thanks
Firstly, if every real number was of the form with then we can easily define a bijection , . As there exists a bijection between and we have that , a contradiction.
Secondly, in both of your problems you know that "most" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.
What is the inverse of with and rationals?
What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take . Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form . That is, every rational number of the form can be written as . As we are, allegedly, in a field we have that is also of this form, with . Thus, every rational number is of the form . This is silly!)