Abel's theorem in problems and solutions - School of Mathematics

148 Problems of Chapter 2
3.2 Problems of Chapter 2
194. a) Answer. No‚ because the integer numbers do not form a
group under addition (cf.‚ 17).
b) Answer. No‚ because the integers numbers without zero do not
form a group under multiplication (all numbers‚ except 1 and –1‚ have
no inverse).
c) Answer. Yes. Use the result of Problem 57.
Solution. Since the real numbers form a commutative group under
addition‚ and without zero also a commutative group under multiplication‚
it suffices to verify that the set of rational numbers form a subgroup
of the group of real numbers under addition‚ and without zero also under
multiplication. One obtains this easily using the result of Problem 57.
Indeed: 1) if and are rational then and are rational as
well; 2) 0 and 1 are rational; 3) if is a rational number‚ then and
(for are also rational. The distributivity is obviously satisfied.
Consequently rational numbers form a field.
d) Answer. Yes. Use the result of Problem 57.
Solution. If where and are rational numbers
and then it is not possible that because is not
a rational number. This means that if and being
rational numbers‚ then All numbers of the form
for different pairs are different‚ because if
then and Let us now prove
that all numbers of the form where and are rational‚ form
a field. To do this‚ we prove that the numbers of the form form
a subgroup under addition in the set of the real numbers‚ and also under
multiplication without zero. Using the result of Problem 57 we obtain:
1) if and then
and and 1 belong to the set
considered‚ because and if
then and (for
Since the distributivity is satisfied by all real numbers the considered set