Please kindly explain in detail. Thanks so much! 1. With usual operations of sum
ID: 3168588 • Letter: P
Question
Please kindly explain in detail. Thanks so much!
1. With usual operations of summation and multiplication, mark those of the following subsets of R which are fields. a) The set Q of all rational numbers b) The set Ui = {a + bvfla, b,E Q)(a, b are rational); c) The set U2 of all rational numbers, which can be presented as an irreducible fraction where a, b are integers, 3 divides a d) The set Us of all real numbers satisfying a 5; e) The set U of all real numbers which can be presented as a su+ b, where a and b are rational.Explanation / Answer
We know that a field must satisfy the following Axioms:
Field Axioms for Addition :
1. The operation of addition is closed, that is x,y F , x+y F (Closure under addition).
2. The operation of addition is commutative, that is x y F, x+y = y+x (Commutativity of addition).
3. The operation of addition is associative, that is x,y z F, x+(y+z)=(x+y)+z (Associativity of addition).
4. The operation of addition has the additive identity element of 0 such that x, x+0 = x (Existence of an additive identity).
5. An element x F has the additive inverse element denoted by -x such that x, x+(x)=0 (Existence of an additive inverse).
Field Axiom for Multiplication:
1. The operation of multiplication is closed, that is x,y F, xy F (Closure under multiplication).
2. The operation of multiplication is commutative, that is x,y F, xy = yx (Commutativity of multiplication).
3. The operation of multiplication is associative, that is x,y,z F, x(yz) = (xy)z (Associativity of multiplication).
4. The operation of multiplication has the multiplicative identity element of 1 such that x F, 1x=x (Existence of a multiplicative identity).
5. An element x F has the multiplicative inverse element denoted by x-1 such that x, x-1 x= xx-1 =1 (Existence of a multiplicative inverse).
Field Axiom for Distributivity:
The operation of multiplication is distributive over addition, that is x,y,z F, x(y+z) = xy+xz (Distributive law).
We will now check whether the given algebraic structures satisfy the field axioms:
a). The set Q of all rational numbers satisfies all the Field Axioms (-p/q is the additive inverse of p/q and q/p is the multiplicative inverse of p/q). Therefore, (Q,+,.) is a field.
b).Let x = a1+b17 , y = a2+b27, and z = a3+b37 be 3 arbitrary elements of U1. Then x+y = (a1+b17) + (a2+b27) = (a1+a2)+(b1+b2) 7 U1 so that U1 is closed under addition ( a1+a2, and b1+b2 are both rational numbers as the set of all rational numbers is closed under addition). Further, since operation of addition is commutative and distributive over the field of real numbers it is commutative and distributive over U1 also. Also, 0 is the additive identity and a-b7 is the additive inverse of a+ b7. Hence U1 satisfies all the field axioms for addition.
Further, xy= (a1+b17)(a2+b27) = (a1a2+7b1b2)+ (a1b2+a2b1) 7 U1 so that U1 is closed under multiplication. ). Further, since operation of multiplication is commutative and distributive over the field of real numbers it is commutative and distributive over U1 also. ). Also, 1 is the multiplicative identity for U1. Further, 1/(a+b7) = (a-b7)/(a2-7b2)= [a/(a2-7b2)]- [b/(a2-7b2)]7 is the multiplicative inverse of a+b7. Hence U1 satisfies all the field axioms for multiplication.
Also, since the operation of multiplication is distributive over addition in the field of real numbers, the distributive law is also satisfied for U1.
Hence, (U1, +, .) is a field.
(c ). the irreducible fractions 3/5 U2 but the multiplicative inverse of 3/5 i.e. 5/3 U2 as 5 is not divisible by 3. Hence U2 is not closed under multiplication, and, therefore, (U2, +, .) is not a field.
(d). The numbers 3, and 5 both U3, but (3)(5) = 15 U3 as 15 > 5. Hence U3 is not closed under multiplication, and, therefore, (U3, +, .) is not a field.
(e). Let x = (a1/2)+b1 , y = (a2/2)+b2, and z = (a3/2)+b3 be 3 arbitrary elements of U4. Then x+y =(a1/2)+b1 + (a2/2)+b2 = (a1+a2)/2 +(b1+b2) U4 so that U4 is closed under addition ( a1+a2, and b1+b2 are both rational numbers as the set of all rational numbers is closed under addition). Further, since operation of addition is commutative and distributive over the field of real numbers it is commutative and distributive over U4 also. Also, 0 is the additive identity and - (a/2)- b is the additive inverse of (a/2)+ b. Hence U4 satisfies all the field axioms for addition.
Further, xy=[(a1/2)+b1][ (a2/2)+b2] = (a1a2/2 +b1 b2) + (a1b2+a2b1)/2 U4 so that U4 is closed under multiplication. Further, since operation of multiplication is commutative and distributive over the field of real numbers it is commutative and distributive over U4 also. Also, 1 is the multiplicative identity for U4. Further, 1/[(a/2)+b] = [(a/2)-b]/[(a2/2-b2) = [a/(a2/2-b2)]/2 – b/(a2/2-b2) is the multiplicative inverse of (a/2)+b. Hence U4 satisfies all the field axioms for multiplication.
Also, since the operation of multiplication is distributive over addition in the field of real numbers, the distributive law is also satisfied for U4.
Hence, (U4, +, .) is a field.