Question

which of the following is not a field. (N,+,.),
a. (Q,,+,.)
b. (Q/(2^(1/2)),+,.)
c. (R,+,.)
d. (N,+,.)​

Answer 1

Answer:

(N,+,.) is not a field.

Step-by-step explanation:

Field:

A field is a set F, containing at least two elements, on which two operations + and. (called and multiplication, respectively) are defined so that for each pair of elements x, y in F there are unique elements x+y and x.y in F for which of the following conditions hold for all elements

x, y, z in F.

• x+y = y +x              (commutative under addition)

• x+(y+z) =  (x+y)+z  (associative under addition)

• There is an element 0 ∈ F, such that x+0 = x. (Identity under addition)

• For each x, there is an element -x ∈ F such that x+(-x) = 0 (Inverse under addition)

• xy = yx                    (commutative under multiplication)

• (x.y).z = x.(y.z)        (associative under multiplication)

• (x+y).z = (x.z)+(y.z)  (Distributive property)

• There is an element 1 ∈ F , such that 1 ≠ 0 and x.1 = x.  

• If x ≠ 0, then there exist an element x⁻¹ ∈ F such that x . x⁻¹ = 1  

By the definition of Field, we can check whether (N,+, .) is a field or not.

• since 0 ∉ N.              

• (N, +) does not have an additive identity.

So, (N, +) is not a Field.  

• All the remaining options (Q, +), (Q/$\sqrt{2}$,+,.), (R,+,.) are field as all these satisfies all the above conditions of the field.

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Answer 2

Answer: Option (d) is the correct option.

(N,+,.) is not a field.

Step-by-step explanation:

Option a has (Q,+,.) which is the field of rational numbers containing fractions of the form $\frac{a}{b}$ , where a, b belongs to Z and $b\neq 0$.

Additive inverse clearly exists such that $-\frac{a}{b} +\frac{a}{b} =\frac{a}{b}+(-\frac{a}{b})=0$.

Multiplicative inverse with $a\neq 0$ is $\frac{b}{a}$ such that $\frac{b}{a} .\frac{a}{b} = \frac{a}{b} .\frac{b}{a} =1$

Option b has .(Q/(2^(1/2)),+,.) is a subfield of (Q,+,.)

Option c has (R,+,.) contains both additive inverse as well as multiplicative inverse, hence it is also a field.

Lastly, in option d, (N,+,.) does not have additive inverse as well as multiplicative inverse since for any a belonging to N, additive inverse $-a$ does not exist and for any element b belonging to N, $\frac{1}{b}$ does not exist in N.

Therefore, (N, +,.) is not a field.