Prove that the multiplicative identity (from (g) in the list in Axiom I) is uniq

Question

Prove that the multiplicative identity (from (g) in the list in Axiom I) is unique (i.e. that there is only one element satisfying (g)).

Axiom I (the algebraic axiom): there is a set R, whose elements are called real numbers, equipped with binary operations so that for every pair c,y of real numbers there is a real number r + y (the sum) and a real number r y (the product). There are real numbers 0, 1 and 0 ^ 1. The following rules are satisfied: (a) x + y = y + x z, y R (commutativity of addition) (b) r + (y +z) (r +y)+zV r,y,z ER (associativity of addition). (c) x + 0 0 + x-x V x ? R (additive identities) (d) if x ER then 3 -rER such that ax + (-r)- 0 (additive inverses) (e) r-y-y r Vr,y ER (commutativity of multiplication) (f) x (y 2) -(x y) z V r,y,z E R (associativity of multiplication). (g) . 1-1 y x E R (Inultiplicative identities). (h) if r E R, 0, the3ER such that 1 (multiplicative inverses)

Explanation / Answer

Let us assume that the multiplicative identity is not unique and that there exists another multiplicative identity ‘e’ (say). Then , for all x ? R, we have x. e = x. Further, since x.1 = x ? x ? R, we have x.1 = x. Hence x.e = x.1 or, x.e-x.1 = 0 or, x.(e-1) = 0 . Now, since x is arbitrary, this means that e-1 = 0 . Therefore, e = 1.

Thus, the multiplicative identity is unique.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.