How do you prove the multiplicative inverse is unique?

Answer 1

You can multiply each side of the multiplicative inverse equation by the other inverse to show that any two multiplicative inverses are equal. Here it is more formally.

Theorem: For all x in R, there exists y in R s.t. x * y = 1. If there is a y' in R such that x * y' = 1, then y = y'.

Proof:

- Start with x * y = 1.

- y * x = 1 (commutative)

- (y * x) * y' = 1 * y' = y'

- y * (x * y') = y' (associative)

- y * 1 = y' (because x*y' = 1)

- y = y'

Answer 2

To prove that the multiplicative inverse is unique, we assume there are two inverses, say (a) and (b), for a given element (x). We then show that (a = b) by using the definition of the multiplicative inverse, which states that (a \times x = b \times x = 1). By multiplying both sides of (a \times x = 1) by the multiplicative inverse of (a), which is (a^{-1}), we get (a^{-1} \times (a \times x) = a^{-1} \times 1). Simplifying the left side gives us (1 \times x = a^{-1}), which is equal to (x = a^{-1}), confirming that the multiplicative inverse is unique.