exists an element b in A such that a ∗ b = e = b ∗ a and b is called the inverse of a and is denoted by a – 1.
Numerical: Show that – a is the inverse of a for the addition operation‘+’ on R and 1 / a is the inverse of a ≠ 0 for the multiplication operation‘×’ on R.
Solution:-a + a = 0 = a +(-a), so – a is inverse of a, where a ∈ R 1 / a * a = 1 = a * 1 / a where a ∈ { R-0 }, for a = 0, this will not hold true Thus-a is additive inverse of a & 1 / a is multiplicative inverse of a.