The following proofs of elementary ring properties use only the axioms that define a mathematical ring: Theorem: 0 ⋅ a = a ⋅ 0 = 0 (Click "show" at right to see the proof of this theorem or "hide" to hide it.) 0 ⋅ a = (0 + 0) ⋅ a = (0 ⋅ a) + (0 ⋅ a) By subtracting (i.e. adding the additive inverse of) 0 ⋅ a on both sides of the equation, we get the desired result. The proof that a ⋅ 0 = 0 is similar. Theorem: A ring (R, +, ⋅) is trivial (that is, consists of precisely one element) if and only if 0 = 1. (Click "show" at right to see the proof of this theorem or "hide" to hide it.) Suppose 1 = 0. Let a be any element in R; then a = a ⋅ 1 = a ⋅ 0 = 0. Therefore, (R, +, ⋅) is the trivial ring. Conversely, if (R, +, ⋅) is trivial, it must contain precisely one element. Therefore, 0 and 1 is the same element, i.e. 0=1. Theorem: (−1)a = −a (Click "show" at right to see the proof of this theorem or "hide" to hide it.) (-1)· a + a = (-1)· a + 1 · a = ((-1) + 1) · a = 0 · a = 0 Therefore (-1) · a = (−1)a + 0 = (−1) · a + (a + (−a))= ((−1) · a + a) + (−a) = 0 + (−a) = (−a). Theorem 3: (−a) ⋅ b = a ⋅ (−b) = −(ab) (Click "show" at right to see the proof of this theorem or "hide" to hide it.) To prove that the first expression equals the second one, (−a) ⋅ b = ((-1)⋅a) ⋅ b= (a⋅ (-1)) ⋅ b = a⋅ ((-1) ⋅ b) = a(−b). To prove that the first expression equals the third one, (−a) ⋅ b = ((-1)⋅a) ⋅ b = (-1) ⋅ (a ⋅ b). A pseudo-ring does not necessarily have a multiplicative identity element. To prove that the first expression equals the third one without assuming the existence of a multiplicative identity, we show that (−a) ⋅ b is indeed the inverse of (a ⋅ b) by showing that adding them up results in the additive identity element, (a ⋅ b) + (−a) ⋅ b = (a − a) ⋅ b = 0 ⋅ b = 0. Template:Algebra-stub
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