PROOF 111
Prove.
If AB = 0, then A = 0 or B = 0
On a field.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
The multiplicative inverse of a is the element a⁻¹ satisfying a·a⁻¹ = 1.
Proof.
(1)Here 1 is the identity of A7.A7
(2)This definition does not ask whether an inverse exists. A8 answers that.convention
∎
Lemma L1
For every a, a × 0 = 0
Proof.
(1)0 + 0 = 0A2
(2)a(0 + 0) = a × 0(1), E
(3)a×0 + a×0 = a×0(2), A4
(4)Add −(a×0) to both sides.(3), E
(5)left side = a×0 + (a×0 + (−(a×0))) = a×0 + 0 = a×0A3, A1, A2
(6)right side = 0A1
(7)a × 0 = 0(4), (5), (6)
∎
Theorem T1
In a field, if AB = 0, then A = 0 or B = 0
Proof.
(1)Suppose AB = 0 and A ≠ 0.assumption
(2)Since A ≠ 0, there is an A⁻¹ with A·A⁻¹ = 1.(1), A8, D1
(3)A⁻¹(AB) = A⁻¹ × 0(1), (2), E
(4)A⁻¹ × 0 = 0L1
(5)A⁻¹(AB) = (A⁻¹A)B = 1 × B = B(2), A5, A6, A7
(6)B = 0(3), (4), (5)
(7)Hence A = 0 or B = 0.(1), (6)
∎
Theorem T2
In arithmetic mod 6, the conclusion of T1 fails.
Proof.
(1)Take 0, 1, 2, 3, 4, 5 as elements, with sum and product defined as remainders upon division by 6.convention, A9
(2)In this arithmetic, A1–A7 are inherited directly from the integers.A9, convention
(3)2 × 3 = 6 and 6 ≡ 0, so in this arithmetic 2 × 3 = 0.A9, A10
(4)2 ≠ 0 and 3 ≠ 0.A12
(5)The product is 0, yet neither factor is 0.(3), (4)
(6)If there were a t with 2t ≡ 1, then 6 would divide 2t − 1.A9
(7)Multiples of 6 are even and 2t − 1 is odd, so no such t exists.(6), A11
(8)2 has no multiplicative inverse. This arithmetic lacks A8.(7), D1
(9)When a product is 0 but neither factor is 0, the two elements are called zero divisors.(5), convention
∎
Corollary C
2×2 square matrices also have zero divisors.
Proof.
(1)Let A be the matrix whose only nonzero entry is 1 at (1,1), and B the matrix whose only nonzero entry is 1 at (2,2).convention
(2)Each of the four entries of AB is a sum of terms of the form 1×0 or 0×0.(1), A13
(3)All those values are 0, so AB = O.(2), L1, A2
(4)A ≠ O and B ≠ O.(1)
(5)Hence A and B are zero divisors.(3), (4), T2
(6)If there were an A⁻¹ with A⁻¹A = I, then A⁻¹(AB) = (A⁻¹A)B = IB = B and A⁻¹O = O.convention, A14, A15
(7)Then B = O, contradicting (4).(3), (4), (6)
(8)Hence A has no inverse matrix.(7)
∎
Remark
Can A8 be removed from this proof?
See what the proof of T1 uses to erase A. Also whether anything else could take its place.