MHB Proving the Equality of A and B: A+B=0 or A=B

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The discussion focuses on proving that if AA equals BB, then either A plus B equals zero or A equals B. It begins by establishing basic algebraic properties such as commutativity, associativity, and the existence of additive and multiplicative identities. The proof utilizes these properties to manipulate the equation AA=BB, ultimately leading to the conclusion that if A plus B is not zero, then A must equal B. The proof is structured logically, relying on axioms and established mathematical principles to reach the final statement. Thus, the conclusion is that the relationship between A and B is defined by either their sum being zero or them being equal.
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Given for all A,B,C...

1) A+B=B+A......AB=BA

2) A+(B+C)=(A+B)+C.......A(BC)=(AB)C

3) A+0=A..........1A=A

4) $$\forall A\exists B(A+B=0)$$.....$$\forall A(\neg(A=0)\Longrightarrow\exists B(AB=1))$$

5) A(B+C)= AB+AC

Then prove: $$AA=BB\Longrightarrow (A+B)=0\vee A=B$$

Note: AB means A.B ...e.t.c ,e.t.c
 
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solakis said:
Given for all A,B,C...

1) A+B=B+A......AB=BA

2) A+(B+C)=(A+B)+C.......A(BC)=(AB)C

3) A+0=A..........1A=A

4) $$\forall A\exists B(A+B=0)$$.....$$\forall A(\neg(A=0)\Longrightarrow\exists B(AB=1))$$

5) A(B+C)= AB+AC

Then prove: $$AA=BB\Longrightarrow (A+B)=0\vee A=B$$

Note: AB means A.B ...e.t.c ,e.t.c

Please post the solution you have ready. :)
 
solakis said:
Given for all A,B,C...

1) A+B=B+A......AB=BA

2) A+(B+C)=(A+B)+C.......A(BC)=(AB)C

3) A+0=A..........1A=A

4) $$\forall A\exists B(A+B=0)$$.....$$\forall A(\neg(A=0)\Longrightarrow\exists B(AB=1))$$

5) A(B+C)= AB+AC

Then prove: $$AA=BB\Longrightarrow (A+B)=0\vee A=B$$

Note: AB means A.B ...e.t.c ,e.t.c

[sp] Proof:

1) AA=BB................Given

2) AA+AB=BB+AB..........additive property of equality

3) A(A+B)=B(B+A)............by axiom 5

4) A(A+B)=B(A+B)............by axiom 1

5) $$A+B\neq 0$$................assumption

6)$$\exists C[C(A+B)=1]$$..............by axiom 4

7) [C(A+B)]=1......................fix C

8) C[A(A+B)]= C[B(A+B)].................by the multiplicative property of equality

9) C[(A+B)A]=C[(A+B)B]..................by axiom 1

10) [C(A+B)]A= [C(A+B)]B..................by axiom 2

11) 1.A= 1.B......................by substituting 7 into 10

12) A=B........................by axiom 3

13) $$A+B\neq 0\Longrightarrow A=B$$..........Closing the assumption that we started at step 5

14) A+B=0 V A=B................This is equivalent to formula (13)Hence: If AA=BB ,then A+B=0 or A=B [/sp]
 
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