The Problems Associated with Division by Zero • by Darryl J. Engler • © April 13, 1972

Theorems

The following theorems can be proved using field postulates and definitions along with the definition:

 a 
0 		=  0  a  
 
 
Theorem 1:
 a 
b 		= a  1 
b 		a  
 
Proof:
Case I (b 0)
(definition of division)
 a 
b 		= a  1 
b
 
Case II (b = 0)
 a 
0 		= 0 = a 0 = a  1 
0
 

 
Theorem 2:
 1 
b 		 1 
d 		= 1
 bd  		  b, d  
 

Proof:
Case I: (b 0, d 0)
1 =  bd  1
 bd 
 1 
b 		=  1 
b 		bd 1
 bd 
 1 
b 		 1 
d 		=  1 
d 		d 1
 bd 
 1 
b 		 1 
d 		= 1
 bd 
 

Case II: (b = 0, d any real number)
0 = 0
0  1 
d 		=  1 
0
 1 
0 		 1 
d 		= 1
 0 d 
 

 
Theorem 3:
 a 
b 		 c 
d 		=  ac 
bd 		  a, b, c, d  
 
Proof:
 a 
b 		 c 
d 		= (a  1 
b 		) (c  1 
d 		) by Theorem 1
= ac  1 
b 		 1 
d 		by Commutative and Associative properties
= ac 1
 bd  		by Theorem 2
=  ac 
bd 		by Theorem 1
 
 
Theorem 4:
 a 
c 		+  b 
c 		=  a + b 
c 		  a, b, c  
 
Proof:
 a 
c 		+  b 
c 		= a  1 
c 		+ b  1 
c 		by Theorem 1
= (a + b)  1 
c 		by Distributive property
=  a + b 
c 		by Theorem 1
 
 
Theorem 5:
   1   
1
x 		= x   x  

 
 
Proof:
Case I (x 0)
   1   
1
x 		=

 

    1   
1
x
 

  x 
x
 

   Identity for Multiplication
 
 
=
 

    x   
x
x 		by Theorem 3
 
=  x 
1 		Identity for Multiplication
 
= x
 
Case II: (x = 0)
   1   
1
0 		=

 

  1 
0

 

 = 0

 
 
 
Theorem 6:
 a 
b 		 c 
d 		=  ad 
bc 		  a, b, c, d  
 
Proof:
 a 
b 		 c 
d 		=    a   
b
c
d
 
= a
b
c  1 
d
      by Theorem 1
 
=    a   
b
c
 

  
   1   
1
d 		  by Theorem 3
 
=    a   
b
c
 
 d   by Theorem 5
 
=   ad  
b
c 		by Theorem 3
 
=  ad 
b 		 1 
c 		  by Theorem 1
 
=  ad 
bc 		by Theorem 3
 
 
It seems interesting to note that some theorems can be simplified both in their statement, i.e., removing restrictions, and in their proofs. Many theorems no longer require two cases, one for zero and one for non-zero values of the variable (theorems 3, 4, and 6 for example.)

Unfortunately, this does not hold for all theorems. Cases must still be used for theorems such as the following:

 
Theorem 7:
 x 
x 		= 1 if x 0,    x 
x 		= 0 if x = 0.
 
 
The rule for addition of rational numbers can now be derived from Theorem 4, and can be shown not to apply to terms such as a/0:

 
Theorem 8:
Case I: Case II:
If b 0, d 0, then  a 
b 		+  c 
d 		=  ad + bc 
bd 		If b = 0, d any real number, then  a 
b 		+  c 
d 		=  c 
d
 
Proof:
Case I: (b 0, d 0)
 a 
b 		+  c 
d 		=  a 
b 		 d 
d 		 +   b 
b 		 c 
d 		  Since b 0, b/b=1 (Theorem 7)
Since d 0, d/d=1 (Theorem 7)
 
=  ad 
bd 		+  bc 
bd 		by Theorem 3
 
=  ad + bc 
bd 		by Theorem 4
 
Case II: (b = 0, d any real number)
 a 
0 		+  c 
d 		= 0 +  c 
d
 
=  c 
d
 
The Problems Associated with Division by Zero • by Darryl J. Engler • © April 13, 1972