Given that a and b are integers such that a = b + 1, Prove: 1 = 0
 
1. a = b + 1 1. Given
2. (a-b)a = (a-b)(b+1) 2. Multiplication Prop. of =
3. a2 - ab = ab + a - b2 - b 3. Distributive Property
4. a2 - ab -a = ab + a -a - b2 - b 4. Subtraction Prop. of =
5. a(a - b - 1) = b(a - b - 1) 5. Distributive Property
6. a = b 6. Division Property of =
7. b + 1 = b 7. Transitive Property of = (Steps 1, 7)
8. Therefore, 1 = 0 8. Subtraction Prop. of =


Proof that zero equals two (Using Algebra)
Given that a and b are integers such that a = b, Prove: 0 = 2
 
1. a = b 1. Given
2. a - b - 2 = a - b - 2 2. Reflexive Prop. of =
3. a(a - b - 2) = b(a - b - 2) 3. Multiplication Prop. of =
4. a2 - ab - 2a = ab - b2 - 2b 4. Distributive Property
5. a2 - ab = ab - b2 - 2b + 2a 5. Addition Property of =
6. a2 - ab = ab + 2a - b2 - 2b 6. Associative Property of +
7. a(a - b) = a(b + 2) - b(b + 2) 7. Distributive Property
8. a(a - b) = (a - b)(b + 2) 8. Distributive Property
9. a = b + 2 9. Division Property of =
10. b = b + 2 10. Transitive Property (steps 1, 9)
11. Therefore, 0 = 2 11. Subtraction Property of =


The difficulty with both "proofs" is a division by zero error.
In the proof that 1 = 0, you divide by zero when you go from step 5 to step 6.
In the proof that 0 = 2, you divide by zero when you go from step 8 to step 9.