Given that a and b are integers such that a = b + 1, | Prove: 1 = 0 |
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1. a = b + 1 | 1. Given |
2. (a-b)a = (a-b)(b+1) | 2. Multiplication Prop. of = |
3. a^{2} - ab = ab + a - b^{2} - b | 3. Distributive Propoerty |
4. a^{2} - ab -a = ab + a -a - b^{2} - b | 4. Subtraction Prop. of = |
5. a(a - b - 1) = b(a - b - 1) | 5. Distributive Propoerty |
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 |
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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. a^{2} - ab - 2a = ab - b^{2} - 2b | 4. Distributive Propoerty |
5. a^{2} - ab = ab - b^{2} - 2b + 2a | 5. Addition Property of = |
6. a^{2} - ab = ab + 2a - b^{2} - 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. Thereforem 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.
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