| 1. | (n+1)2 = n2+2n+1 | Binomial Expansion | |
| 2. | (n+1)2-(2n+1) = n2 | Euclid's Common Notion (3) | |
| 3. | (n+1)2-(2n+1)-n(2n+1) = n2-n(2n+1) | ditto | |
| 4. | (n+1)2-(n+1)(2n+1) = n2-n(2n+1) | Factorization | |
| 5. | Euclid's Common Notion (2) | ||
| 6. | [(n+1)-(2n+1)/2]2 = [n-(2n+1)/2]2 | Completing the square | |
| 7. | (n+1)-(2n+1)/2 = n-(2n+1)/2 | Taking square roots of both sides | |
| 8. | n+1 = n | Euclid's Common Notion (3) | |
| 9. | 1 = 0 | That's also nice, right? |
This proof is harder than the previous one, If you want to find out the error, at least level of F4 Additional Mathematics is required, It is conspicuous that 1 does not equal 0. If 1 were to equal 0, computers would never be invented. However, what's wrong with the above proof? Don't you discover that?
Actually, it is testing us the basic concept of taking square roots. The above proof is all correct until step 7. If two real numbers equal each other, the square of them must equal each other but we should notice that the converse of this is not absolutely true.
From step 6, we know that L.H.S = R.H.S. Once taking square roots of both sides, there should be two answers. One is L.H.S = R.H.S and the another one is L.H.S = -R.H.S. In this case, impossible answers will be rejected.
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