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| 1. | a,b>0 | that's given |
| 2. | a = b | that's also given |
| 3. | ab = b2 | this is step 2 times b |
| 4. | ab-a2 = b2-a2 | Euclid's Common Notion (3) |
| 5. | a(b-a) = (b+a)(b-a) | distributive law |
| 6. | a = b+a | dividing by (b-a) |
| 7. | 0 = b | Euclid's Common Notion (3) |
| 8. | b = 2b | Euclid's Common Notion (2) |
| 9. | 1 = 2 | Nice, right? |
Ha,,,the above analysis seems to be right but what's wrong with the calculations? It is undeniable that 1 does not equal 2 but what 's wrong ? Do you know that? Actually the above proof is playing a trick on you and me. It has performed a division of zero on step 6. Don't you discover that? On step 6, both sides are divided by (b-a) but b = a according to step 2 which is given , indicating that (b-a) equals 0. Therefore, the above proof has collapsed totally beyond step 6. Thus, the conclusion 1 = 2 must be wrong.
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since step 4 we have 0 in the both sides [ 0=0 ]
and what more can be proven here? nothing.
this must be why high school math's junk compare with what you learn from 1st year Eng. math.
在第5步驟(b-a)不能消除
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