1. (n+1)2 = n2+2n+1Binomial Expansion
2. (n+1)2-(2n+1) = n2Euclid'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. (n+1)2-(n+1)(2n+1)+(2n+1)2/4 = n2-n(2n+1)+(2n+1)2/4Euclid's Common Notion (2)
6. [(n+1)-(2n+1)/2]2 = [n-(2n+1)/2]2Completing the square
7. (n+1)-(2n+1)/2 = n-(2n+1)/2Taking square roots of both sides
8. n+1 = nEuclid's Common Notion (3)
9. 1 = 0That'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.
 

Posted by antoniosehk at 痞客邦 PIXNET Comments(3) Trackback(0) Hits(574)


open trackbacks list Trackbacks (0)

Comments (3)

Post Comment
  • if n is (1/h) , and h is so tiny that can be ignore sometimes. then n+1 = n can still be true since the difference in between is very slim.
  • before i go to sleep heres a paradox we play in the old days.

    can you prove 2 * 2 != 4?

    solution:

    by irrational numbers. square root of 2 is 1.414 (can go as detail as you want, result will still be the same)

    1.414*1.414 + 1.414*1.414 =3.998792 < 4

    done. lol
  • 步驟七 錯了 開根號後 應該是有絕對值

Comment Permissions: Allow commenting

Leave Comment

*Name/Nickname
E-mail
Personal Website
Comment Title
*Comment
* Private Comment