Is 2 Really Equal to 1? A Mathematical Paradox

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SUMMARY

The forum discussion centers on a mathematical paradox that claims to prove that 2 equals 1 through a series of algebraic manipulations. The key steps involve setting a equal to b, leading to the erroneous conclusion that (a+b)(a-b)=b(a-b) allows for division by zero. The discussion highlights the critical error in dividing by zero, which invalidates the proof. Additionally, it references the misuse of the square root of negative numbers, specifically using "i" as the square root of (-1), to further illustrate the fallacy.

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Himal kharel
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let a=b
a2=ab
a2-b2=ab-b2
(a+b)(a-b)=b(a-b)
a+b=b
a+a=a<from above>
2a=a
2=1
PROVED
 
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When you divide by zero you can 'prove' all sorts of nonsense.
 
Here's another classic, using "i" as the square root of (-1).
Can you spot the error?

1=\sqrt{1}=\sqrt{((-1)*(-1))}=\sqrt{(-1)}*\sqrt{(-1)}=i*i=-1
That is:
1=-1
 
Himal kharel said:
let a=b
a2=ab
a2-b2=ab-b2
(a+b)(a-b)=b(a-b)
a+b=b
a+a=a<from above>
2a=a
2=1
PROVED

The part in red

(a+b)(a-b)=b(a-b)

Remember we defined a=b, therefore (a-b) must equal zero. You can't divide zero as it is a division error.
 

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