Brain Teaser Solution: ImpossibleMath2


Brain teaser:
Reference
[ImpossibleMath2]
Submitter
Jack Tan (jahk@uiuc.edu)
Submission date
14 April 1997
Solution:
Submitter
Jack Tan (jahk@uiuc.edu)
Submission date
20 April 1997

Solution

The trick in this problem lies between lines 4 and 5:

 4   (Sqrt(-1))^2 = Sqrt(-1)*Sqrt(-1)  Expand the defintion of i
 5   (Sqrt(-1))^2 = Sqrt(-1 * -1)      Rule:  Sqrt(x)*Sqrt(y) = Sqrt(x*y)
In algebra, there is a rule which states that the product of the square roots of two non-negative integers is the square root of their product. That is,

Sqrt(a) * Sqrt(b) = Sqrt(a*b)
for non-negative a, b.

The key here is the word "non-negative." In line 4, the rule is applied to a=b=-1, which are certainly negative. The abuse of the rule allows the problem to conclude that i^2 equals 1, which is false (by definition).

Moral: the qualifiers in theorems are very important.


Jack Tan and Emmie Chen / jahk@uiuc.edu

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