A student approaches you and shares the following argument... how do you respond? Is the student wrong in any of those lines? sqrt = square root

I can prove that 2 = 0  Follow my logic and see if you disagree with me with any of these statements.

A. 2 = 1 + 1

B. 2 = 1 + sqrt ( 1 )

C. 2 = 1 + sqrt [ (-1) * (-1) ]

D. 2 = 1 + sqrt (-1) * sqrt (-1)

E. 2 = 1 + i * i

F. 2 = 1 + i2

G. 2 = 1 + (-1)

H. 2 = 0

Comments (5)

misschristine's picture

The error comes when you take the square root.  There are two possibilities for the square root of 1:  +1, -1.  This proof only takes one of the possibilities into consideration.  This omission leads to the error in the proof.

Edward Latham's picture

Student response to the two possibilities...

"OK ... two possibilities ... that just means that if we go with +1 we end up with 2 = 2, but if we deal with the -1, which so many refuse to deal with , we end up with 2 = 0. So I should always ignore negatives because they don't make sense or they just make things difficult on math teachers?"

How do we respond to that confusion?

Rachel Baron's picture

A student at this level can likely understand that some equations have two solutions. (I always picture a quadratic curve, and how it goes through two x-values for certain y-values.)

The equation equals both 2 and 0, but that doesn't mean that the two solutions are equal. The square root of 16 could be 4 or -4, but that doesn't mean that 4 = -4. The word "mouse" means both an animal and a computer accessory, but that doesn't mean that the two are interchangeable.

Still, I'd certainly express my enthusiasm and excitement that the student is interested and advanced enough in math to find this interesting/funny. And any time a student makes me think, I make sure they feel good about it! :)

S Jones's picture

With the definition of square roots.   The square root of negative one times the square root of negative one ** has to be ** negative one because that's the definition of square root.  (I remember tripping over this contradiction and figuring this was why when you are given a problem "square root of 36"  the *only* answer is 6... yet when the problem is "x^2 = 36" then you have two solutions.) 
     i is not a real number.   You can't casually switch from real to imaginary and then switch back... it's like running into yourself when you go back in time... 

fsiegeltuch's picture

When the square root symbol is used the convention is that it is asking for the principal square root and so it is a function and mapping to a unique value. That is why when you press the sqrt key on your calculator it always gives you a non-negative number for a non-negative input. See: http://mathworld.wolfram.com/PrincipalSquareRoot.html  Hence it is true that sqrt(1) = 1 and that is not the error.

The error occurs in going from Line C to Line D. sqrt(a x b) = sqrt(a) x sqrt(b) it is not true if both a & b are negative. 

Hope this helps,