# The square of the square root of -1????

Discussion in 'Recreational Math' started by Zanziba, Mar 4, 2004.

1. ### ZanzibaGuest

A friend set me a paradox puzzle that I cannot fault his logic on. HELP please

You take the square root of -1 and square ir, the answer must be -1, however....

He says, write it as root -1 multiply root -1

Then place them all under the same root (Root (-1 times -1))

Now multiply -1 by -1 to get +1

Hence root -1 squared = root 1 = 1

Which is obviously pish posh, but i can't fault the algebra logic...

Steve

Zanziba, Mar 4, 2004

2. ### Ken OliverGuest

The property sqrt(a)*sqrt(b) = sqrt(a*b) holds for non-negative numbers
only. Thus the following line of demonstration is invalid.

Ken Oliver, Mar 4, 2004

3. ### philippe 92Guest

Hello,

Also for sqrt(a)/sqrt(b).
Both a and b should be non negative real numbers, otherwise :
i=sqrt(-1) = sqrt(1/-1) = 1/sqrt(-1) = 1/i
so i^2 = 1 and 1 = -1

which was the variant of the joke I knew.

philippe 92, Mar 4, 2004
4. ### DaveGuest

You're both only telling half the story. Square rooting creates TWO
solutions, not one, so square root of -1 is not i, but +/-i:

-i*-i=(minuses cancel)i*i=-1. (constitutes proof at least in the form of
a counterexample to the hypothesis "sqrt(-1)!=-i")

Not sure what Ken's going on about ("sqrt(a)*sqrt(b) = sqrt(a*b) holds
for non-negative numbers only"). That's probably true if you assume
positive numbers only have positive square roots:

a=-9; b=-16
sqrt(a)*sqrt(b)=i.sqrt(9)*i.sqrt(16)
=i.(+/-3).i.(+/-4)
=-1*3*4 OR -1*-3*4 OR -1*3*-4 OR -1*-3*-4
=-12 or +12 or +12 or -12.
=+/-12.

a=9; b=16
sqrt(a)*sqrt(b)=sqrt(9)*sqrt(16)
=(+/-3).(+/-4)
=3*4 or -3*4 or 3*-4 or -3*-4
=+12 or -12 or -12 or +12
=+/-12

QED

You could create more "paradoxes" using the fact that the nth root of
something produces n solutions. What are the three cube roots of -1?
(Hint: two of them are complex, and IIRC the angles between the three
vectors to each root are equal (if so that's .5+/-i.sin(pi/3) but I
could be miles off; however (.5+i*.866)^3 is fairly close...))

Dave.

Dave, Mar 4, 2004
5. ### philippe 92Guest

Hello,
[...]
"square root of" may be considered either as a multivalued function
(what you do), or a convention as "principal root of", which is usually
done.
that is square root of 4 is +2, so we write the quadratic solutions as
(-b +/- sqrt(delta) )/2a, with the +/- inside the formula.
With a multivalued function we could write just +sqrt(delta) which would
include the two values.
Here some paradox which may be related with cube roots of 1 :
Let x solution of x^2 + x + 1 = 0
x is not 0, so dividing by x : x + 1 + 1/x = 0
But from original equation : x + 1 = -x^2, Then -x^2 + 1/x = 0
That is x^2 = 1/x or x^3 = 1 and x = 1.
And from original equation : 3 = 0 !
(of course the bug is that x^3 = 1 as _three_ solutions, x=1, j and j^2.
j = -0.5 + i*sqrt(3)/2, j^2 = -0.5 - i*sqrt(3)/2.
j and j^2 are the solutions of the original equation, x = 1 has been

philippe 92, Mar 4, 2004
6. ### DaveGuest

Ok, so what's the principal root of -1?

If an assuption X (in this case "'square root of' means only 'the
principal root of'") leads to a paradox, that generally constitutes
proof that X is false (reductio ad absurdum, IIRC). The OP's paradox
("sqrt(-1).sqrt(-1)=-1 BUT sqrt(-1).sqrt(-1)=sqrt(-1.-1)=sqrt(1)=1")
stems from the use of principal roots, and the solution to the paradox
is to point out that principal roots are not the only roots, and that
there is in fact no paradox.

With false assumptions you can create paradoxes to your heart's content,
e.g.:

1+2=3
assume 1+1!=2

(IAACPPP (C++ programmer), hence != for "doesn't equal". =/= always
throws me, and there's no exception handler big enough for that Dave.

Dave, Mar 4, 2004
7. ### philippe 92Guest

IMHO +i, by definition of i, which is _one_ number in |C (not a couple
of numbers)
if X (principal root) _and_ Y (sqrt(a)*sqrt(b)=sqrt(a*b) for any a,b)
result in a contradiction, so _either_ X or Y is false...
Use of principal root (X) => Y false :
sqrt(a)*sqrt(b) != sqrt(a*b) if a or b <0 or not real

If we want to keep the usefull property sqrt(a)*sqrt(b)=sqrt(a*b), I
agree with you, it's impossible to use "principal roots" and we have to
list all cases as you did to solve the "paradox".

It's exactly what I said :
Either you use "principal roots" and you forbide a multiplicative sqrt()
_or_ you list all roots as you did (multivalued function).
In both cases the paradox is solved.
Of course...
The false assumptions may be explicit as in your example
or implicit when you apply badly some rules like dividing by 0 or
extracting only one square root or extending a function's property
defined for positive real numbers to |C and s.o.

philippe 92, Mar 4, 2004
8. ### VirgilGuest

There is no such thing as "the" square root of -1, since there are two
such square roots, not just one.

Also, sqrt(a)*sqrt(b) = sqrt(a*b) only holds when a and b are both
non-negative and "sqrt" means the principle square root of a
non-negative real number.

One has to be very careful when doing root extractions on negatives or
complexes.

Virgil, Mar 5, 2004
9. ### ZanzibaGuest

Here some paradox which may be related with cube roots of 1 :
Now that's one I can "Get him back with"

Thanks guys. I understand most of what you have said. ANyone know of
an interesting bookob complex numbers that I may like to read? Or
interesting websites for beginners to the subject?

Zan

Zanziba, Mar 5, 2004