Question

Spot the error. Surely there must be an error somewhere, 1 = -1 is absurd! so im having trouble with this

1 = -1 * -1 = sqrt (-1*-1) = sqrt (-1) * sqrt (-1) = i * i = i^2 = -1

Answer 1

The problem is that \(\sqrt{1}=\pm1\).

Answer 2

no, sqrt(1) = 1

Answer 3

check your calculator

Answer 4

By definition, \(\sqrt{x}=\{y\;|\;y^2=x\}\). When \(x=1\), both \(1,\) and \(-1\) are in the set.

Alternatively, you can say that \(\sqrt{x}=|x|\), and for \(x=1\), then \(\sqrt{1}=1\).

Answer 5

The fallacy is that the rule √xy = √x√y is generally valid only if at least one of the two numbers x or y is positive.

Answer 6

If you use the second way, then \(\sqrt{-1}\cdot\sqrt{-1}=|-1|\cdot|-1|=1\cdot1=1\).

Answer 7

Although there are problems when you do what I just did.

Answer 8

\[\sqrt{-1}\sqrt{-1} \neq \sqrt{-1 \times -1}\]

Answer 9

but sqrt(-2) = sqrt(-1 * 2) = sqrt(-1) * sqrt(2)

Answer 10

"1 = -1 * -1 = sqrt (-1*-1) = sqrt (-1) * sqrt (-1) = i * i = i^2 = -1"

This is just wrong (and it won't work if you try and use plus/minus either)

Answer 11

so sqrt(a*b) = sqrt(a)*sqrt(b) is only valid for a,b > 0
but what about sqrt(-1 *2) = sqrt(-1)*sqrt(2) ?

Answer 12

oh, at least one of a,b must be positive, ok

Answer 13

i sqrt 2 is fine

Answer 14

\[\sqrt{-2} = \sqrt{-1 * 2} = \sqrt{-1} * \sqrt{2}\] because one of them is positive so it is valid.

√xy = √x√y is generally valid only if at least one of the two numbers x or y is positive.

Answer 15

micah has it !!
The fallacy is that the rule √xy = √x√y is generally valid only if at least one of the two numbers x or y is positive.

Answer 16

convert to i wherever possible instead of messing with sqrt signs

Answer 17

sorry i can only give one medal

Answer 18

micah, then you have no choice but to convert two negative signs to positive in a product

Answer 19

king george, i see your response, its a bit complicated. the function sqrt(x) is not multivalued, but i guess it could be

Answer 20

but if we use king georges idea, thatn sqrt(-1) = + - i

Answer 21

because (-i)^2 = -1, (+i)^2 = -1

Answer 22

there are 2 roots of -1, i and -i

Answer 23

right, but what does sqrt(-1) equal to, is it multivalued or not?

Answer 24

i

Answer 25

in context

Answer 26

scroll above, and see king georges response

Answer 27

, king george, sqrt( x^2) = |x| , so i think you are wrong there

Answer 28

sqrt (-1*-1) = sqrt (-1) * sqrt (-1)
---------------------------
In the set of Real Numbers, the square root of a product is the product of the square roots. That property does not hold true within the set of Complex Numbers as illustrated by the posted example.

Product of Square Roots Theorem
If m and n are not negative and are real numbers, then SQRT(m) * SQRT(n) = SQRT(mn).

Answer 29

ok so in the real numbers this is not valid

Answer 30

Ugh....i^2 = -1 and x^2 = -1 has 2 solutions

Can you get abiguity, yes - so be careful

Answer 31

Oops. I totally was wrong there :(

Answer 32

directrix, this is not true what you said
"In the set of Real Numbers, the square root of a product is the product of the square roots."
if the numbers are both negative , or even one negative

Answer 33

directrix, this is very confusing, and i think wrong. points deducted

// sqrt (-1*-1) = sqrt (-1) * sqrt (-1)
---------------------------
In the set of Real Numbers, the square root of a product is the product of the square roots. That property does not hold true within the set of Complex Numbers as illustrated by the posted example.

Product of Square Roots Theorem
If m and n are not negative and are real numbers, then SQRT(m) * SQRT(n) = SQRT(mn). //

Answer 34

Correction:
In the set of Real Numbers, the square root of a product is the product of the square roots *provided that the two numbers are both nonnegative*. In the complex numbers the square root of a product is the product of the square roots *provided that at least one is positive*

Answer 35

sqrt -1 is either principal sqrt (x>0) or principal branch of the complex square root.

Don't mess with radicals, convert to i and to -1 pronto.

Answer 36

, what is principal branch of the complex square root

Answer 37

You don't want to know....

Answer 38

Edited correction:
In the set of Real Numbers, the square root of a product xy is equal to the product sqrt(x)*sqrt(y) provided that both x and y are nonnegative. In the complex numbers the square root of xy is equal to the product sqrt(x)*sqrt(y) provided that at least one of x and y is positive.

Answer 39

estudier, i think i can handle it :)
please send me a link

Answer 40

http://en.wikipedia.org/wiki/Branch_point

Answer 41

Edited correction 3:
In the set of real numbers, sqrt(x*y) = sqrt(x)*sqrt(y) provided that both x and y are nonnegative. In the complex numbers we have sqrt(x*y) = sqrt(x)*sqrt(y), provided that at least one of x or y is positive.

Answer 42

Edited correction 3:
In the set of real numbers, sqrt(x*y) = sqrt(x)*sqrt(y) provided that both x and y are nonnegative. In the complex numbers we have sqrt(x*y) = sqrt(x)*sqrt(y), provided that at least one of x or y is positive.
Otherwise it is not equal. and in the reals we have sqrt(-2) is undefined, for example

Answer 43

Fallacy
-----------
Sqrt(-1) * Sqrt(-1) = Sqrt ( -1 * -1) = Sqrt(1) = 1.
=====
Sqrt(-1) * Sqrt(-1) = -1 by definition

Consider the following assertion:

(-1)^2 = (1)^2
log (-1)^2 = log (1)^2
2 log (-1) = 2 log (1)
log(-1) = log(1)
Therefore, -1 = 1

Where lies the error or misconception?