## perl 3 years ago 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

1. KingGeorge

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

2. perl

no, sqrt(1) = 1

3. perl

4. KingGeorge

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 $$\sqrt{1}=1$$.

5. micahwood50

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.

6. KingGeorge

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

7. KingGeorge

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

8. micahwood50

$\sqrt{-1}\sqrt{-1} \neq \sqrt{-1 \times -1}$

9. perl

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

10. estudier

"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)

11. perl

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) ?

12. perl

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

13. estudier

i sqrt 2 is fine

14. micahwood50

$\sqrt{-2} = \sqrt{-1 * 2} = \sqrt{-1} * \sqrt{2}$ 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.

15. perl

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.

16. estudier

convert to i wherever possible instead of messing with sqrt signs

17. perl

sorry i can only give one medal

18. perl

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

19. perl

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

20. perl

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

21. perl

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

22. estudier

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

23. perl

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

24. estudier

i

25. estudier

in context

26. perl

scroll above, and see king georges response

27. perl

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

28. Directrix

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).

29. perl

ok so in the real numbers this is not valid

30. estudier

Ugh....i^2 = -1 and x^2 = -1 has 2 solutions Can you get abiguity, yes - so be careful

31. KingGeorge

Oops. I totally was wrong there :(

32. perl

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

33. perl

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). //

34. perl

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*

35. estudier

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.

36. perl

, what is principal branch of the complex square root

37. estudier

You don't want to know....

38. perl

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.

39. perl

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

40. estudier
41. perl

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.

42. perl

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

43. Directrix

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?