Question

what is wrong here?

Answer 1

\[-1=(-1)^1 = (-1)^{\frac{2}{2}}=\sqrt{ (-1)^2}=\sqrt{1} = 1\]

Answer 2

this seems to be saying there is an order in which we evaluate the numerator and denominator in fractional exponents

Answer 3

\[\sqrt{x}=\pm x\]

Answer 4

still, then -1 = -1 and -1=1

Answer 5

I know something like this was asked here a while ago, but I don't think anyone got to the bottom of it.

Answer 6

I guess it would then be -1 = -1 or -1 = 1, and this is of course true, but still....

Answer 7

does this have something to do with absolute value or something?

Answer 8

yeah

Answer 9

\(\sqrt{x^2}= |x|\)

Answer 10

correct, even that suggest an order

Answer 11

what is |x| when x is negative ?
-x, right ??

Answer 12

yes
still, if we define |x| = sqrt(x^2) = x^(2/2) (which I don't think we do) it still suggest an order

Answer 13

there has always just been some weirdness with the notations...

Answer 14

The way my book defines |x| has nothing to do with exponents, its the piece wise definition that is real, that other one is just to make that a little more simple ...or something I don't know why people use it.

Answer 15

\((-1)^{\dfrac{2}{2}}=(\sqrt{-1})^2\)

Answer 16

right, again order....

Answer 17

yes, the order is important, square root is evaluated first....

Answer 18

so, what are the real definitions for this stuff, I never here anyone talk about fractional exponents having order.

Answer 19

so they are not really fractions in the exponent, or they would not have order, since exponents are repeated multiplication....bah this is gonna bug me:)

Answer 20

so maybe there is order when the fractions are not in there reduced forms

Answer 21

so if (top part of fraction, bottom part of fraction) = 1, we have no order:)

Answer 22

whiskey and math might not be the best combo:)