Question

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



Go!

Answer 1

an openstudy oldie but goodie

Answer 2

so wat's da question?

Answer 3

do you see a problem with what I did?

Answer 4

theres nothing wrong with that

Answer 5

so -1 = 1?

Answer 6

god I hope something is wrong, or I just broke all of math

Answer 7

Thanks, now let's see what fun consequences we can derive from this awesome result.

0=1-1=1+1

0=2

Yay

Answer 8

lol

Answer 9

thus 1/2 = \(\infty\)

Answer 10

It's hard to say whether \[
(-1)^{\frac 11} = (-1)^{\frac 22}
\]or \[
(-1)^{\frac 22} = \sqrt{ (-1)^2}
\]is the problem.

Answer 11

By induction, since 0=2, and 2=2+0, then...

Answer 12

I think its the first step

Answer 13

I believe zepp had asked this question a few years ago.

Answer 14

\( (-1)^\frac{1}{1}\neq (-1)^\frac{2}{2} \)
we should have respect to even odd :P

Answer 15

fractions mean something different when raised to an exponent

Answer 16

the equivalence is changed in this meaning

Answer 17

Yeah, and you had answered it lol

http://openstudy.com/updates/5014e187e4b0fa2467312775

Answer 18

\[\large -1 = e^{i \pi}\]\[\large \sqrt{e^{i 2\pi}}= \pm 1\]

Answer 19

so what you are saying then @ikram002p is that \(1=\frac{2}{2}\ne \frac{1}{1}=1\)?

Answer 20

hehe well consedering the power which have its own properties

Answer 21

I dont really know the answer, and when I asked my professor for a good explination he laughed at me. I am not quite sure what to take away from that

Answer 22

The fact that this hurts my head reading it means I probably should get some sleep soon...

Answer 23

This has the whole problem of the fact that any time you take the root we cut it into two parts and just take the real positive part when taking the square root. Damn principal square root.

It's like the 4th root could really be a 1-to-4 mapping haha.

Answer 24

its good stuff:)

Answer 25

Perhaps fractional exponents are just bad practice?

Answer 26

\[\sqrt[4]{16}=2=-2=2i=-2i\] The world is broked

Answer 27

I think so @wio its bad notation that we need. The better of two devils i suppose.

Answer 28

i love u, kainui

Answer 29

I guess I have to give you a medal now.

Answer 30

another good one. how is it that addition is commutative yet we get a different infinite sum when we move around terms in many infinite sums...

Answer 31

Consider \[
(-1)^{3/2}
\]We have \[
((-1)^{3})^{1/2} = (-1)^{1/2} = i
\]and \[
((-1)^{1/2})^{3} = (i)^{3} = -i
\]

Answer 32

Oh ya, that classic one.\[S = 1 - 1 + 1 -1 \cdots\]

Answer 33

right...

Answer 34

\[\Large \sqrt[1]{-1}=-1\]???

@zzr0ck3r I think that's because those aren't convergent sums.

Answer 35

but even some when we change finite amount of terms will change the outcome...

Answer 36

The problem is that order of operations MATTER and we are trying to suggest that they somehow don't with such notation.

Answer 37

even on converging sums

Answer 38

Give an example of a convergent infinite sum that doesn't work. I'm curious, this is my groove.

Answer 39

I thought, maybe I am remembering something wrong.

Answer 40

sec ill look

Answer 41

http://math.stackexchange.com/questions/417280/continued-fraction-fallacy-1-2/417322#417322 Relevant.

Answer 42

for example \[
(-1) = (\sqrt{-1})^2 = -1
\]But \[

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

Answer 43

http://math.stackexchange.com/questions/646665/why-does-commutativity-of-addition-fail-for-infinite-sums

Answer 44

@Kainui

There is a famous theorem of Riemann that says that for a conditionally convergent series, we can re-order the terms so that the series converges to any limit or even diverges.

Answer 45

this is not my bag, I am an algebraist, but its still fascinating

Answer 46

Well I guess there is that thing I heard of, where if you have a sphere and cut it up into a bunch of pieces then you can get two spheres of exactly the same size; which seems ridiculous and related.

Answer 47

yeah CRAZY!

Answer 48

Banach-Tarski?

Answer 49

I cant wait till topology next term!

Answer 50

Did you see that thing I was talking about a month ago with that geometric series where it converged to the average?

Answer 51

go read on the Cantor set/function...

Answer 52

a continuous step function....

Answer 53

ffs what is going on !!!!!!!!!!!!!!!!!

Answer 54

lol

Answer 55

Yeah that sounds exciting, there's a lot of interesting stuff in math that I'll have to get around to teaching myself since I won't be taking it haha. What are either of those?

Answer 56

its hard to explain but pretty easy to read

Answer 57

actually pretty good sumation of the cool properties here
http://en.wikipedia.org/wiki/Cantor_function

Answer 58

Well I have some suspicions about how topology can apply to organic chemistry and knots that I'd like to understand more.

Answer 59

sweet, you should explore that. what a great MS research topic for someone in both fields

Answer 60

when my professor did his lecture on the Cantor set, he said "today we step into the dark side" (the same guy that laughed at me for the question in the topic here)

Answer 61

I mean, knots and molecules are both three dimensional structures that we try to represent in two dimensions. I have a feeling that some organic synthesis problems might be worth trying to decompose like "skein relations" in knot theory. But I don't really know enough about that sort of stuff or group theory.

Answer 62

sounds believable to me!

Answer 63

Hahaha why did he say that?

Answer 64

because its very counter intuitive

Answer 65

you take a set, and keep removing things from it, and in the end the set has the same length as when you started...

Answer 66

by length i mean measure...

Answer 67

all of dis is interesting.

but its still theoretical.

Answer 68

I think if money was not an issue, I'd try to really understand the structure of DNA and cells and try to find the algorithm for how life unfolds. We're symmetrical. We're all "even" functions in a sense haha.

Answer 69

Wait, do you mean like because the set is infinite? For instance I think Galileo said something like there are an infinite number of integers and an infinite number of squares. But it seems like there are fewer squares. But both are infinite, which is interesting.

Answer 70

@kainui read that link, its to much to explain, and I would have to reference that anyway...

Answer 71

...wow.

Answer 72

I will be taking my medals after you watch that(if you dont die from laughter).

Answer 73

Lol mean jerk time.

Answer 74

lol

Answer 75

i... no comments.

Answer 76

This reminds me of when I was teaching my friend mechanics and was telling him that you can find the initial velocity of... if you time your squirt and measure the distance and height lol.

Answer 77

is it abnormal to puke and laugh at the same time?

Answer 78

What's Silicon Valley I feel like I would love this movie/show.

Answer 79

its on hbo, its about a bunch of coders creating a startup business and getting funding

Answer 80

Um, not going to acknowledge the fact that: \[
(\sqrt{x})^n \neq \sqrt{x^n}
\]And thus \(x^{n/2}\) because it ignores this distinction?

Answer 81

This is why I originally said what I said before. Either we impose the rule \[
x^{n/2} = (\sqrt{x})^2
\]And the second step is wrong, otherwise \[
x^{1/1} =x^{2/2}
\]must be wrong.

Answer 82

yeah I always figured it was that step that was the problem because there are counter examples to that claim, I just never had an "axiomatic" reason for it.

Answer 83

Wolfram actually imposes that rule:
http://www.wolframalpha.com/input/?i=%28-1%29%5E%282%2F2%29

Answer 84

@wio yeah, when I think of numbers I basically see every number as being of the form: \[\Large re^{i \theta}\] So you have an angular part and a scalar part. So this means, -1, i, -1, and 1 are really like the "sign" part of your number and can be considered separately. So if you look at -1 as being rotated by 180 degrees then you can interpret the powers as being rotation.

Then the question becomes, are you counter rotating as much as you're rotating or have you tricked yourself?

Answer 85

If you impose that rule, then the notation has meaning. Otherwise the notation itself is unacceptable.

Answer 86

But god, can you imagine if we tried to teach that when we learn things like \(\sqrt[a]{b^c}=b^{\frac{c}{a}}\).

Answer 87

this is my favorite part about this site...

Answer 88

That rule is false when \(a\equiv 0\pmod 2\). It's true only half the time and it gets taught. It really needs to be fixed.

Answer 89

Or actually, the problem of that rule is only when \(b\) is negative.

Answer 90

I think if we only reduce/scale fractions when there denominator is coprime with the numerator we are ok.

Answer 91

when in lowest terms....

Answer 92

wait

Answer 93

that makes no sense....

Answer 94

You don't want 2s canceling?

Answer 95

right

Answer 96

2/4 cant go to 1/2 but 3/6 can, I think

Answer 97

I think rule should be \[
\sqrt[a]{x^b} = (x^b)^{1/a}
\]Then it should be taught \[
(x^b)^c = x^{bc} \iff x\geq 0
\]