Why 1=-1

  • Thread starter Dark Fire
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Main Question or Discussion Point

Just kidding, I know there's a flaw in my calculation (as a matter of fact, I'm quite sure what).
Now, with that said;

1^0.25=i, because i^4=1
Based on my formula: a^x=b & b^y=a ***
Where: x=1/y & y=1/x

Further on, if:
1^0.25=i
then:
1^4=i=1
So:
i^2 & 1^3=-1
But 1^2 & 1^3=1
So 1=-1

***Possibly only works for real numbers, not Complex nor Imaginary numbers, who probably has different rules set, which I'm not taught nor read about

And I've got the impression of this forum lacking flamers; thank god for that; meaning I don't need to explain myself further more :D
Thanks <3
Just in case: this is really just a question: where's my flaw.
Is it my formula who is wrong, or does it just only work for real numbers?
 

Answers and Replies

  • #2
D H
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HallsofIvy already addressed this in your other thread on basically the same issue:
In general, the "law" axbx= (ab)x is true for real values of a and b, not complex values.
 
  • #3
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Oh, I'm sorry, didn't pay attention.
I'm "polen" allergic (idk the English word.. like grass and stuff), and I've been working for 9h just now, and I've been up for a while, from like 6 am, so I'm... totally... gone? Wasted? xD
Thanks for your reply anyways.
So my formula is correct for all real numbers then?

Btw: how come it does not apply to complex and imaginary numbers?
Oh.. and I don't get the similarity of a^x*b^x=(ab)^x and my formula..
3^2*3^2=(3*3)^2 ?

Edit: hahahahaha, I've made a thread of almost same content xD
Sorry, sorry, sorry, delete thread plx.
 
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  • #4
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The word is "pollen", in case you're still wondering.
 
  • #5
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Isn't the error rather assuming the solutions of x^4=1 are all equal?
 
  • #6
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*-<|
That made absolutely no sense.
 
  • #7
tiny-tim
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I'm "polen" allergic
"I have hayfever" …

Bless you! :smile:

(try loratadine … should be less than 20¢ a day … non-drowsy :wink: )
 
  • #8
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*-<|
That made absolutely no sense.
I have that effect on many people, but still 1 has 4 quartic roots (1,-1,i and -i), none of which are equal to eachother. To demonstrate, we have i^4 = 1^4, take quartic root on each side, you get 1=i, which is false.
 
  • #9
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I have that effect on many people, but still 1 has 4 quartic roots (1,-1,i and -i), none of which are equal to eachother. To demonstrate, we have i^4 = 1^4, take quartic root on each side, you get 1=i, which is false.
You're a real quicky, aren't you..
My question isn't whether or not they're equal, my question is why!

@Topic
I wrote this today, to figure out if a^x*b^x=(ab)^x only works for real numbers.
The conclusion was: no, they work for complex numbers too:
(i+1)^2(i/2)^2=((i+1)(i/2))^2
-0,5i=-0,5i

Though, I still don't get why that formula has anything to do with my:
a^x=b
a=b^y
x=1/y
y=1/x

Which at least works for real numbers:
a=7
x=3
b=7^3=343
343^y=343^(1/x)=343^(1/3)=7=a

Give me more than ignorant words to describe why it does not apply to complex values.
 
  • #10
D H
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It doesn't even necessarily apply to real numbers. Just because (-1)2=(1)2 does not mean that -1=1.
 
  • #11
matt grime
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Dark Fire: just because you verified something for one case does not mean it is true.

Look, raising to the n'th power viewed as a function from C to C, generically, is an n to 1 map. It doesn't have an inverse. One can take a choice of root, a so-called branch, but it won't behave exactly as you seem to wish it to (this choice is what you did when you said the 1^0.25=i). It is up to you to justify why something should be true, since a priori there is no justifiable reason why it should.


If you want to "see" what is going on, then you need to visualize the corrresponding Riemann surface. Imagine taking the complex plane and cutting along the positive real axis, then 'stretching and rotating' so you get something that looks a little like a helical screw, with four turns. This is the way to make sense of the 1 to n "function" that is raising to the power 0.25.
 
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  • #12
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First of all, x^(1/n) has a name. It is called the "nth root of x". For example, x^(1/7) is the 7th root of x. The thing you apparently don't completely understand is that there is more than one nth root of a number. In fact, there are exactly n (if you don't mind complex numbers). 1^(1/2) is either 1 or -1 (your choice), and 1^(1/4) is 1, -1, i, or -i. This doesn't mean they are equal.

Another example would be if someone asked you where you live, but you have two houses. Either house would be a correct answer, but that doesn't make them the same house.

The fundamental problem is that you seem to think that all answers to a question must be the same. Actually, the rule is: all answers to a question must be the same if the question is known to have only one answer. You are asking a question with more than one answer, so setting all of its answers equal to one another is pointless.
 
  • #13
D H
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Xezlec, nice answer --- but, if you look at any post by the OP, you will see that the OP's name is stricken (has a horizontal line through the middle). This means the OP is taking a respite from PF for a while, and answering the thread is a bit moot.
 
  • #14
Garth
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Well then, as this thread exists I cannot resist starting another discussion about -1 = 1 !

How about:

1 = 1
1 = [itex]+\sqrt{(+1)}[/itex]

But +1 = (-1).(-1), therefore

1 = [itex]+\sqrt{(-1).(-1)}[/itex]
1 = [itex]+\sqrt{(-1)}.\sqrt{(-1)}[/itex]

put i = [itex]+\sqrt{-1}[/itex]

Therefore from above

1 = i.i
1 = i2
1 =-1.

Just food for thought - something we resolved in the Sixth Form (High School). :wink:

Garth
 
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  • #15
D H
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But +1 = (-1).(-1), therefore

1 = [itex]+\sqrt{(-1).(-1)}[/itex]
1 = [itex]+\sqrt{(-1)}.\sqrt{(-1)}[/itex]
The function
[tex]f(x) = \sqrt x[/tex]
is defined for real, non-negative x only. The last step is not valid.
 
  • #16
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sqrt(a*b)=sqrt(a)*sqrt(b) only for non-negative reals a,b
 
  • #17
Garth
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The function
[tex]f(x) = \sqrt x[/tex]
is defined for real, non-negative x only. The last step is not valid.
Are you saying
[tex]i = \sqrt{-1}[/tex]
is not valid, i..e not allowing the whole of complex analysis?

Garth
 
  • #18
Garth
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sqrt(a*b)=sqrt(a)*sqrt(b) only for non-negative reals a,b
That is correct, in fact if a < 0 and b < 0 then a.b > 0 but

[tex]\sqrt{a.b} = [+\sqrt{a}].[-\sqrt{b}][/tex]

but can you say why?

It is instructive to do it on the Argand diagram.

It is a useful insight into a potential 'sign' trap in algebra.

Garth
 
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  • #19
D H
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Are you saying
[tex]i = \sqrt{-1}[/tex]
is not valid, i..e not allowing the whole of complex analysis?

Garth
Of course not. It is the square root sign that is the problem, as this is short for the square root function. Saying [itex]i = \sqrt{-1}[/itex] is abuse of notation. Abuse of notation is fine so long as you acknowledge it as such.

A simple way around this "paradox" is to note that [itex]a^rb^r = (ab)^r[/itex] is valid only for two sets of cases: (1) r is a positive integer, or (2) a and b are positive reals and r is real.

This answer begs the question: Why is that relation only valid for those limited cases? A better answer in my opinion is that you are asking for trouble any time you use a multivalued function in an equality and don't pay attention to the fact that the "function" is multivalued. For example, using the identity [itex]\sin^{-1}(\sin x) = x[/itex], [itex]\sin 2\pi = 0[/itex], and [itex]\asin 0 = 0[/itex] lets me conclude [itex]0=2\pi[/itex]. Saying
[tex]1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot\sqrt{-1} = i^2 = -1[/tex]
leads to a contradiction because you are using a multivalued function and pretending it is a true function.
 
  • #20
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Xezlec, nice answer --- but, if you look at any post by the OP, you will see that the OP's name is stricken (has a horizontal line through the middle). This means the OP is taking a respite from PF for a while, and answering the thread is a bit moot.
Oh.

*shrug*
 
  • #21
Garth
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Of course not. It is the square root sign that is the problem, as this is short for the square root function. Saying [itex]i = \sqrt{-1}[/itex] is abuse of notation. Abuse of notation is fine so long as you acknowledge it as such.

A simple way around this "paradox" is to note that [itex]a^rb^r = (ab)^r[/itex] is valid only for two sets of cases: (1) r is a positive integer, or (2) a and b are positive reals and r is real.

This answer begs the question: Why is that relation only valid for those limited cases? A better answer in my opinion is that you are asking for trouble any time you use a multivalued function in an equality and don't pay attention to the fact that the "function" is multivalued. For example, using the identity [itex]\sin^{-1}(\sin x) = x[/itex], [itex]\sin 2\pi = 0[/itex], and [itex]\asin 0 = 0[/itex] lets me conclude [itex]0=2\pi[/itex]. Saying
[tex]1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot\sqrt{-1} = i^2 = -1[/tex]
leads to a contradiction because you are using a multivalued function and pretending it is a true function.
Absolutely, you can illustrate the same point on the Argand diagram where multivalued functions can be visualised.

Now [tex]+1 = \exp{0.i} = \exp{2\pi i} = \exp{4\pi i}[/tex],

so taking the square root, we divide the argument angle by two:

[tex]\sqrt{+1} = \exp{\pi i}[/tex] or [tex]\sqrt{+1} = \exp{2\pi i}[/tex],

where the first equality gives the negative root and the second the positive root.

As we use the positive root in the 'paradox' it is the '[itex]4\pi[/itex]' argument that has to be divided in two thus:

[tex]\sqrt{-1.-1} = \exp{\frac{4\pi i}{2}} = \exp{\frac{\pi i}{2}}.\exp{\frac{3\pi i}{2}}[/tex],

where [tex]\exp{\frac{\pi i}{2}} = +i = +\sqrt{-1}[/tex]

but [tex]\exp{\frac{3\pi i}{2}} = -i = -\sqrt{-1}.[/tex]

Hence

[tex]+1 = \sqrt{-1} = \sqrt{-1.-1} = [+\sqrt{-1}].[-\sqrt{-1}] = +i.-i = -i^2 = +1[/tex].

A simple and baffling 'paradox' that illustrates algebraic traps when using multivalued 'functions'.

Garth
 
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  • #22
Garth
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[tex]\sqrt{-1.-1} = \exp{\frac{4\pi i}{2}} = \exp{\frac{\pi i}{2}}.\exp{\frac{3\pi i}{2}}[/tex],

where [tex]\exp{\frac{\pi i}{2}} = +i = +\sqrt{-1}[/tex]

but [tex]\exp{\frac{3\pi i}{2}} = -i = -\sqrt{-1}.[/tex]

Garth
In the above division of
[tex]\exp{\frac{4\pi i}{2}} = \exp{\frac{\pi i}{2}}.\exp{\frac{3\pi i}{2}}[/tex]

you might wonder why [itex]4\pi[/itex] was divided into [itex]\pi[/itex] and [itex]3\pi[/itex] rather than simply dividing [itex]2\pi [/itex] into [itex]\pi [/itex] and [itex]\pi [/itex], which then leads to the paradox.

The answer is seen on the Argand diagram.

As I explained above for [itex]\sqrt{+1} = +1[/itex] we have to take the value for +1 on the unit circle in the complex plane
[tex]\exp{2\pi i}[/tex].

This is then divided into two. However the first half of that [itex]2\pi[/itex] angle (argument) is from 0 to [itex]\pi[/itex] and when that is divided it produces
[tex]\exp{\frac{\pi i}{2}}[/tex]
which is ,
but the second half of that [itex]2\pi[/itex] angle (argument) is from [itex]\pi[/itex] to [itex]2 \pi[/itex] and when that is divided into two it produces
[tex]\exp{\frac{3\pi i}{2}} [/tex]
which is [-i].

This 'paradox' not only exposes the problems with multivalued functions such as [itex]\sqrt{x}[/itex] but also, once we allow imaginary numbers, the problems with multi-'valued' integers: such as the [itex]\pm[/itex]unit number [itex]\exp{[2n\pi i]}[/itex], and [itex]\exp{[(2n+1)\pi i]}[/itex] for all n.

Garth.
 
  • #23
Garth
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This 'paradox' not only exposes the problems with multivalued functions such as [itex]\sqrt{x}[/itex] but also, once we allow imaginary numbers, the problems with multi-'valued' integers: such as the [itex]\pm[/itex]unit number [itex]\exp{[2n\pi i]}[/itex], and [itex]\exp{[(2n+1)\pi i]}[/itex] for all n.

Garth.
Of course I meant to say "once we allow imaginary numbers, the problems with multi-'valued' integers: such as the [itex]\pm[/itex]unit number [itex]\exp{[2n\pi i]}[/itex], and [itex]\exp{[(2n+1)\pi i]}[/itex] for all integer n."

The time lock on the edit function beat me to it!

Garth
 

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