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Topic: i^2 = 1 (Read 2772 times) |
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dante
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i^2 = 1
« on: Jun 21st, 2007, 9:20pm » |
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This is what I found when I was doing my high school .. Till now no one answered ..
-1 ^ -1 = -1
-1 ^ (-1/2) = -1 ^ (1/2)
1/(-1^(1/2))=-1 ^(1/2)
1/i = i
i^2 =1
Sorry if someone has already said this
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| « Last Edit: Jun 21st, 2007, 9:50pm by dante » |
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Sir Col
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Re: i^2 = 1
« Reply #1 on: Jun 22nd, 2007, 4:30am » |
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(-1)1/2 = i
(-1)-1/2 = ((-1)1/2)-1 = i-1 = 1 / i = i / i2 = i / -1 = -i
So your second line, (-1)-1/2 = (-1)1/2, is not true.
[edit]Tidied up sloppy notation; thanks, FiBsTeR.[/edit]
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| « Last Edit: Jun 23rd, 2007, 5:18am by Sir Col » |
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Grimbal
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Re: i^2 = 1
« Reply #2 on: Jun 22nd, 2007, 5:27am » |
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but
(-1)-1 = -1
is true, ok? So take the square root
left: ((-1)-1)1/2 = (-1)-1·1/2 = (-1)-1/2
right: (-1)1/2
So, (-1)-1/2 = (-1)1/2
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| « Last Edit: Jun 25th, 2007, 4:53am by Grimbal » |
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dante
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Re: i^2 = 1
« Reply #3 on: Jun 22nd, 2007, 6:53am » |
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Thanx grimbal ...
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rmsgrey
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Re: i^2 = 1
« Reply #4 on: Jun 22nd, 2007, 7:05am » |
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But:
(-1)2=1=12
((-1)2)1/2=(12)1/2
(-1)2/2=12/2
(-1)1=11
-1=1
so:
i2=-1
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FiBsTeR
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Re: i^2 = 1
« Reply #5 on: Jun 22nd, 2007, 7:38am » |
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I would like to clear one thing up first:
I've been taught that, for examle:
-1½  (-1)½, since the exponentiation occurs before the multiplication.
In the above lines, am I to assume that this is being abided by? There are many instances above where this occurs, and I want to be sure I am interpreting them correctly.
EXAMPLE:
on Jun 22nd, 2007, 4:30am, Sir Col wrote:
To me, this is not true, because the square root of 1 would be taken before being multiplied by -1, so:
-11/2 = -1.
EDIT: My original post had a "does not equal sign" that didn't show up.
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| « Last Edit: Jun 22nd, 2007, 8:41am by FiBsTeR » |
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FiBsTeR
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Re: i^2 = 1
« Reply #6 on: Jun 22nd, 2007, 7:54am » |
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Going back to the original problem, I think the flaw comes in taking the square root and assuming that there is still equality.
For example, suppose I have a=3 and b=-3:
a2=9 and b2=9, but I would be wrong to assume that a=b.
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| « Last Edit: Jun 22nd, 2007, 8:42am by FiBsTeR » |
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dante
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Re: i^2 = 1
« Reply #7 on: Jun 22nd, 2007, 9:25am » |
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on Jun 22nd, 2007, 7:05am, rmsgrey wrote:But:
(-1)2=1=12
((-1)2)1/2=(12)1/2
(-1)2/2=12/2
(-1)1=11
-1=1
so:
i2=-1 |
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I wish to know what is your conclusion
if x = y then
x^1/2 != y^1/2
or
x^1/2 not neccessarily equal to y^1/2
is there any rule of maths behind it .. am just curious
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| « Last Edit: Jun 22nd, 2007, 9:28am by dante » |
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Sameer
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Re: i^2 = 1
« Reply #8 on: Jun 22nd, 2007, 9:52am » |
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ok look at it this way
if x2 = 1, what are the possible values of x?
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Sir Col
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Re: i^2 = 1
« Reply #9 on: Jun 23rd, 2007, 5:16am » |
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Sorry, FiBsTeR, you're right. I was being sloppy with my notation: I should have written (-1)1/2.
on Jun 22nd, 2007, 5:27am, Grimbal wrote:but...
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Thank you, Grimbal! 
The point I was trying to make, dante, is that your first line is valid, whereas the second line is invalid. So the error must have occurred in raising both sides to the power of 1/2.
It is my understanding that when we raise a positive base to a unit fraction we focus on what is called the principle value.
For example, 641/2 = 8. However, there are two square roots of 64: 8 and -8.
Similarly if we wished to evaluate 641/3 we get the principle value 4, but there are another two roots: -2+2sqrt(3)i and -2-2sqrt(3)i.
Now when we evaluate (-64)1/2 we have the two roots: 8i and -8i. Hence it is not clear which of these should be the principle value.
In general application of the third power law, (ab)c = abc, is only valid when a is positive.
I hope that helps.
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| « Last Edit: Jun 23rd, 2007, 6:28am by Sir Col » |
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rmsgrey
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Re: i^2 = 1
« Reply #10 on: Jun 23rd, 2007, 7:05am » |
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on Jun 22nd, 2007, 9:25am, dante wrote:
I wish to know what is your conclusion
if x = y then
x^1/2 != y^1/2
or
x^1/2 not neccessarily equal to y^1/2
is there any rule of maths behind it .. am just curious |
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Taking a square root involves making a choice. For positive numbers, the default is to take the positive root, but that doesn't work so well when you start extending into other numbers.
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FiBsTeR
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Re: i^2 = 1
« Reply #11 on: Jun 23rd, 2007, 7:31am » |
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on Jun 23rd, 2007, 5:16am, Sir Col wrote:| Now when we evaluate (-64)1/2 we have the two roots: 8i and -8i. Hence it is not clear which of these should be the principle value. |
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Now my question is this: given any equation, not limited to positive quantities on both sides, can you raise both sides to the ½ power and still have equality?
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towr
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Re: i^2 = 1
« Reply #12 on: Jun 24th, 2007, 8:02am » |
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on Jun 23rd, 2007, 7:31am, FiBsTeR wrote:| Now my question is this: given any equation, not limited to positive quantities on both sides, can you raise both sides to the ½ power and still have equality? |
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You can always have equality after applying a mapping to equal values. But, you have to map the equal values to equal results.
The problem is not whether you can have equality, but whether you might not also have something else if you choose differently.
If you have an equality between two 'quantities', they are wholly exchangable; they wouldn't be equal if they weren't.
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FiBsTeR
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Re: i^2 = 1
« Reply #13 on: Jun 24th, 2007, 1:11pm » |
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on Jun 24th, 2007, 8:02am, towr wrote:
The problem is not whether you can have equality, but whether you might not also have something else if you choose differently. |
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Well, if I understand you correctly, the problem in the proof comes in choosing the wrong value in the second line after taking the square root.
(-1)-1 = -1
[(-1)-1]½ = (-1)½
From here, there are two "choices" for the left-hand side: (-1)½ or -(-1)½. Using the first value, the equality does not hold, so the error in the proof (I'm assuming) is that the negative value should have been "chosen":
-(-1)-½ = (-1)½
-1/i = i
i2 = 1, which is true.
A simpler way of seeing this is given the equation:
i4 = 1
i2 = 1/(i2)
After taking the square root, the left-hand side is either: i or -i. Again, the negative i must be chosen to maintain a true statement:
-i = 1/i
i2 = -1, which is true.
Someone please tell me if there is a mistake here, because I've never encountered this kind of thing before where you have to "choose" a correct value in order to maintain equality.
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towr
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Re: i^2 = 1
« Reply #14 on: Jun 24th, 2007, 2:53pm » |
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I think the short answer is that powers (and also logarithms) work differently on complex numbers than what you might expect from how they work on real numbers. This is briefly mentioned on wikipedia here. Of course, they seem to fail to mention how to deal with this problem properly.
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Sir Col
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Re: i^2 = 1
« Reply #15 on: Jun 25th, 2007, 12:36am » |
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I have been led to believe that equality does not hold when raising to non-integer powers. The example FiBsTeR gave is a good one: a2 = b2 does not imply that a = b. That is, finding the nth root creates n "branches of equality". So in the example given here, a = b or a = -b. Similarly if x3 = y3 then x = y, (-1+sqrt(3)i)y/2, (-1-sqrt(3)i)y/2.
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towr
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Re: i^2 = 1
« Reply #16 on: Jun 25th, 2007, 1:23am » |
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on Jun 25th, 2007, 12:36am, Sir Col wrote:| I have been led to believe that equality does not hold when raising to non-integer powers. |
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It has to hold for some choice of roots. If two expressions are equal, then they are different representations of the same value, but they're still the same value.
It's the rewriting of the expressions that has to be the problem, if you rewrite it in an inappropriate way, you change the value of the expression (and obsviously lose equality by it).
Quote:| The example FiBsTeR gave is a good one: a2 = b2 does not imply that a = b. |
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a2 = b2 does imply that (a2)1/2 = (b2)1/2, but this isn't equivalent to a=b, we lose something when we multiply the exponents.
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| « Last Edit: Jun 25th, 2007, 1:24am by towr » |
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Obob
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Re: i^2 = 1
« Reply #17 on: Jun 25th, 2007, 1:40am » |
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But saying what the two sides of (a2)1/2=(b2)1/2 mean is tricky. For the only consistent way of defining the square root of an arbitrary complex number z is that it is the set of all complex numbers w with w2=z. So this equality expresses an equality of sets, not one of numbers. And, indeed, if a2=b2 then (a2)1/2 = {a,-a} = {b,-b} = (b2)1/2
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| « Last Edit: Jun 25th, 2007, 1:42am by Obob » |
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srn437
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Re: i^2 = 1
« Reply #18 on: Sep 8th, 2007, 8:24pm » |
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(a^b)^c=a^bc doesn't work if a is negative and b and/or c is/are fractions(or are complex numbers).
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