Solving the Square Root of -1

In summary, the conversation discusses the square root of -1 and a proof that 1=-1. However, this proof is incorrect as it does not take into account the different branches of the square root function and leads to a contradiction. It is important to understand that the square root function can only be applied to non-negative real numbers and cannot be used for negative numbers.
  • #1
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hi
i was trying to determine the sqrt of -1,and here is wat i have done.
sqrt of -1=-1^2/4,this further gives [[-1]^2]^1/4, which further gives 1^1/4
therefore the sqrt of -1 is 1^1/4 which is 1
have u any objections,let me know
 
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  • #2
square 1, what is the answer? is it -1?
 
  • #3
but at least u saw my proof ,is there any mistake
 
  • #4
of course there is. you said that the square root of -1 is 1, if so squaring 1 should give you -1, it doesn't therefore there is a mistake, and a rather obvious one: you're reasoning is wrong, simple as that.

If you square a number, then square root it do you necessarily get what you started with? No, because you need to choose a branch of the sqaure root, or the 4'th root in this case, taking 4th powers is a 4-1 function, so taking 4th roots doesn't uno taking 4th powers.
 
  • #5
I can't find where his proof breaks any math rules.

Aside from the obvious distinction matt put out, I still don't see a reason why his proof is incorrect (although I know that [itex] i != 1 [/itex]

Is it the 4th root?
 
  • #6
[tex]a^{bc}=\left(a^{b}\right)^c[/tex]

is only valid in general for non-negative, real a, where roots are defined to always be non-negative real numbers. In other words, you simply can't prove that this equality holds when [itex]a[/itex] is negative (and you define roots appropriately). Notice that [itex]\pm i[/itex] are fourth roots of [itex]1[/itex] - but that doesn't mean that every fourth root of 1 is a square root of -1.

All abia ubong has done is give an example to show precisely what I said - that you can't (necessarily) express powers in that way - it gives you a wrong answer!

You really shouldn't think about it as "the square root of -1," anyway. The square root operation can be defined for negative numbers - but this is a definition. For example, you can define [itex]\sqrt{a} = ib[/itex] where [itex]\mathbb{R} \ni b > 0, \ b^2 = -a[/itex] when [itex]a<0[/itex]. You restrict its range appropriately so that it gives you exactly one answer. This does not guarantee that your new operation keeps all the nice properties that it had when it was only defined for positive numbers.
 
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  • #7
Ok, I didnt know that. thanks.
 
  • #8
abia
this is a easy way to think about it

(-a)^1/2 * (-b)^1/2 does not equal (ab)^(1/2)

unless a=-1 and b is positive

make sense (:
 
  • #9
pls take a look at this i/i=sqrt of (-1/-1)which eventually gives 1
also i/ican be expressed as 1/i *i now 1/i equals sqrt of (1/-1)which is sqrt of -1
that is i .now 1/i *i =i *iwhich is -1
what i have done now means that i/i=1=-1. i hope this answers matt question that 1=-1 which makes sqrt of -1 to be 1
 
  • #10
abia ubong said:
pls take a look at this i/i=sqrt of (-1/-1)which eventually gives 1
also i/ican be expressed as 1/i *i now 1/i equals sqrt of (1/-1)which is sqrt of -1
that is i .now 1/i *i =i *iwhich is -1
what i have done now means that i/i=1=-1. i hope this answers matt question that 1=-1 which makes sqrt of -1 to be 1


So you are telling us that you do believe that 1= -1?

For those of us who do not believe that 1= -1, (1)2= 1, not -1 and so 1 is not the square root of -1!
 
  • #11
pls take a look at this i/i=sqrt of (-1/-1)which eventually gives 1
also i/ican be expressed as 1/i *i now 1/i equals sqrt of (1/-1)which is sqrt of -1
that is i .now 1/i *i =i *iwhich is -1
what i have done now means that i/i=1=-1. i hope this answers matt question that 1=-1 which makes sqrt of -1 to be 1

This thread addresses this issue. Essentially, as matt explained, you seem to be confused about the fact that you can't mix and match different branches of the square root function and expect things to work out nicely.
 
  • #12
abia you really need to think about what your saying.
according to your proof the roots of x^2+1 are -1 and 1.
such function has no REAL ROOTS, but does have imaginary roots i,-i.
make sense?
 

1. How do you solve the square root of -1?

The square root of -1, also known as "i", is an imaginary number and cannot be solved in the traditional sense. It is a fundamental concept in complex numbers and plays a crucial role in various mathematical equations.

2. Can you explain the concept of imaginary numbers?

Imaginary numbers are numbers that cannot be expressed as a real number. They have the form "bi", where "b" is a real number and "i" is the imaginary unit, equal to the square root of -1. These numbers are crucial in solving problems that involve negative numbers under a square root.

3. How are imaginary numbers used in real-world applications?

Imaginary numbers are used in a variety of fields, including engineering, physics, and electronics. They are used to represent quantities that involve phase shifts, such as alternating currents and electromagnetic fields. They are also used in signal processing and financial modeling.

4. What is the difference between real and imaginary numbers?

Real numbers are numbers that can be expressed on a number line and include all rational and irrational numbers. Imaginary numbers, on the other hand, cannot be expressed on a number line and are typically used to represent quantities that involve negative numbers under a square root.

5. Are there any rules for operating with imaginary numbers?

Yes, there are rules for operating with imaginary numbers, just like there are rules for operating with real numbers. Some of these rules include the distributive property, the commutative property, and the associative property. It is important to note that the product of two imaginary numbers is a real number, while the sum of two imaginary numbers is also an imaginary number.

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