- #1
Ali Asadullah
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1^2=(-1)^2
taking square root on B.s
1=-1..?
taking square root on B.s
1=-1..?
Edi said:square root of (-1)^2 (witch is really same as sqrt 1 )= -1 and 1
and sqrt 1^2 (witch is .. you know.. same as sqrt 1 )= 1 and -1 ..
sooo... something to do with this.
but its quite fascinating.
Then Sir what will be the square of i?micromass said:Note, that it is however true that [tex](\sqrt{x})^2=x[/tex], but you can only do this when x is positive.
Ali Asadullah said:Then Sir what will be the square of i?
Ali Asadullah said:Bur Sir, Sqrt(x^2)=x is valid only when x is positive. How can it be valid for -1?
micromass said:Note, that it is however true that [tex](\sqrt{x})^2=x[/tex], but you can only do this when x is positive.
olivermsun said:I believe that it's conventionally taken to hold for negative numbers as well, since for positive x,
[tex]\sqrt{-x} = i\sqrt{x}[/tex]
so that
[tex](\sqrt{-x})^2 = (i\sqrt{x})^2 = -1 \cdot x = -x.[/tex]
It isn't and he did not say it was. Sqrt(x^2)= |x| which is equal to x only when x is positive.Ali Asadullah said:Bur Sir, Sqrt(x^2)=x is valid only when x is positive. How can it be valid for -1?
Ali Asadullah said:Another Question Sir,
Is it right that sqrt(x^2)=(sqrtx)^2 for all real x?
But if you view -x in the complex plane, then there is a well-defined way to choose a principal square root, and hence it's convenient to interpret [tex]\sqrt{-x}[/tex] in this way. If you're interested, I suppose you might find this in Brown and Churchill, Complex Variables and Applications, or something similar (but I'd have to double check).micromass said:In fact, the square root is only defined for positive real numbers. The square root is not defined for negative numbers. I.e. the value [tex]\sqrt{-1}[/tex] does not exist. The problem is of course that both i and -i qualify as being the square root, and there is not good reason to choose one above the other.
olivermsun said:But if you view -x in the complex plane, then there is a well-defined way to choose a principal square root, and hence it's convenient to interpret [tex]\sqrt{-x}[/tex] in this way. If you're interested, I suppose you might find this in Brown and Churchill, Complex Variables and Applications, or something similar (but I'd have to double check).
micromass said:I've checked the entire book of Brown and Churchill, and it appears that they never define something called [tex]\sqrt{-1}[/tex] or something similar.
micromass said:This is not true. The square root only yields ONE answer. Thus the square root of (-1)^2 is 1. It is NOT 1 and -1.
The square root of a number x is defined as the unique POSITIVE number y such that [tex]y^2=x[/tex]. It is not a multi-valued operation.
micromass said:This is not true. The square root only yields ONE answer. Thus the square root of (-1)^2 is 1. It is NOT 1 and -1.
The square root of a number x is defined as the unique POSITIVE number y such that [itex]y^2=x[/itex]. It is not a multi-valued operation.
No. Assuming the function evaluates to a positive real number, the square root of that number is a single positive real number, just as micromass says above.dalcde said:The square root of a function has two values.
Yes.dalcde said:The square root function yields only one unique number.
No, you are wrong on this one.dalcde said:The square root of a function has two values. The square root function yields only one unique number.
To find the solutions to this equation, you can take the square root of both sides. This will give you two solutions: x = √a and x = -√a, since the square root of a number can be positive or negative.
No, it depends on the value of a. If a is a positive number, there will always be two solutions. However, if a is a negative number, there will be no real solutions. If a is equal to 0, there will be one solution, which is x = 0.
No, there can only be a maximum of two solutions to this equation. This is because a quadratic equation, like x^2 = a, can have a maximum of two real solutions. If there are any additional solutions, they will be complex numbers.
The solutions to this equation can be graphed on a coordinate plane as points (x, y). Since we are solving for x, we can substitute different values for a to find the corresponding values of x. For example, if a = 4, the solutions would be (2, 0) and (-2, 0). These points can then be plotted on the graph.
Yes, the solutions to this equation can be irrational numbers. This means that they cannot be expressed as a fraction of two integers. For example, if a = 2, the solutions would be (√2, 0) and (-√2, 0), which are irrational numbers.