- #1
Zorodius
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(-1)^2 = 1.
1 ^ (1/3) = 1.
so why is (-1)^(2/3) a horrible complex number?
1 ^ (1/3) = 1.
so why is (-1)^(2/3) a horrible complex number?
Zorodius said:(-1)^2 = 1.
1 ^ (1/3) = 1.
so why is (-1)^(2/3) a horrible complex number?
ObsessiveMathsFreak said:[tex]-1=e^{i \pi}[/tex]
[tex](-1)^{\frac{2}{3}}=\left(e^{i \pi}\right)^{\frac{2}{3}}=e^{\frac{2 i \pi}{3}}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}[/tex]
Zorodius said:(-1)^2 = 1.
1 ^ (1/3) = 1.
so why is (-1)^(2/3) a horrible complex number?
Zorodius said:(-1)^2 = 1.
1 ^ (1/3) = 1.
so why is (-1)^(2/3) a horrible complex number?
ObsessiveMathsFreak said:The power rule only works if the base number is positive.
D H said:That's only one of the roots. There are two others
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The power rule still does work in a sense, as one of the solutions to (-1)^(2/3) is one.
mathwonk said:the problem with saying sqrt(-1) = i and not -i, is that there is no such thing as i. i.e. there is no way to choose a particular square root of -1.
Both imaginary numbers [i and -i] have equal claim to being the number whose square is −1.
D H said:The standard meaning for [itex]\sqrt x[/itex] is indeed the positive square root of [itex]x[/itex] for positive [itex]x[/itex]. People should rightfully complain if you use this notation to include the negative root. By extension, it is better to say [itex]\sqrt{-1}=i[/itex] rather than [itex]\pm i[/itex]. Further extending the convention, the square root symbol means the principal root for any complex number. Some people will rightfully complain that this extension is carrying things a bit too far.
If your intent is to allow the negative root it is better to say something like [itex]1^{1/2}[/itex], [itex](-1)^{1/2}[/itex], or [itex]i^{1/2}[/itex]. By convention, this notation denotes all roots, not just the principal root. For example, the nth roots of unity are simply [itex]1^{1/n}[/itex].
mathwonk said:the problem with saying sqrt(-1) = i and not -i, is that there is no such thing as i. i.e. there is no way to choose a particular square root of -1. there are two of them and no way to prefer one over the other. so there is no way to say which number the letter i represents.
it is different for reals as you can take sqrt(1) as the one that itself has a real square root, i.e. the positive one.
put more abstractly, the complex numbers have a non trivial automorphism which interchanges i and -i, but the reals have none interchanging 1 and -1.
The reason for this is due to the laws of exponents. When we raise a number to a power, we are essentially multiplying that number by itself a certain number of times. In the case of x^(ab), we are multiplying x by itself ab times. However, when we have (x^a)^b, we are first raising x to the power of a, and then taking that result and raising it to the power of b. This results in a different answer than x^(ab) because we are only multiplying x by itself a times and then multiplying that result by itself b times.
Sure. Let's take x = 2, a = 3, and b = 2. When we calculate x^(ab), we get 2^(3*2) = 2^6 = 64. However, when we calculate (x^a)^b, we get (2^3)^2 = 8^2 = 64. While the final result is the same, the process to get there is different, resulting in x^(ab) not being equal to (x^a)^b.
Yes, there is one exception to this rule. When both a and b are integers, then x^(ab) = (x^a)^b. This is known as the power of a power property in the laws of exponents.
Understanding the difference between x^(ab) and (x^a)^b is important in simplifying and manipulating exponential expressions. It also helps to avoid common errors in calculations and ensures that the correct answer is obtained. Additionally, understanding the laws of exponents is crucial in many areas of mathematics and science, as exponential functions are used to model various real-world phenomena.
One way to remember the difference is to think about the order of operations. In x^(ab), we first raise x to the power of a, and then raise that result to the power of b. This can be represented as x^a * x^b. On the other hand, in (x^a)^b, we first raise x to the power of a, and then raise that result to the power of b. This can be represented as (x^a) * (x^a) * (x^a)... b times. Remembering this can help us avoid confusing the two expressions.