- #1
εllipse
- 197
- 0
Can anyone show me the reason [itex](x^a)^b = x^{ab}[/itex]? Is this something that always works or is it just a general rule.
By continuity means that the exponential f:Q->R from rationals to reals is a continuous function. The rationals are dense in the reals, so only one continuous map f:R->R exist that agrees with the rational exponetnial. To put things more simply, and assuming additonally that for a positive real aquasar987 said:I am interested in this also. And specifically, what is a number raised to an irrational number? Wiki says "Exponentiation to an arbitrary real exponent can then be defined by continuity." What is meant by 'continuity' exactly? Is it that even though it is impossible to give a clear meaning to a^b (where b is irrationnal), we define a^b as having the same properties as exponentiation to a rational number?
The reason it breaks down is exp is many to one.εllipse said:Well here's what has me confused.
[tex]((-1)^2)^\frac{1}{2}=(-1)^{2\frac{1}{2}}=(-1)^1=-1[/tex]
vs
[tex]((-1)^2)^\frac{1}{2}=(1)^\frac{1}{2}=1[/tex]
So either the powers can't be multiplied in this case, or I'm mistaken and [itex]1^\frac{1}{2}={\pm}1[/itex]. I've been told the latter is false, so I was wondering where the multiplication of powers came from. I was assuming it would be easy to show, but maybe not.
It is true that [itex]1^{\frac{1}{2}}=\pm 1[/itex], but not true that [itex]\sqrt{1}=\pm 1[/itex]. This is because of the definition of the square root symbol. The definition of fractional, irrational and complex powers is as follows: In general, [itex]x^a[/itex] is defined as [itex]e^{a\log{x}}[/itex]. log is a multiple-valued function: [itex]\log{z}=\ln{|z|}+i arg(z)[/itex], where arg z is the argument of z and can take on any value of the form [itex] arg z = Arg z +2\pi k, k \in \mathbb{Z}[/itex] This means thatεllipse said:Well here's what has me confused.
[tex]((-1)^2)^\frac{1}{2}=(-1)^{2\frac{1}{2}}=(-1)^1=-1[/tex]
vs
[tex]((-1)^2)^\frac{1}{2}=(1)^\frac{1}{2}=1[/tex]
So either the powers can't be multiplied in this case, or I'm mistaken and [itex]1^\frac{1}{2}={\pm}1[/itex]. I've been told the latter is false, so I was wondering where the multiplication of powers came from. I was assuming it would be easy to show, but maybe not.
What is the definition of the square root symbol if not "raised to the power 1/2" ?!LeonhardEuler said:It is true that [itex]1^{\frac{1}{2}}=\pm 1[/itex], but not true that [itex]\sqrt{1}=\pm 1[/itex]. This is because of the definition of the square root symbol.
The definition is the positive root. That's why it is often written as [itex]\pm \sqrt{x}[/itex] when you want to indicate either possibility. It is convenient because if you write [itex]\sqrt{16}[/itex] everyone will know you mean 4, while if you write [itex]-\sqrt{16}[/itex] everyone knows you mean -4.quasar987 said:What is the definition of the square root symbol if not "raised to the power 1/2" ?!
James R said:Consider [itex](a^4)^3[/itex], for example.
This means:
[tex](a^4)^3 = a^4 a^4 a^4 = a^{12} = a^{(4 \times 3)}[/tex]
So, it seems sensible that, in general
[tex](x^a)^b = x^{ab}[/tex]
This is, of course, nowhere near a proof, but it's at least an intuitive argument for the natural numbers.
quasar987 said:What is the definition of the square root symbol if not "raised to the power 1/2" ?!
This is because according to the power rule of exponents, when we raise a power to another power, we multiply the exponents. In this case, (x^a)^b is the same as x^(a*b), which is equivalent to x^{ab}.
A real-life example of this rule can be seen in compound interest calculations. When we compound interest annually, we raise the initial amount to the power of the number of years. If we then compound the interest monthly, we raise the result to the power of 12 (number of months in a year), which is equivalent to raising the initial amount to the power of 12 times the number of years.
No, this rule holds true for all real numbers and variables, as long as x is not equal to 0.
This rule is related to the distributive property because both involve the multiplication of terms. In this case, we are multiplying exponents, while in the distributive property, we are multiplying coefficients and variables.
Yes, this rule can be generalized to any power raised to another power. For example, (x^a)^b can also be written as (x^b)^a, which is equivalent to x^{ba}. This is known as the commutative property of exponents.