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*always*works or is it just a general rule.

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where a and b are what? Natural numbers? Reals?

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- #4

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To get a proof, you will have to start with natural numbers, then rational, then real, and finally complex. In each subsequent case, I think you boil it down to the previous case.

Going from rational to real will be the hardest case, I think.

In Natural numbers, seems like you can use induction.

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MathematicalPhysicist

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quasar987

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- #7

lurflurf

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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:definea^b as having the same properties as exponentiation to a rational number?

a^x<a^y <-> x<y

we have for p and q rational x real

p<x<q <-> a^p<a^x<a^q

thus for any real number (having defined a rational exponential) we can trap a^x in as small an interval as we like by traping x between two rationals that are sufficiently close. Thus it is not impossible to give a clear meaning to exponentiation of a real number, but it is convient to give that meaning in terms of exponentiation of a rational number.

thus

[tex]a^x:=\lim_{r\rightarrow x}a^r[/tex]

where the limit is taken with r constrained to rational numbers

This can be avoided by defining exponential and log functions then (for a>0)

a^x:=exp(x*log(a))

for this to work log and exp much be defined, my perfered definition being

exp is the function (which exist and is unique) having

exp(x+y)=exp(x)*exp(y) for all real x and y

[tex]\lim_{x\rightarrow 0}\frac{\exp(x)-1}{x}=1[/tex]

and

log is the function (which exist and is unique) having

log(x*y)=log(x)+log(y) for all real x and y

[tex]\lim_{x\rightarrow 0}\frac{\log(1+x)}{x}=1[/tex]

This method has the problem of making exp and log appear a bit mysterious.

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[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.

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lurflurf

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The reason it breaks down is exp is many to one.εllipse said:

[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.

exp(x)=exp(y) <-> 2pi*i|(x-y)

We chose a principle value for log, but the rule

(x^a)^b=x^(a*b)

cannot hold for all values

in principle value

(x^a)^b=exp(b*log(x^a))=exp(b*log(exp(a*log(x))))

x^(a*b)=exp(a*b*log(x))

which is true when

log(exp(a*log(x)))=a*log(x)

which is true when

-pi<Im[a*log(x)]<=pi

other wise the rule fails

the given example

(-1)^(.5*2)=(-1)^2=exp(2*log(-1))=exp(2*pi*i)=1

((-1)^.5)^2=exp(2*log((-1)^.5))=exp(2*log(exp(.5*log(-1))))=exp(2*log(exp(.5*pi*i)))=exp(2*log(i))=exp(2*pi*i/2)=exp(pi*i)=-1

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HallsofIvy

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That is for m,n positive integers. The two laws: a

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LeonhardEuler

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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:

[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.

[tex]x^a=e^{a\log{x}}=e^{a(\ln{|x|} + i arg x)}=e^{a\ln{|x|}}e^{ia Arg x +2ia\pi} = e^{a\ln{|x|}}e^{ia Arg x}e^{2iak\pi}[/tex]

If x is positive real, then Arg x=0, if negative then Arg x= [itex]\pi[/itex]. [itex]e^{2k\pi}[/itex] is 1 for any integer k. This means that if a is also an integer, the function will be single-valued. If a is a fraction then ka will only be an integer for some values of k. If a=1/2, then there will be two possible values of [itex]e^{2aik\pi}[/itex], hence two values of [itex]x^a[/itex].

This definition also provides an easy answer to εllipse's question:

[tex](x^a)^b=(e^{a\log{x}})^b[/tex]

For convenience, let [itex]w = e^{a\log{x}}[/itex]

[tex](x^a)^b=w^b=e^{b\log{w}}[/tex]

Now, since log(e^(z))=z+2kpi,

[tex](x^a)^b=\exp{[b\log{e^{(a\log{x})}]}=e^{b(a\log{x}+2k\pi)} = e^{ba\log{x}}e^{2kb\pi}[/tex]

On the other hand,

[tex]x^{ab}=e^{ab\log{x}}[/tex]

So, if b is an integer:

[tex](x^a)^b=e^{ba\log{x}}e^{2kb\pi}=e^{ba\log{x}}= e^{ab\log{x}} \rightarrow (x^a)^b=x^{ab}[/tex]

As was to be shown. Note that if b is not an integer, then the function is multiple-valued and the relation need not hold if the wrong branch cut is chosen, as in the example εllipse gave. The fact that k can be chosen to be zero, though, guarantees the existance of a branch cut that will cause the relation to hold.

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quasar987

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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.

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James R

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This means:

[tex](a^4)^3 = a^4 a^4 a^4 = a^{12} = a^{(4 \times 3)}[/tex]

So, it seems

[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.

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LeonhardEuler

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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" ?!

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jcsd

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so:

a^2 = a*a

a^3 = a*a*a

a^n = a*a*....*a (where n is a ntural number)

are just definitions

From the basic properties of multplication, we can obtain a^n*a^m = a^(n+m) which allows us to dfeine integer powers and (a^n)^m which allows us to define rational powers.

So the relationship (a^n)^m comes from the basic properties of multplication and is used to extend our defitnion into rational powers.

The defitnion of expoentaion can be extended to real and complex powers as has already been discussed above.

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so whats the answer to that question

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or to write it another way, (a^4)^3 = (a*a*a*a)(a*a*a*a)(a*a*a*a)

This means:

[tex](a^4)^3 = a^4 a^4 a^4 = a^{12} = a^{(4 \times 3)}[/tex]

So, it seemssensiblethat, 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.

Since everything is multiplication where a(bc)=(ab)c and ab=ba you can ignore the brackets.... so a is multiplied by itself 12 times or a^12

.. or 3 groups of (a*a*a*a), 3 x 4 = 12

Edit:.... wow, this was a really old thread!

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what about (-1)^ [tex]\frac{1}{\sqrt{2}}[/tex] ?

is it a real number ?

is it a real number ?

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disregardthat

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Yes, that is the definition of the square root symbol.What is the definition of the square root symbol if not "raised to the power 1/2" ?!

We can define a^r for any positive real a and real r in the following naive way (it can be done more effectively by starting out with the log and exp-functions as power series):

For natural n we define a^n as multiplication n times (recursively, if you insist). That (a^n)^m=a^(nm) is trivially proved. To prove the order-preserving properties of a^n as a function in n is also trivial. (that is, a^n<=a^m <--> n<=m for a>=1, and oppositely for a<1). Extending to all negative integers is easy, and order will still be preserved.

a^(1/n) is defined as the positive zero of the polynomial x^n-a (which exists by the intermediate theorem and is unique as seen by factoring). We define a^(p/q) as (a^(1/q))^p which is well-defined by the previous definitions. This definition obviously coincide on the common domain of the previous definition. Again, that (a^(p/q))^(n/m)=a^(pn/qm) is trivially proved. That a^(p/q) is order-preserving is also easy to prove.

Now, for real r, let r_n be a increasing sequence of rational numbers converging to r. We define a^r as the limit the sequence a^(r_n). For a>1, this is an increasing sequence bounded above by a^k for some rational k>r, which can be shown using the order-preserving properties we arrived at previously. For a<1 it is a decreasing sequence similarly bounded below. For a = 1 it is constant, so the limit exists for any real r, and can trivially be shown unique for any such converging sequence. Similarly this definition coincide with the previous ones.

It remains to prove that (a^r)^s=a^(rs), but this is now trivial by using converging sequences for s and r.

For negative a you immediately run into trouble in the second step, when you want to define a^(1/n) by x^n-a which has no roots for even n.

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