i know what the meaning of [tex]a^p[/tex] is when p is an integer or rational. e.g., [tex]a^3 = a.a.a[/tex] or [tex]a^{\frac{1}{5}}[/tex] is such a number that when multiplied five times gives the number a. but what is the menaing of [tex]a^p[/tex] when p is an irrational number?
same thing just a irrational amount of times instant of a integer/rational times. like a^3,14 (simple part of pi) its the same as a^3*a^0,14=a^3*a^(7/50)=a^3*(a^(1/50))^7 pi and such can be explained as a infinite amount of sums like this and by that a infintie amount of parts like this
This shows how bad things are in some schools. If you are going to consider irrational exponents then a better approch is needed. The modern way is: 1. Define the Natural logarithm using an integral, viz, [tex]ln(x)[/tex]. This is continuous and continuously differentiable for [tex]x>0[/tex] 2. Its inverse will be [tex]e^x[/tex] for all real [tex]x[/tex]. 3. This means that [tex]a^p = e^(pln(a))[/tex]. This is then taken to be the definition of [tex]a^p[/tex]. 4. This is or was A-level in England till the year 2000 but you don't need to know that, these days, as things are dummed down.
Well, it depends. For example, the value 2^pi does not mean anything, meaning that all it represent is another irrational number, untill you give pi a rational approximation. However, something like 10^log 2 has a meaning as it is another expression of 2. Here it is important to notice that ln 2 is a limit, as in the more digits you assign to log 2, the closer the expression 10^log 2 is to 2.
I am tempted to echo eds' statement, "This shows how bad things are in some schools"! Are you claiming that irrational numbers do not exist or "do not mean anything"? Your last statement "ln 2 is a limit, as in the more digits you assign to log 2, the closer the expression 10^log 2 is to 2." is correct but that means that "10^log 2" does in fact exist and "mean something"! You certainly can define, as eds said, 2^pi as e^(pi ln 2) but since pi ln 2 is an irrational number itself, you still haven't answered the original question: If p is an irrational number, then there exist a sequence of rational numbers {r_{i}} converging to p. We define a^{p} to make the function a^{x} continuous: [itex]a^p= \lim_{i\rightarrow \infty} a^{r_i}[/itex].
so, here r_{i} is a closer and closer approximation to the irrational number p as i becomes larger and larger. am i right? and can we say that [itex]r_{i} \rightarrow p[/itex] as [itex]i \rightarrow \infty[/itex]?
Werg, what do you mean by 'meaning'. pi, is just as meaningful as 1/2, mathematically, and if you want to discuss the philosophy of it there is another forum entirely devoted to that. Of course, mathematicians bandy about these tongue in cheek statements, but I wouldn't condone doing so here where the opportunity for misapprehension is so large.
There is a way to prove that the product of the irrational number gives another irrational number provided that we are dealing with roots of powers (such as 2^1/2 and 2^1/2).
:tongue: Also inverse operations don't count. I admit that I am not sure about the validity of what I said, but I remember seeing a problem that asked to prove that the product of irrational numbers, provided certain conditions, is also irrational. I'll check it right now, I'll get back at you as soon as I find it.
Here is one definition: [tex]a^p[/tex] is the unique number y>0 such that [tex]\int_{1}^{y} \frac{dt}{t}=p\int_{1}^{a} \frac{dt}{t}[/tex]
I prefer the integral definitions. A good way to approximate them by hand if you ever lose your calculator :P Its also interesting when your a is not equal to 0 or 1, you get a nice transcendental number. Look up Gelfond-Schnieder Theorem
O sorry I forget to mention a has to be algebraic as well. Applying the theorem again, we can see if p is irrational, then 1/p is also irrational, and then a is transcendental, not algebraic.