Something that I've never been taught

[epkid08]

Probably not that important...

3^n = k

What does k equal, expressed through integers, when n is greater than or less than, but not equal to any integer?

Example:

3^4 = 3*3*3*3
3^2.3499 = ?

 

[arildno]

VERY good question, epkid!

As you can readily see, the standard way of defining INTEGER exponents is just meaningless for non-integer exponents! (You CAN'T multiply a number a non-integral number of times with itself!)

However, there IS a way to define such numbers rigourously, but that takes a bit too long to expound.

I'll therefore argue for this in a rather intuitive way:

Now, you know that if we multiply powers with the same base, that is equal to retaining that base and adding the exponents together, that is:
$$a^{n}*a^{m}=a^{n+m}$$
All right?

Let's play a bit with this, and see if we can attach a WELL-KNOWN meaning to the symbol $2^{\frac{1}{2}}$

If our basic rule for multiplying powers are to hold, we must have:
$$2^{\frac{1}{2}}*2^{\frac{1}{2}}=2^{\frac{1}{2}+\frac{1}{2}}=2^{1}=2$$

That is, we must have $2^{\frac{1}{2}}=\sqrt{2}$!

Raising a number in the power of one-half is the same as taking the square root of that number!

And similarly, raising a number to the power of 1/3 is the same as taking the third root of the number, and so on.

But then, we can give meaning to RATIONAL exponents.

FOr, writing: $a^{\frac{p}{q}}=(a^{\frac{1}{q}})^{p}$
(using the rule $(a^{m})^{n}=a^{mn}$!), we see that
$a^{\frac{p}{q}}$ is just the q-th root of "a" multiplied with itself p times!

Okay about that?

 

[HallsofIvy]

Probably not that important...

3^n = k

What does k equal, expressed through integers, when n is greater than or less than, but not equal to any integer?

Example:

3^4 = 3*3*3*3
3^2.3499 = ?

What do you mean by "expressed through integers"? Of course, in this case, when n is not an integer itself, neither is k. In fact, it is not too difficult to prove that k is not rational number. That means that it cannot be expressed as a fraction of two integers. Now, if n itself is a fraction, say n= 3/2, then 3^n= 3^(3/2) is the squareroot of 27, $\sqrt{27}$. Would you consider that to be "expressed through integers"?

 

[epkid08]

What do you mean by "expressed through integers"? Of course, in this case, when n is not an integer itself, neither is k. In fact, it is not too difficult to prove that k is not rational number. That means that it cannot be expressed as a fraction of two integers. Now, if n itself is a fraction, say n= 3/2, then 3^n= 3^(3/2) is the squareroot of 27, $\sqrt{27}$. Would you consider that to be "expressed through integers"?

Well I already knew that, but I'm talking way back to grade school, when they defined 3^3 = 3*3*3. I guess I was thinking maybe there is a way to define 3^2.23 through real integers, such as 3*3*3*2.23 (obviously not).

On another note, would it be excepted notation to leave an irrational number as an exponant (or even to have one in the first place)?

 

[Crosson]

This is a frequent question with good university students who are re-examining their fundamentals!

Setup the problem:

3^2.3499 = 3^(23499/100000) = x

implies

x^100000 = 3^23499

Clearly we have given a relationship, solely in terms of integers, that determines the value of x (now the problem is reduced to finding the root of a polynomial).

The principle that we used to determine x, and to interpret the rational exponent, was that we wanted the algebraic properties of integer exponents to carry over to rational ones.

If the exponent b is irrational in a^b, then we must define a^b as the limit of a sequence of rational approximations, or some other equivalent.

 

[arildno]

On another note, would it be excepted notation to leave an irrational number as an exponant (or even to have one in the first place)?

As for irrational exponents, that can well be defined, though the mathematical drudgery involved in the rigorous definition of it precludes my answer.

 

[Werg22]

I believe the easiest way is to define exponentiation with base e then proceed to show that e^x = y has a solution for every y > 0 (which is rather trivial result when e is considered in terms of the integral of 1/x).