Advanced Power Taking, II
This is part two a translation of an article that appeared in the January 2006 issue of Pythagoras.
What is 2√3 exactly? In this post we shall give an exact definition of the exponential function 2x for all real numbers x. We also look at other bases for exponential functions.
We learned what what 2q means when q is a rational number; what about 2√3?. Because we cannot write √3 as a fraction we will have to come up with something really new.
We put the algebra aside and concentrate on an other property of the function q→2q: if p<q then 2p<2q; in words: taking powers of 2 is monotone.
This follows from the addition property: we have q=p+(q-p) and so 2q=2p×2q-p. Because 2q-p>1 we find that 2q>2p.
Exercise Show that, indeed, 2r>1 if r is a positive rational number.
We now intend to ensure that the function x→2x becomes monotone. For √3 this means two things: if p<√3 then 2p<2√3 and if p>√3 then 2p>2√3. The question then becomes whether there is a real number that meets those requirements, and preferably exactly one because then we do not have to choose: that one number will be the value of 2√3.
At least one
We are going to prove that there is at least one number x with the property that 2p<x<2q for all rational numbers with p<√3<q. For this we use the fundamental property of the number line that we used to prove that numbers like ∛2 exist:
Theorem Every nonempty set of numbers that has an upper bound also has a least upper bound.
In a few more words: if A is a subset of R for which there is an x such that a≤x for all a∈A (x is an upper bound) then there is an upper bound that is smaller than all other upper bounds.
We apply this to the set A={2p:p<√3}. This set does have an upper bound, 22=4, for example. So there is an upper bound, α, that is the smallest among all upper bounds. For this α we know that 2p≤α when p<√3, because α is an upper bound for A. Also, if q>√3 then α≤2q, because every such 2q is an upper bound for A (and α is the smallest). We want to improve the ≤ to <. For this we need another important property of R: between any two real numbers there is a rational number. (We will prove this later.)
We use that property as follows: if p<radic;3 then there is a rational number r such that p<r<√3. For that r we have 2p<2r≤α and so 2p<√3. In exactly the same way you can show that α<2q whenever q>√3.
At most one
Now we must show that only α meets our requirements. We do that by showing that the diameter of the set of suitable numbers is equal to 0; and that means that only one number will fit in the set.
Take two rational numbers p and q in the interval (1,2) with p<√3<q. The difference 2q-2p is equal to 2p(2q-p-1). On the interval (0,1) we have 2r≤1+r (we will show this later). We can use this to bound 2q-2p:
The last inequality is because p<2. Now we can show that the diameter of the set of suitable numbers is equal to 0.
To see that the diameter is less than 10-6 we calculate the first seven digits of √3 after the decimal point. Then we see that p<√3<q, where p=1.7320508$ and q=1.7320509. We have q-p=10-7 so that
By doing this for ever better approximations of √3 we find that the diameter is smaller than every positive number that we can specify. The diameter is therefore equal to 0.
Properties of 2x
In exactly the same way as for √3 we can determine what 2x is for every real number x that is not rational: 2x is the unique number y with the property that 2p<y<2q for all rational numbers p and q that satisfy p<x<q.
The function that we construct in this way has all properties that we may expect of an exponential function.
Exercise Verify that 2x×2y=2x+y for all real numbers x and y.
We can also prove that 2x<2y when x<y (we only know this if one of x and y is rational). First take a rational number p with x<p<y; then we know 2x<2p<2y
Many rational numbers
Take two real numbers x and y with x<y. Then the difference y-x is positive., hence there is a natural number n such that 1/n<y-x. Then we also have 1<nx-ny; that means there is a whole number m between nx and ny. But then we see that x<m/n<y. So there is a rational number between x and y.
An inequality
We still need to show that 2r<1+r when r is an rational number in the interval (0,1).
We use the AGM-inequality again.
Write r=k/n. And apply the AGM-inequality
to the following n numbers: x1=…=xk=2 and xk+1=…=xn=1. We get
or 2k/n≤1+k/n.
Other bases
For every number a>1 we can define the function ax in exactly the same way as for 2. The proof that there is exactly one possibility for ax when x is not rational needs a few modifications.
If you replace every 2 by an a in the section on the inequality then you will get
Simplification of the right-hand side yields
When we determine a√3 we take p<q in the interval (1,2) and then we get
and this helps in showing that there is only one number that is suitable to be a√3.
If 0<a<1 we define ax via 1/a: ax=1/(1/a)x.
