Real powers
The question is the following: we know
what 22 or 23 means, or even 22/3 , the number whose
√
2
2
cube equals 2 . But what does 2 mean? We will give the definition Rudin has in the
exercises of Chapter 1. We will only use the following facts for r, s > 0, n, m ∈ Z:
• rn+m = rn rm .
• (rn )m = rmn .
• (rs)n = rn sn .
• if r > 1 and m ≥ n then rm ≥ rn . If r < 1 and m ≥ n then rm ≤ rn .
• if s < r and n > 0 then sn < rn . If s < r and n < 0 then sn > rn .
Natural roots. We first define the n-th root of a real number, for n ∈ N.
Theorem 0.1. For any r > 0 and n ∈ N there exists a unique positive real number y such
that y n = r.
Proof. The proof is Theorem 1.21 in Rudin. The idea is to construct the set
S = {x > 0 : xn ≤ r}
and to show that S is nonempty, bounded above, and thus has a supremum. Calling y this
supremum, he then shows y n = r. The proof of this is somewhat involved and is similar to
our proof (from the first lecture) that {a ∈ Q : a2 < 2} does not have a greatest element.
To show there is only one such y, we note that 0 < y1 < y2 implies that y1n < y2n and so
if y1 6= y2 are positive then y1n 6= y2n .
This definition extends to integer roots.
Definition 0.2. If r > 0 and n ∈ N we define r−1/n as the unique positive real number y
such that y n = 1/r.
Rational powers. The above definitions allow us to define rational powers.
Definition 0.3 (Preliminary definition of rational powers). If r > 0 and m, n ∈ N we define
rm/n to be the unique positive real number y such that y n = rm . Also r−m/n is defined as
(1/r)m/n .
Because a rational number can have more than one representation m/n we need to show
this is well defined.
Proposition 0.4. If a positive a ∈ Q can be represented by m/n and p/q for m, n, p, q ∈ N
then for all r > 0,
rm/n = rp/q .
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Proof. First note that (rm/n )nq = ((rm/n )n )q = rmq and (rp/q )nq = ((rp/q )q )n = rpn . However
as m/n = p/q we have pn = mq and so these numbers are equal. There is a unique nq-th
root of this number, so rm/n = rp/q .
Note that the above proof applies to negative rational powers: suppose that r > 0 and
a ∈ Q is negative such that a = −m/n = −p/q. Then
r−m/n = (1/r)m/n = (1/r)p/q = r−p/q .
Definition 0.5 (Correct definition of rational powers). If r > 0 and a > 0 is rational we
define ra = rm/n for any m, n ∈ N such that a = m/n. If a < 0 is rational we define
ra = (1/r)−a .
Properties of rational powers. Let a, b ∈ Q and r, s > 0.
• If a = m/n for m ∈ Z and n ∈ N then ra is the unique positive number such that
(ra )n = rm .
Proof. For m ≥ 0 this is the definition. For m < 0, this is because (ra )n = ((1/r)−a )n =
(1/r)−m = rm and if s is any other positive number satisfying sn = rm then uniqueness
of n-th roots gives s = ra .
• ra+b = ra rb .
Proof. Choose m, p ∈ Z and n, q ∈ N such that a = m/n and b = p/q. Then a + b =
mq+np
and therefore ra+b is the unique positive number such that (ra+b )nq = rmq+np .
nq
But we can just compute
(ra rb )nq = ((ra )n )q ((rb )q )n = rmq rnp = rmq+np .
And by uniqueness we get ra rb = ra+b .
• (ra )b = rab .
Proof. Write a = m/n and b = p/q for m, p ∈ Z and n, q ∈ N. Then rab is the unique
positive number such that (rab )nq = rmp . But
((ra )b )nq = (((ra )b )q )n = ((ra )p )n = ((ra )n )p = (rm )p = rmp ,
giving (ra )b = rab .
• (rs)a = ra sa .
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Proof. Again write a = m/n for m ∈ Z and n ∈ N. Then (rs)a is the unique positive
number such that ((rs)a )n = (rs)m . But
(ra sa )n = (ra )n (sa )n = rm sm = (rs)m .
• If r > 1 and a ≥ b then ra ≥ rb . If r < 1 and a ≥ b are rational then ra ≤ rb .
Proof. Suppose first that r > 1 and a ≥ 0 with a = m/n for m, n ∈ N. Then if ra < 1,
we find rm < 1n = 1, a contradiction, as rm > 1. So ra > 1. Next if a ≥ b then
a − b ≥ 0 so ra−b ≥ 1. This gives ra = ra−b rb ≥ rb .
If r < 1 then ra (1/r)a = 1a = 1, so r−a = (1/r)a . Similarly r−b = (1/r)b . So since
1/r > 1 we get r−a = (1/r)a ≥ (1/r)b = r−b . Multiplying both sides by ra rb we get
ra ≤ rb .
• If s < r and a > 0 then sa < ra . If s < r and a < 0 then sa > ra .
Proof. Let a = m/n with m, n ∈ N. Then if sa ≥ ra we must have sm = (sa )n ≥
(ra )n = rm . But this is a contradiction since sm < rm . This proves the first statement.
For the second, 1/s > 1/r so sa = (1/s)−a > (1/r)−a = ra .
Real powers. We define a real power as a supremum of rational powers.
Definition 0.6. Given r > 1 and t ∈ R we set
rt = sup{ra : a ∈ Q and a ≤ t} .
If 0 < r < 1 then define rt = (1/r)−t .
Proposition 0.7. If a ∈ Q then for r > 0, the definition above coincides with the rational
definition.
Proof. For this proof, we take ra to be the defined as in the rational powers section.
Suppose first that r > 1. Clearly ra ∈ {rb : b ∈ Q and b ≤ a}. So to show it is the
supremum we need only show it is an upper bound. This follows from the fact that b ≤ a
implies rb ≤ ra (proved above).
If 0 < r < 1 then ra (r−1 )−a = (1/r)−a so the definitions coincide here as well.
Properties of real powers. Let t, u ∈ R and r, s > 0.
• rt+u = rt ru .
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Proof. We will use the following statement, proved on the homework. If A and B are
nonempty subsets of [0, ∞) which are bounded above then define AB = {ab : a ∈
A, b ∈ B}. We have
sup(AB) = sup A sup B .
(1)
It either of the sets consists only of 0, then the supremum of that set is 0 and both
sides above are 0. Otherwise, both sets (and therefore also AB) contain positive
elements. For any element c ∈ AB we have c = ab for some a ∈ A, b ∈ B. Therefore
c = ab ≤ sup A sup B and therefore this is an upper bound for AB. As sup(AB) is the
least upper bound, we get sup(AB) ≤ sup A sup B. Assuming now for a contradiction
that we have strict inequality, because sup A > 0 we also have sup(AB)/ sup A < sup B.
Thus there exists b ∈ B such that sup(AB)/ sup A < b. As b must be positive, we also
have sup(AB)/b < sup A and there exists a ∈ A such that sup(AB)/b < a, giving
sup(AB) < ab. This is clearly a contradiction.
Now to prove the property, suppose first that r > 1. By the statement we just proved,
we need only show that
{rb : b ∈ Q and b ≤ t + u} = AB ,
where A = {rc : c ∈ Q and c ≤ t} and B = {rd : d ∈ Q and d ≤ u}. (This is because
these are sets of non negative numbers.) This holds because each rational b ≤ t + u
can be written as a sum of two rationals c, d such that c ≤ t and d ≤ u.
For 0 < r < 1 we have rt+u = (1/r)−(t+u) = (1/r)(−t)+(−u) = (1/r)−t (1/r)−u = rt ru .
• (rs)t = rt st .
Proof. We first note that
r−t = 1/rt .
(2)
This is true because r−t rt = r0 = 1, so r−t = 1/rt .
For the property, if r, s > 1 then we can just use equation (1), noting that {(rs)a :
a ∈ Q and a ≤ t} = AB, where A = {ra : a ∈ Q and a ≤ t} and B = {sa : a ∈
Q and a ≤ t}. If 0 < r < 1 but s > 1 with rs > 1 we get (rs)t /rt = (rs)t (1/r)−t .
We now use equation (1) again, noting that {sa : a ∈ Q and a ≤ t} = AB, where
A = {(rs)a : a ∈ Q and a ≤ t} and B = {(1/r)a : a ∈ Q and a ≤ −t}. This gives
(rs)t /rt = st . This same proof works if r > 1 but 0 < s < 1 with rs > 1. If 0 < r < 1
but s > 1 with rs < 1 we consider st /(rs)t = st (1/(rs))−t and use the above argument.
This also works in the case r > 1 but 0 < s < 1 with rs < 1. Finally, if 0 < r < 1 and
0 < s < 1 then (rs)t = (1/(rs))−t = ((1/r)(1/s))−t = (1/r)−t (1/s)−t = rt st .
• (rt )u = rtu .
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Proof. We will first show the equality in the case r > 1 and t, u > 0. We begin with
the fact that (rt )u is an upper bound for {ra : a ∈ Q and a ≤ tu}. So let a ≤ tu be
rational and assume further that a > 0. In this case we can write a = bc for b, c ∈ Q
and b ≤ t, c ≤ u. By properties of rational exponents, we have ra = (rb )c . As rb ≤ rt
(by definition) we get from monotonicity that (rb )c ≤ (rt )c . But this is an element of
the set {(rt )d : d ∈ Q and d ≤ u}, so (rt )c ≤ (rt )u . Putting these together,
ra = (rb )c ≤ (rt )c ≤ (rt )u .
This shows that (rt )u is an upper bound for {ra : a ∈ Q and 0 < a ≤ tu}. For the case
that a < 0 we can use monotonicity to write ra ≤ r0 ≤ (rt )u . Putting this together
with the case a > 0 gives that (rt )u is an upper bound for {ra : a ∈ Q and a ≤ tu}
and therefore rtu ≤ (rt )u .
To prove that (rt )u ≤ rtu we must show that rtu is an upper bound for {(rt )a : a ∈
Q and a ≤ u}. For this we observe that rt > 1. This holds because t > 0 and therefore
we can find some rational b with 0 < b < t. Thus rt ≥ rb > r0 = 1. Now let a be
rational with 0 < a ≤ u; we claim that (rt )a ≤ rtu . Proving this will suffice since if
a < 0 then (rt )a < (rt )0 = 1 ≤ rtu . To show the claim, note that if we show that
rt ≤ (rtu )1/a we will be done. This is by properties of rational exponents: we would
then have
a
(rt )a ≤ (rtu )1/a = rtu .
So we are reduced to proving that
sup{rb : b ∈ Q and b ≤ t} ≤ (rtu )1/a ,
which follows if we show that for each b ∈ Q such that b ≤ t, we have rb ≤ (rtu )1/a .
Again, this is true if rab ≤ rtu because then rb = (rab )1/a ≤ (rtu )1/a . But a ≤ t and
b ≤ u so rab ≤ rtu . This completes the proof of (rt )u = rtu in the case r > 1 and
t, u > 0.
In the case r > 1 but t > 0 and u < 0, we can use (2):
(rt )u = 1/(rt )−u = 1/r−tu = rtu .
If instead r > 1 but t < 0 and u > 0,
(rt )u = (1/r−t )u = 1/(r−t )u = 1/r−tu = rtu .
Here we have used that for s > 0 and x ∈ R, (1/s)x = 1/sx , which can be verified as
1 = (s(1/s))x = sx (1/s)x . Last if r > 1 but t < 0 and u < 0, we compute
(rt )u = ((1/r)−t )u = 1/(r−t )u = 1/r−tu = rtu ,
completing the proof in the case r > 1.
If 0 < r < 1 then
(rt )u = ((1/r)−t )u = (1/r)−tu = rtu .
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• If r > 1 and u ≤ t then ru ≤ rt . If 0 < r < 1 and u ≤ t then ru ≥ rt .
Proof. Assume r > 1. If u = 0 and t > 0 then we can find a rational b such that
0 < b ≤ t, giving rt ≥ rb > r0 = 1. For general u ≤ t we note 1 ≤ rt−u , so multiplying
both sides by the (positive) ru we get the result.
If 0 < r < 1 then ru = (1/r)−u ≥ (1/r)−t = rt .
• If s < r and t > 0 then st < rt . If s < r and t < 0 then st > rt .
Proof. First consider the case that s = 1. Then r > 1 and for any t > 0 we can find a
rational b such that 0 < b < t. Therefore rt ≥ rb > r0 = 1. For general s < r we write
rt = st (r/s)t > st . If t < 0 then st = (1/s)−t > (1/r)−t = rt .
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