§l. The intermediate value theorem
203
words that for all e lying between a and b the value f(e) lies between f(a)
and f (b). ("Between" means either a ::; e ::; b or b ::; e ::; a.)
Suppose, for definiteness, that a < band f(a) < f(b) - equality is excluded by (i); it reduces to showing that
a < e < b ==> f(a) < f(e) < f(b).
If f(e) < f(a) < f(b) then the image of [e, bj, which is an interval since f is
continuous, contains f(a), whence there is a u E [e, bj such that f(a) = f(u),
contrary to (i). If f(a) < f(b) < f(e) then f(b) is in the image of [a, e], a new
contradiction by the same argument.
Having established this preliminary point, and with a and b as above, let
us show that f is increasing, i.e. that x < y implies f(x) < f(y) since f is
injective. It is simplest to examine the various possible positions of the pair
x, y relative to the pair a, b. If for example x < y < a < b, then a is between
x and b, so f(a) is between f(x) and f(b), so f(x) < f(a) since f(a) < f(b),
and since y is between x and a, f(y), which is between f(x) and f(a), is
> f(x). If, another possible case, a < x < b < y, one already knows that
f(a) < f(x) < f(b), or f(b) is between f(x) and f(y) since b is between x
and y, whence f(b) < f(y) and thus f(x) < f(y). Etc.
Example 1. Let us work on the interval I = 1R+ and consider the function
f(x) = x n , where n is an integer 2:: 1. It is continuous, strictly increasing,
with values 2:: 0, takes the value 0 for x = 0 and tends to +00 when x
increases indefinitely. The interval J = f(I) must therefore be 1R+. We thus
find again, and much more easily, a result we have already proved (Chap. II,
nO 10, example 3), and even more: Every real positive number possesses one
and only one positive nth root for any integer n 2:: 1 and this determines a
continuous function ofx. Since x p / q = (X P )l/ q , one concludes, more generally,
that the function x p / q is continuous for x > 0, for all p, q E Z, q =f. o.
Example 2. The same argument applies to the function exp(x). If one works
on R the function is continuous (sum of a power series) and has values 4 > o.
On the other hand it takes values as close to 0 or as large as one desires:
one has exp(n) = exp(l)n > 2 n for n E Nand exp( -n) = 1/ exp(n) < 1/2n.
It follows from this that the image of IR must be the set R+. of all strictly
positive real numbers. The function being strictly increasing, hence injective,
the equation
x = expy
thus has, for all x > 0, one and only one solution y E lR. As we have already said on several occasions, the inverse map is precisely the function
logy (example 1 of nO 2). It is common to define the logarithm function in
this way.
4 We have seen in Chap. II, nO 22, that exp(x + y) = exp(x) exp(y); it follows
immediately that exp(x) = exp(x/2)2 > 0 for any x E IR. Since it is clear that
expy> 1 for y > 0, the function is, moreover, strictly increasing.
203
words that for all e lying between a and b the value f(e) lies between f(a)
and f (b). ("Between" means either a ::; e ::; b or b ::; e ::; a.)
Suppose, for definiteness, that a < band f(a) < f(b) - equality is excluded by (i); it reduces to showing that
a < e < b ==> f(a) < f(e) < f(b).
If f(e) < f(a) < f(b) then the image of [e, bj, which is an interval since f is
continuous, contains f(a), whence there is a u E [e, bj such that f(a) = f(u),
contrary to (i). If f(a) < f(b) < f(e) then f(b) is in the image of [a, e], a new
contradiction by the same argument.
Having established this preliminary point, and with a and b as above, let
us show that f is increasing, i.e. that x < y implies f(x) < f(y) since f is
injective. It is simplest to examine the various possible positions of the pair
x, y relative to the pair a, b. If for example x < y < a < b, then a is between
x and b, so f(a) is between f(x) and f(b), so f(x) < f(a) since f(a) < f(b),
and since y is between x and a, f(y), which is between f(x) and f(a), is
> f(x). If, another possible case, a < x < b < y, one already knows that
f(a) < f(x) < f(b), or f(b) is between f(x) and f(y) since b is between x
and y, whence f(b) < f(y) and thus f(x) < f(y). Etc.
Example 1. Let us work on the interval I = 1R+ and consider the function
f(x) = x n , where n is an integer 2:: 1. It is continuous, strictly increasing,
with values 2:: 0, takes the value 0 for x = 0 and tends to +00 when x
increases indefinitely. The interval J = f(I) must therefore be 1R+. We thus
find again, and much more easily, a result we have already proved (Chap. II,
nO 10, example 3), and even more: Every real positive number possesses one
and only one positive nth root for any integer n 2:: 1 and this determines a
continuous function ofx. Since x p / q = (X P )l/ q , one concludes, more generally,
that the function x p / q is continuous for x > 0, for all p, q E Z, q =f. o.
Example 2. The same argument applies to the function exp(x). If one works
on R the function is continuous (sum of a power series) and has values 4 > o.
On the other hand it takes values as close to 0 or as large as one desires:
one has exp(n) = exp(l)n > 2 n for n E Nand exp( -n) = 1/ exp(n) < 1/2n.
It follows from this that the image of IR must be the set R+. of all strictly
positive real numbers. The function being strictly increasing, hence injective,
the equation
x = expy
thus has, for all x > 0, one and only one solution y E lR. As we have already said on several occasions, the inverse map is precisely the function
logy (example 1 of nO 2). It is common to define the logarithm function in
this way.
4 We have seen in Chap. II, nO 22, that exp(x + y) = exp(x) exp(y); it follows
immediately that exp(x) = exp(x/2)2 > 0 for any x E IR. Since it is clear that
expy> 1 for y > 0, the function is, moreover, strictly increasing.