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Math ASSIGNMENT 9 SOLUTIONS
1. Let f . (a, b) > R be continuous, with (a, b)
R. Show that if f(r) = 0 for each rational
number r ∈(a, b), then f(x) = 0 for all x ∈(a, b).
Every real number x ∈(a, b) can be written as a limit of a sequence of
rational numbers,
{r_{n}}. Thus, since f is continuous,
Thus, f(x) = 0 for all x ∈(a, b).
2. Let f . (a, b) > R and g . (a, b) > R be continuous, with (a, b) R, so
that f(r) = g(r) for
each rational number r ∈(a, b). Prove that f(x) = g(x) for all x ∈(a, b).
We will use the previous problem . We know that f(r) = g(r) for all rational
numbers in
(a, b). Thus, define h(x) = f(x)g(x). Then, h is continuous and h(r) = 0 for
every rational
number in (a, b). Thus, h(x) = 0 for all x ∈(a, b). Therefore, f(x)g(x) = 0 for
all x ∈(a, b)
and the conclusion follows.
3. Define the function f by
Show that f is discontinuous at every x ∈R.
We will show that it is discontinuous at each rational and at each
irrational. Note that each
rational number can be written as the limit of a sequence of rational numbers,
for n sufficiently large. Likewise , it can be written as the limit of a sequence
of irrational
numbers, Thus, if f is continuous at r we
would have to have that
which cannot be, so f cannot be continuous at any rational number .
The similar argument will work to show that f is not continuous at any
irrational number.
4. Define the function h by
Show that h is continuous at x = 0 and at no other point.
Let {x_{n}} be any sequence that converges to 0. Then given any
> 0 there is
an so
that if n > N then
. When we apply h to this sequence, we get
a sequence
that is x_{n} if x_{n} is irrational and 0 if x_{n} is rational. However, since {x_{n}}
converges to 0, we
can show that for any > 0 we can find an N
∈N so that if
so the sequence converges and f is continuous at x = 0.
If x ≠ 0, then let
If x ∈Q then h(x) = 0, but there exists a
sequence of
irrational numbers converging to x and the sequence
will also converge
to x but
so that h cannot be continuous at x. If x ∈R\Q, there is a
sequence
of rationals that converges to x but whose functional values are all 0. Again,
the function
does not take a sequence convergent to x to a sequence convergent to h(x), so it
cannot be
continuous at any x ∈R\Q.
5. For each rational number x, write x as p/q where p and q are integers with
no common factors
and q > 0. Define the function g by
Thus, g(x) = 1 for all integers, Show
that g is continuous at each
irrational and discontinuous at each rational.
First, note that for any x ∈R there is an n ∈Z and a ∈(0, 1) so that x = n +
a and
g(x) = g(n + a). If x is irrational, then so is a and we have that g(x) = g(a) =
0. If x is
rational, then so is a and g(x) = g(a). Thus, to understand g it suffices to look
at the values
of g for x ∈(0, 1).
Any irrational number, a, has a nonrepeating decimal and we can consider
that decimal
expansion as one sequence converging to a. Given any > 0 there is an so
that
Thus, the sequence
converges to 0.
Now we need to show this for any sequence {x_{n}} converging to a. Let > 0 and
a ∈(0, 1)\Q.
Let so that x_{n} > a. We can write
with and
relatively
prime. We need to show that as n goes to infinity. If that is true, then
will go to 0 = f(a).
To show that , we will assume not. Thus, there is an M ∈N so
that
for k = 1, 2, 3, . . . . This means then that there are only a finite
number of fractions
of the form
in (0, 1) with
relatively prime. This means then
that the set
is finite. Thus, the limit of this sequence is a member of
the sequence,
i.e., a = p/q so that q < M. Thus, a ∈Q. This is a contradiction. Thus, the
denominators
of the fractions must go to infinity and for any sequence {x_{n}} converging to an
irrational a,
the sequence
converges to 0 = f(a) and the function is continuous at a.
If r ∈Q and r = p/q , then g(r) = 1/q. Choose
Now, for any sequence converging to r
that contains irrational numbers
so there is no way to guarantee that
for all n sufficiently large the values of g are close to g(r). Thus, g is not
continuous at r.
6. Let f and g be continuous functions on [a, b] such that f(a) ≥ g(a) and
f(b) ≤ g(b). Prove
that
Define a function h(x) = g(x)  f(x) on [a, b]. Then note that h is continuous
and h(a) =
g(a)  f(a) ≤ 0 and h(b) = g(b)  f(b) ≥ 0. If h(a) = 0, then f(a) = g(a) and we
are done.
Likewise , if h(b) = 0 then f(b) = g(b) and we are done. Thus, assume that h(a)
≠ 0 (so
h(a) < 0) and assume that h(b) ≠ 0 (so h(b) > 0). Therefore, by the
Intermediate Value
Theorem, there is an x_{0} ∈(a, b) so that h(x_{0}) = 0 and this implies that f(x_{0}) =
g(x_{0}).
7. Prove that for some x ∈(0, 1).
Define a function f on the interval [0, 1] by f(x) =
Then note that f
is continuous
and f(0) = 1 < 0 and f(1) = 1 > 0. Therefore, by the Intermediate Value
Theorem , there
is an x ∈(0, 1) so that f(x) = 0. At that point we have
for some x ∈(0, 1).
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