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18.100B Problem Set 1 Solutions
1) The proof is by contradiction. Assume ∃r ∈ Q such that
r^{2} = 12. Then we may write r as
with a, b ∈ Z and we can assume that a and b have no common factors . Then
so 12b^{2} = a^{2}.
Notice that 3 divides 12b^{2} and hence 3 divides a^{2}. It follows that 3 has to
divide a (one way
to see this: every integer can be written as either 3n,3n+1,
or3n+2 for some integer n. If you
square these three choices, only the first one
gives you a multiple of three.)
Let a =3k, for k ∈ Z. Then substitution yields 12b^{2} = (3k)^{2} =9k^{2}, so dividing by
3 we have
4b^{2} =3k^{2}, so 3 divides 4b^{2} and hence 3 divides b^{2}. Just as for a, this
implies that b has to divide
b. But then a and b share the common factor of 3, which contradicts our choice
of representation
of r. So there is no rational number whose square is 12.
2) S ⊆ R, S ≠ Ø, and u = sup S. Given any n ∈ N, ∀s ∈ S,
, so is
an upper
bound for S. Assume is also an upper bound for S. Since
, u would not
be the
least upper bound for S, which is a contradiction. Therefore
is not an upper
bound for S.
3) Recall that a subset of the real numbers , A ⊆ R, is bounded if there are real
numbers a and a'
such that
t ∈ A ⇒ a' ≤ t ≤ a.
Since A, B ⊆ R are bounded, they have upper bounds a and b respectively, and
lower bounds a'
and b'. Let α = max (a, b) and β = min (a', b'). Clearly,
t ∈ A ⇒ β ≤ a' ≤ t ≤ a ≤ α
t ∈ B ⇒ β ≤ b' ≤ t ≤ b ≤ α,
hence any t ∈ A ∪ B satisfies β ≤ t ≤ α and A ∪ B is bounded.
Notice that, in particular, this shows that max{sup A, supB} is an upper bound
for A∪B, so
we only have to show that it is the least upper bound. Suppose γ< max{sup A, supB}. Then
without loss of generality, γ< supA. By definition of
supremum, γ is not an upper bound of A,
so ∃a ∈ A with γ<a. But a ∈ A ⇒ a ∈ A ∪
B, so γ is not an upper bound of A ∪ B. Therefore
max{sup A, supB} = supA ∪ B.
4) Start by noting that, if n, m ∈ N then
from which it follows that
for
n, m ∈ Z (why?). Similarly, you can show that
for n, m
∈ Z. Recall that, if x> 0,
then is defined to be the unique positive real number such that
.
a ) We have that m/n = p/q so mq = pn. Notice that
and
that
, which is also equal to
. But we know that there is a
unique real
number y satisfying hence the two numbers we
started with have to be equal, i.e.,
Notice that if this equality didn't hold, then we could not make sense of the
symbol b ^{r} for
r ∈ Q, because the value would change if we wrote the same number
r in two different ways.
b) Let r, s ∈ Q with and
. Since nq is an integer we know that
but and similarly
. Since mq and np
are
integers we can conclude
.
But there is a unique positive real number, y, such that
, so we
know that
c) Now with b> 1, given r, s ∈ Q, s ≤ r we want to show b^{s} ≤ b^{r}. Let
,0
<n,0 ≤ m
since s ≤ r. Then , and it is easy to see that 1 ≤ b^{m}, since 0 ≤ m
and 1 <b.
Thus a positive power of b ^{rs} is greater than or equal to 1, which
implies 1 ≤ b^{rs}. Multiplying
by b^{s} gives , so b^{s} ≤ b^{r}
. Hence for any ,
so b^{r} is an upper bound for B(r).
Since b^{r} ∈ B(r), b^{r} must be the least upper bound, so
b^{r} = supB(r).
d) So let x, y ∈ R. If r, s ∈ Q are such that r ≤ x, s ≤ y, then r + s ≤ x + y so
∈ B(x + y)
and . Keeping s fixed, notice that for any r ≤ x we
have
thus is an upper bound for B (x) which implies
. We rearrange
this to
and conclude that or
.
Suppose the inequality is strict . Then ∃t ∈ Q, t<x + y, such that
.
We will find
r, s ∈ Q, with r ≤ x, s ≤ y and t<r+s<x+y. First, find N ∈ N so that
N (x + y  t) > 1,
then find r ∈ Q so that and s ∈ Q such that
(the existence
of N, r, s follow from the Archimedean property of R as shown in class). Now,
notice that
hence we have t<r + s<x + y just like we wanted .
But now we have
which is a contradiction because, since r<x and s<y, we
have b^{r} <b^{x} and b^{s} <b^{y}!
5) We know that in any ordered field , squares are greater than or equal to zero .
Since i^{2} = 1, this
means that 0 ≤1. Butthen1=0+1 ≤1+1=0 ≤ 1 which implies 0 =
1, a contradiction!
6) I'll write for this relation on C to distinguish it from the normal order on
R. To show that
is an order on C, we must show both transitivity and totality
(or given x, y ∈ C, exactly one of
the following is true: x y, y
x, or x = y).
First for transitivity, let x, y, z ∈ C, x = a + bi,
y = c + di, z = e + fi such
that x y z. Therefore a ≤ c ≤ e, so a ≤ e by the transitivity of
the order on R.
If a<e, then x z, so we are done. If a = e, then a = c = e so we have from
the
definition of that b<d<f, so once again by the transitivity of the order on R,
b<f.
Now a = e and b<f ⇒ x z, so we have shown transitivity.
Now to show
totality. Consider x, y ∈ C, x = a + bi, y = c + di. Without loss of generality,
let a ≤ c. Suppose a = c. Then b<d ⇔ x
y, b>d ⇔ y
x, and b = d ⇔ x = y, so by
the
totality of the order on R, we have the totality of
on C in the case of a =
c. Suppose instead
that a<c. Then we know x y, and it is not the case that y
x
or x = y, so we have totality
in this case as well. Thus we have proven that
is
an order on C.
This order does not have the leastupperbound property. Consider
the set of complex numbers
with real part less than or equal to zero:
S = {a + bi : a ≤ 0,b ∈ R}.
S is bounded above, for instance by the number 1, but it is not possible for any
number z = a+bi
to be the supremum of S. If a ≤ 0, then a + bi
a +(b + 1)i ∈ S,
so a + bi is not an upper
bound for S. If a> 0, then a +(b  1)i a + bi, and a
+(b  1)i is also an upper bound for S,
so a + bi is not the least upper bound.
Therefore S has no least upper bound, even though it is
bounded above.
The geometric interpretation comes from looking at the parallelogram whose
vertices are the
points 0, x, x + y and y. Then the equation states that the sum
of the squares of the lengths
of the two diagonals (the vectors x + y and x  y)
is the same as the sum of the squares of the
lengths of the four sides.
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