MATH 20700
Oct 5
Homework 1
1. Problem 9
(a) Let x = A|B, x′ = A′ |B ′ be cuts in Q. We know that A and B must both be
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disjoint, as well as A′ and B ′ . Suppose that there exists q ∈ A A′ and q ∈ B B ′ .
This implies that there exists a, a′ , b, b′ such that q = a + a′ = b + b′ . However, we
know that all a must be less than all b, and similarly all a′ must be less than all
b′ . Thus, a + a′ < b + b′ . Therefore, we have a contradiction, which shows that
A + A′ and B + B ′ must be disjoint.
√
√
Now let us consider x = − 2 = A|B and x′ = 2 = C|D. Note that x and
x′ are irrational, therefore they are not contained either of the sets defining the
dedekind cuts. This implies that the sum x + x′ defined by (A + C)|(B
√ + D)√does
not contain the number zero, because neither of the sets contain − 2 and 2 to
form the number. Thus Q is not the union of A + C and B + D.
(b) Therefore, defining x + x′ = (A + A√′ )|(B + B ′ ) is incorrect. We can consider the
same case as above, where x = − 2 and x = 2. Adding these together gives
0, defined by the dedikind cut shown in part (a) not to include 0. However, 0
is a rational number, and therefore should be included in the second set of the
dediking cut. Thus, the original definition is incorrect.
(c) Let us consider x = −1 and x′ = −1. Then we have that x · x′ = −1 ∗ −1 = 1,
which implies that the lower set, A · A′ should contain all numbers less than 1.
However, this set does not contain 0, as neither A nor A′ contains 0. Thus, we
have a hole in the set A · A′ , showing that definition of x · x′ does not work.
2. Problem 11
Let x = A|B, y = C|D such that x · y = 1.
(a) If x > 0, then we can define C to be the set of all q ∈ Q|q ≤ 0
and we can define D to be q ∈ Q| 1q ∈ A.
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q ∈ Q| 1q ∈ B,
(b) If x < 0, then we can define C to be the set of all q ∈ Q such that 1q ∈ B, and we
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can define D to be the set of all q ∈ Q|q > 0 q ∈ Q| 1q ∈ A.
(c) Suppose that there exists a y and y ′ such that y ̸= y ′ and x ∗ y = x ∗ y ′ = 1. Note
that multiplication within R is both associative and commutative. We then have
the following:
xy = 1 =⇒ (xy) ∗ y ′ = 1 ∗ y ′ =⇒ x(yy ′ ) = y ′
=⇒ x(y ′ y) = y ′ =⇒ (xy ′ )y = y ′ =⇒ y = y ′
Thus, this forms a contradiction that y ̸= y ′ . Therefore we can conclude that any
x uniquely determines y.
Homework 1
MATH 20700
Oct 5
3. Problem 13
Let b be the least upper bound of a bounded non-empty subset S of R.
(a) Let us pick an ϵ > 0 and consider b − ϵ. This is clearly within the subset S, and
because b is a least upper bound, we have that b − ϵ ∈ S and b − ϵ < b. Suppose
now that there exists no s ∈ S such that b−ϵ ≤ s ≤ b. This implies therefore that
b − ϵ would be a least upper bound for S, but b − ϵ is less than b, which implies
that b is no longer the least upper bound for S. Thus we have a contradiction.
Therefore, there must exist as s such that b − ϵ ≤ s ≤ b.
(b) No. Let us consider the set S = {1}. The least upper bound b for S is equal to
1. Now if we consider ϵ = 1/2, there is no element s ∈ S such that b − ϵ < s < b.
(c) Consider x = A|B. By the definition of the Dedikind cut, we know that x is at
least an upper bound for A. Now let us consider u < x to be an upper bound of
A. Because u < x, this implies that u ∈ A. However, since A defines a dedikind
cut, we know that A does contain it’s maximum value. Therefore, there must
exist a ∈ A such that u < a. Thus, u is not an upper bound of A. Therefore, x
is the least upper bound of A.
4. Problem 14
Let x = A|B with A = {q ∈ Q|q ≤ 0} {q ∈ Q|q 2 < 2}. Now let us consider
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x · x = C|D. We can define C to be {q ∈ Q|q ≤ 0} {q ∈ Q|q = a · b∀a, b ∈ A}. The
least upper bound of C therefore must be 2. Suppose that there exists c ∈ C such that
c > 2. From the construction of c, we know that c = a · b for some a, b, ∈ A. However,
we know that a · bmax(a, b) · max(a, b) < 2. Therefore, c cannot be greater than 2. So
we 2 is a valid upper bound for C.
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Suppose there exists another upper bound c < 2. Then this implies that c ∈ C by A
containing all q ∈ Q such that q 2 < 2. However, C does not contain it’s maximum
value, therefore it cannot contain an upper bound. Thus, c cannot be the upper bound.
So we have that the lowest upper bound of C is 2. From Problem 13, we can conclude
that x · x must therefore be equal to 2.
5. Problem 16
Let x > 0 and n ∈ N, and consider y to be the least upper bound of S = {q ∈ Q|q n ≤
x}. S is clearly nonempty with one positive element, min(1, x). Furthermore, we can
clearly see that S is bounded above by 1 + x.
Let us consider y n < x. We can set ϵ = x − y n , and from Problem 15 we can set a
δ. Thus we can pick u to be y n + 2δ which implies that |u − y| < δ which implies that
un < ϵ + y n = x. However this means that un must be in S, but u > y, which therefore
contradicts that y is an upper bound.
Now let us consider y n > x. Now let us set ϵ = y n − x and set a δ. Consider any s ∈ S,
so s ≤ y. If s > y −δ, then |s−y| < δ which implies that sn > y n −ϵ = x, contradicting
Homework 1
MATH 20700
Oct 5
that s ∈ S. Therefore every s ≤ y − δ, thus y − δ is an upper bound which is smaller
than y, however that is a contradiction because y is a least upper bound.
Therefore, we have that y n must be equal to x. Now let us consider there exists two y1
and y2 such that y1n = y2n = x. Without loss of generality, let 0 < y1 < y2 . Then from
Problem 15 we have that y2n − y1n = (y2 − y1 )(y2n−1 + y2n−2 y1 · · · + y1n−1 ) > 0. Clearly
this forms a contradiction, meaning that y must be unique.
6. Problem 19
Let S ⊂ R such that S is bounded below and non-empty. We can then define a lower
bound of S to be a number b ∈ R S such that ∀s ∈ S, b ≤ s. The greatest lower bound
of S is defined to be a a lower bound bx such that for any other lower bound by , bx ≥ by .
One property of the greatest lower bound is that for any S, the greatest lower bound
b must be unique. Suppose, for the sake of contradiction, that there exists another
different greatest lower bound b̃ of S. Because b is a greatest lower bound, this implies
that b ≥ b̃. However, b̃ is also a greatest lower bound, which implies that b̃ ≥ b. Thus,
this implies that b = b̃, thus contradicting our proposition of two unique greatest lower
bounds.