T. Barthelmé, N. Paulet 2026
MATH/MTHE 281
Homework # 1
Date Due: Friday Jan. 16 at 11:59pm, No late penalties until Wednesday Jan. 21st at 11:59pm
E1.1: Let [(j1 , k1 )], [(j2 , k2 )] be two elements of Z. Show that addition
[(j1 , k1 )] + [(j2 , k2 )] = [(j1 + j2 , k1 + k2 )]
is well defined, i.e., prove that for any (j1′ , k1′ ) ∈ [(j1 , k1 )] and (j2′ , k2′ ) ∈ [(j2 , k2 )], we
have that (j1′ + j2′ , k1′ + k2′ ) is equivalent to (j1 + j2 , k1 + k2 ).
Solution: Let (j1′ , k1′ ) ∈ [(j1 , k1 )] and (j2′ , k2′ ) ∈ [(j2 , k2 )]. By definition of being equivalent,
we have that
j1′ + k1 = j1 + k1′ and j2′ + k2 = j2 + k2′ .
Using this, we get
j1 + j2 + k1′ + k2′ = (j1 + k1′ ) + (j2 + k2′ ) = (j1′ + k1 ) + (j2′ + k2 ) = j1′ + j2′ + k1 + k2 .
Thus (j1′ + j2′ , k1′ + k2′ ) and (j1 + j2 , k1 + k2 ) are equivalent.
E1.3: Show that the relations < and ≤ on Z have the following properties:
1. [(0, j)] < [(0, 0)] for all j ∈ Z>0 ;
2. [(0, j)] < [(k, 0)] for all j, k ∈ Z>0 ;
5. [(j, 0)] < [(k, 0)], j, k ∈ Z≥0 , if and only if j < k;
Solution: The definition of “<” in Z is [(j, k)] < [(l, m)] if and only if j + m < l + k.
1. Let j ∈ Z>0 , then 0 + 0 < j + 0, so [(0, j)] < [(0, 0)].
2. Let j, k ∈ Z>0 . Then 0 + 0 < k + j, so [(0, j)] < [(k, 0)].
5. Let j, k ∈ Z≥0 . If j < k, then j + 0 < k + 0 so [(j, 0)] < [(k, 0)].
Conversely, if [(j, 0)] < [(k, 0)], then j + 0 < k + 0 so j < k.
E1.4: Show that a subset A ⊂ Q is bounded if and only it is has a lower bound and an
upper bound.
Solution: Suppose that A is bounded. Thus there exists M ∈ Q>0 such that |q| ≤ M for
every q ∈ A. We claim that M is an upper bound and −M is a lower bound. Indeed, let
q ∈ A. Then
|q| ≤ M =⇒ −M ≤ q ≤ M,
immediately giving the assertion.
Next suppose that A has an upper bound u ∈ Q and a lower bound ℓ ∈ Q. We claim that
|q| ≤ max{|u|, |ℓ|} for every q ∈ A. Indeed, first suppose that q ∈ A is positive. Then
q≤u =⇒ |q| ≤ |u| ≤ max{|u|, |ℓ|}.
Next suppose that q ∈ A is negative. Then
ℓ≤q =⇒ −q ≤ −ℓ =⇒ |q| ≤ |ℓ| ≤ max{|u|, |ℓ|}.
as desired. •
MATH/MTHE 281
Homework # 1
Date Due: Friday Jan. 16 at 11:59pm, No late penalties until Wednesday Jan. 21st at 11:59pm
E1.1: Let [(j1 , k1 )], [(j2 , k2 )] be two elements of Z. Show that addition
[(j1 , k1 )] + [(j2 , k2 )] = [(j1 + j2 , k1 + k2 )]
is well defined, i.e., prove that for any (j1′ , k1′ ) ∈ [(j1 , k1 )] and (j2′ , k2′ ) ∈ [(j2 , k2 )], we
have that (j1′ + j2′ , k1′ + k2′ ) is equivalent to (j1 + j2 , k1 + k2 ).
Solution: Let (j1′ , k1′ ) ∈ [(j1 , k1 )] and (j2′ , k2′ ) ∈ [(j2 , k2 )]. By definition of being equivalent,
we have that
j1′ + k1 = j1 + k1′ and j2′ + k2 = j2 + k2′ .
Using this, we get
j1 + j2 + k1′ + k2′ = (j1 + k1′ ) + (j2 + k2′ ) = (j1′ + k1 ) + (j2′ + k2 ) = j1′ + j2′ + k1 + k2 .
Thus (j1′ + j2′ , k1′ + k2′ ) and (j1 + j2 , k1 + k2 ) are equivalent.
E1.3: Show that the relations < and ≤ on Z have the following properties:
1. [(0, j)] < [(0, 0)] for all j ∈ Z>0 ;
2. [(0, j)] < [(k, 0)] for all j, k ∈ Z>0 ;
5. [(j, 0)] < [(k, 0)], j, k ∈ Z≥0 , if and only if j < k;
Solution: The definition of “<” in Z is [(j, k)] < [(l, m)] if and only if j + m < l + k.
1. Let j ∈ Z>0 , then 0 + 0 < j + 0, so [(0, j)] < [(0, 0)].
2. Let j, k ∈ Z>0 . Then 0 + 0 < k + j, so [(0, j)] < [(k, 0)].
5. Let j, k ∈ Z≥0 . If j < k, then j + 0 < k + 0 so [(j, 0)] < [(k, 0)].
Conversely, if [(j, 0)] < [(k, 0)], then j + 0 < k + 0 so j < k.
E1.4: Show that a subset A ⊂ Q is bounded if and only it is has a lower bound and an
upper bound.
Solution: Suppose that A is bounded. Thus there exists M ∈ Q>0 such that |q| ≤ M for
every q ∈ A. We claim that M is an upper bound and −M is a lower bound. Indeed, let
q ∈ A. Then
|q| ≤ M =⇒ −M ≤ q ≤ M,
immediately giving the assertion.
Next suppose that A has an upper bound u ∈ Q and a lower bound ℓ ∈ Q. We claim that
|q| ≤ max{|u|, |ℓ|} for every q ∈ A. Indeed, first suppose that q ∈ A is positive. Then
q≤u =⇒ |q| ≤ |u| ≤ max{|u|, |ℓ|}.
Next suppose that q ∈ A is negative. Then
ℓ≤q =⇒ −q ≤ −ℓ =⇒ |q| ≤ |ℓ| ≤ max{|u|, |ℓ|}.
as desired. •
