Elliot Paquette
The Ohio State University
September 16, 2016
HW3 (partial) Solutions
Problem 1. Prove the following arithmetic facts about Z.
(a) The additive identity is unique: (∃a ∈ Z, a + b = a) ⇒ b = 0. Proof. Suppose
a, b ∈ Z have the property that a + b = a. By the existence of the additive inverse
and the element 0 ∈ Z, there is an element c ∈ Z so that a + c = 0. As a + b = a
then adding c to both sides, a+b+c = a+c = 0. Using commutativity, associativity,
and finally the definition of 0,
a + b + c = b + (a + c) = b + 0 = b.
Combining what we have shown, b = 0.
(b) The multiplicative identity is unique: (∃a ∈ Z, a · b = a, a 6= 0) ⇒ b = 1.
(c) ∀a ∈ Z, −a = (−1) · a.
(d) (−1) · (−1) = 1. Proof. By definition, −1 is the integer so that 1 + (−1) = 0. For
clarity, we switch to symbols. Let a = 1, and let b be the additive inverse of a, so
a + b = 0. Then a · (a + b) = a · 0 = 0 and b · (a + b) = b · 0 = 0. Combining these
facts,
a · (a + b) = b · (a + b).
Expanding each side of the equation,
a · a + a · b = b · a + b · b.
By commutativity, a · b = b · a, and so we can subtract it from both sides (i.e. add
the additive inverse to both sides). Then
a · a = b · b.
Since a · a = 1 · 1 = 1, by definition of the multiplicative identity, we are done.
(e) ∀a, b ∈ Z, −(a + b) = −a + −b.
Problem 3. Prove: for all a, b, c ∈ Z if a < b and c < 0 then a · c > b · c.
Problem 5. Prove the following consequences of the well ordering principle that hold for all
nonempty sets S ⊂ Z
Math 3345 HW3 (partial) Solutions
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Fall 2016
(a) If there exists an a ∈ Z so that for all s ∈ S, a ≤ s, then there exists an a ∈ S so
that for all s ∈ S, a ≤ s. Proof. Let S ⊂ Z and a ∈ Z be like in the hypothesis of
the problem. Define a set of integers T ⊂ Z by shifting all the elements of S by
−a + 1, that is
T = {z ∈ Z : z + a − 1 ∈ S} .
Then if t ∈ T, t + a − 1 ∈ S, and therefore t + a − 1 ≥ a. Rearranging, we conclude
t ≥ 1. Therefore T ⊆ N.
By the well ordering principle, T has a smallest element x, i.e. all elements t ∈ T
have t ≥ x. Now, if u ∈ S, then u + 1 − a ∈ T, since (u + 1 − a) − 1 + a = u ∈ S.
By definition of x, we therefore have u + 1 − a ≥ x. Rearranging, u ≥ x − 1 + a.
As this holds for all u ∈ S, we have that x − 1 + a is a lower bound for S. Better
still, by definition of T, x − 1 + a is in S, and so we have shown that which we
wished to prove.
(b) If there exists an a ∈ Z so that for all s ∈ S, a ≥ s, then there exists an a ∈ S so
that for all s ∈ S, a ≥ s.
Problem 6. Prove that for all a, b ∈ N there exists an n ∈ N so that a · n ≥ b. This could be
proved by the WOP, but there is also a very simple direct demonstration.
Problem 7. (b) ∀d, n, m, a, b if d|n and d|m then d|(an + bm). By definition of divisibility,
there exist k, ` ∈ Z so that dk = n and d` = m. Then
an + bm = adk + bd` = d(ak + b`).
So by definition of divisibility, d divides an + bm.
(c) ∀a, d, n if d|n then ad|an.
(d) ∀a, d, n if a 6= 0 and ad|an then d|n.
Problem 9. Prove that for all n ∈ N
1 + 3 + 5 + · · · + (2n − 1) = n2 .
The proof is by induction on n. The statement P (n) is that the displayed formula
holds. As we want to prove something for all n ∈ N, the base case is P (1). This is
true since the sum on the left is 1 and the right hand side is 12 = 1. For the inductive
hypothesis, we want to show P (n) ⇒ P (n + 1). Rewriting the sum in the case n + 1,
1 + 3 + 5 + · · · + (2n − 1) + (2n + 1) = (1 + 3 + 5 + · · · + (2n − 1)) + (2n + 1).
Using P (n),
(1 + 3 + 5 + · · · + (2n − 1) = n2 .
So,
1 + 3 + 5 + · · · + (2n − 1) + (2n + 1) = n2 + (2n + 1).
Since n2 + (2n + 1) = (n + 1)2 , the induction is complete.
Math 3345 HW3 (partial) Solutions
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Fall 2016
Problem 11. Let a(1), a(2), · · · be a sequence having the properties:
(a) a(1) = 1.
(b) (∀n ∈ N)a(2n) ≤ 3a(n).
(c) (∀n ∈ N)a(2n + 1) ≤ 3a(n).
Prove that a(n) ≤ n2 .
Math 3345 HW3 (partial) Solutions
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Fall 2016