MATH 2200 Homework #5 Solutions
Thomas Goller
February 25, 2013
General Comments:
• The notation “a | b” does not mean usual division! It makes no sense to write “2 | 6 =
3” or something of the sort. “a | b” is a statement, not a number ! If you are confused
by this, then you absolutely must see me – misinterpreting this notation will seriously
undermine your ability to do well on the coming learning celebration!
• It also doesn’t make sense to write something like “6 | 2 · 3 = 6 | 6”, because we
have not defined a notion of “equality” for statements. Instead, write something like
“6 | 2 · 3, namely 6 | 6”.
• If you were able to prove some of the false claims, then you need to be more careful!
Make sure you understand the reasoning behind every step of the proofs I include in
the homework solutions, and justify every step in the proofs you write. A common and
devastating mistake is to misinterpret definitions; make sure you completely understand
all major definitions! If you’re struggling with a definition, just come ask me. The
definition is probably not as bizarre as it seems.
Exercise 4.1.5
Proposition (4.1.4). Let a, b, c ∈ Z, with a "= 0. Then
(i) If a | b and a | c, then a | (b + c);
(ii) If a | b, then a | bc for all integers c;
(iii) Suppose b "= 0. If a | b and b | c, then a | c.
Proof. (i): In the notes.
(ii): Since a | b, there is s ∈ Z such that b = as. Then bc = asc, so a | bc.
(iii): Since a | b and b | c, there are s, t ∈ Z such that b = as and c = bt. Then c = ast,
so a | c.
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Exercise 4.1.16
Claim (a). Let m, n ∈ Z and d ∈ Z≥1 . If d ! n, then d ! mn.
This claim is false! Counterexample: let d = 2, n = 1, and m = 2. Then 2 ! 1, but
2 | 2 · 1.
Claim (b). Let m, n ∈ Z and d ∈ Z≥1 . If d | n, then d | mn.
Proof. This is proposition 4.1.4 (ii).
Claim (c). Let m, n ∈ Z and d ∈ Z≥1 . If dn | mn and n "= 0, then d | m.
Proof. Since dn | mn, there is s ∈ Z such that mn = dns. Then n(m−ds) = 0, so m−ds = 0
since n "= 0. Thus m = ds, so d | m.
Comments:
• It is okay if you went from mn = dns straight to m = ds. In this special case where n
is on both sides of the equation, you really can divide away the n, but for that you need
the assumption n "= 0. You should be clear about where you are using this assumption
in the proof, so it would be good to at least say “... mn = dns. Then since n "= 0,
we see that m = ds”. At any rate, the reason “dividing by n” works in this case is
because of the argument I included in the above proof.
Claim (d). Let m, n ∈ Z and d ∈ Z≥1 . If d | mn, then d | m or d | n.
This claim is false! Counterexample: let d = 6, m = 2, and n = 3. Then 6 | 2 · 3, but
6 ! 2 and 6 ! 3.
Comments:
• Note that to construct a counterexample, we need to choose d, m, n such that “d | m
or d | n” is false, namely such that d ! m and d ! n. Note that the “or” changes to an
“and”!
Claim (e). Let m, n ∈ Z≥2 and let p ∈ Z≥2 be prime. If p | mn, then p | m or p | n.
Proof. Since p | mn, the prime factorization of mn is
mn = p · p1 · · · pr .
Let m = q1 · · · qs and n = !1 · · · !t be the prime factorizations of m and n. Then
mn = q1 · · · qs · !1 · · · !t
is another prime factorization of mn. By the uniqueness of the prime factorization of mn, p
must be one of the qi ’s, in which case p | m, or one of the !i ’s, in which case p | n.
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Comments:
• I have cheated by adding the condition that m, n ≥ 2. This ensures that the FTA
applies as stated in the notes, and makes the claim much less cumbersome to prove.
The claim is still true if we allow m, n to be any integers. The case when m = 0 or
n = 0 is easy to check directly, as is the case when |m| = 1 or |n| = 1. When |m| ≥ 2
and |n| ≥ 2, the only trouble is if m or n is negative. This isn’t really a big deal: an
integer a ≤ −2 still has a “prime factorization”: just take the prime factorization of |a|
and add a negative sign in front. For example, −12 = −2·2·3. With this generalization
of the notion of “prime factorization” to negative integers, the proof goes through as
above. Yuck! Sorry if you were annoyed by all of these cases! Next time a proof gets
that nasty, send me an e-mail – I’ve probably made a mistake!
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