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Assignment 10
MATH 1200
SOLUTION
(1) Problem 6, Chapter 11. Let E be the set of positive even integers. We call
a number e ∈ E a “prima” if e cannot be expressed as a product of two
other members of E.
(a) Show that 6 is prima but 4 is not.
SOLUTION: Clearly 4 is not prima since 4 = 2 · 2, a product of two
members of E.. However, 6 is prima, since as a product, 6 can only
be written as 6 = 1 · 6 = 2 · 3. In both cases only one of these factors
is in E.
(b) What is the general form of a prima in E.
SOLUTION: An element e ∈ E is a prima if and only if e = 2n where
n is odd. To see this, suppose that e ∈ E. Then e is even so that
e = 2n. If e is prima, then we claim that n must be odd. If not, n is
even, so n = 2k. Then e = 2 · (2k) a product of two elements of E,
but this is impossible since we assume that e is prima. On the other
hand, if e is not prima, then e = a · b where both a and b are even. So
a = 2k and b = 2m for some k and m. So e = (2k)(2m) = 2 · (2km).
Therefore n = 2km so n is even.
(c) Every element of E is either a prima or a product of primas.
SOLUTION: Let e ∈ E. So e is even. So e = 2n for some n. We
prove by induction on n that 2n is either prima or a product of primas.
Base case. n = 1. Then e = 2 and 2 is a prima.
Inductive step. Suppose that n > 1 and that for all k < n we have that
e = 2k is either prima or a product of primas. Now consider e = 2(n).
If n is odd, then e is a prima. Otherwise, n = 2k for some k < n. So
e = 2 · (2k). By our inductive hypothesis, k is either prima, so that
e = 2 · 2 · k is a product of primas, or k itself is a product of primas
k = e1 · ... · ek so also e = 2 · 2 · e1 · ... · ek a product of primas.
(d) Show that E does not satisfy a “unique prima factorization theorem”
SOLUTION: For example 60 = 6 · 10 a product of primas, and 60 =
2 · 30 also a product of primas. (2 = 2 · 1, 6 = 2 · 3, 10 = 2 · 5 and
30 = 2 · 15 all of the form 2 times an odd number, so all are primas).
(2) BONUS QUESTION: Problem 10, chapter 2: Suppose that a and b are
positive integers and that d = hcf (a, b). Prove that there are s and t
positive such that d = sa − tb.
SOLUTION: We know that there are s, t ∈ Z such that
(1) d = sa + tb.
Since all these numbers are positive and d ≤ a and d ≤ b, either s > 0
and t ≤ 0 or alternately s ≤ 0 and t > 0. If s > 0 and t < 0 then we are
done. If on the other hand s ≤ 0 and t ≥ 0 we can find alternate coefficients
by first noting that for any n,
(2) 0 = (nb)a − (na)b =
We can then add equation 1 to equation 2 to get
1
2
(3) d = (nb + s)a − (na − t)b
And we can choose n large enough so that nb + s > 0 and na − t > 0.
For example, if n > − sb then nb + s > 0 and if n > at then na − t > 0 so
choosing n > max{− sb , at } would give the required coefficients.