Math 215: Homework 4
February 20, 2013
Proposition 2.E: Suppose n, m Z and n > m. Then n < m.
Proof . Let n, m Z such that n > m. So n m N. Observe that
n m = n + (m)
= m + n
= m + (n)
= m (n)
(denition of subtraction)
(Proposition 1.22(i)
(denition of su
Math 215: Homework 2
February 4, 2013
Solutions
Proposition 1.16: If m and n are even integers, then so is m + n.
Proof . Let m and n be even integers. To show that m + n is even we must show that
there exists an integer j such that
m + n = j 2.
(1)
Since
Math 215: Homework 8a
March 19, 2012
Proposition 4.5: For all k Z0 , k! N.
Proof . We proceed by induction on k 0. When k = 0, k! = 0! = 1 N. This
establishes the base case. Suppose for some k Z0 that k! N. Since k 0,
k + 1 1 N. Since k! N as well,
(k + 1
Math 215: Homework 7 Solutions
February 24, 2012
Project 2.35: Compute gcd(4, 6), gcd(7, 13), gcd(4, 10) and gcd(5, 15). You do NOT
have to prove that you have found the gcd. But you do have to exhibit the integers x and y
in the denition of the gcd.
Solu
Math 215: Homework 5 Solutions
February 24, 2012
Proposition HW5.1: The integer 1 is not divisible by 2. That is, 2 1.
Proof. Suppose to produce a contradiction that 2 | 1. Then there is an integer j such that
1 = 2 j. Since 1, 2 N, Proposition 2.11 impli
Math 215: Homework 3 Solutions
February 13, 2013
Proposition 1.22:
(i) For all m Z, (m) = m.
Proof . Let m Z. Notice that
(m) + m = 0
by Proposition 1.8. Also,
(m) + (m) = 0
by the denition of additive inverses. Hence
(m) + m = (m) + (m).
Proposition 1.9
Math 215: Homework 7 Solutions
March 8, 2013
Proposition 2.33: Let A be a nonempty subset of Z. Suppose for some b Z that b a
for all a A. Then A has a least element.
Proof . Let A be a nonempty subset of Z that admits a lower bound b Z. So b a
for all a
Math 215: Homework 9
March 30, 2012
Your name here
Proposition 4.30: For all k, m N, where m 2,
fm+k = fm1 fk + fm fk+1 .
Proposition 4.30: For all k, m N, where m 2,
fm+k = fm1 fk + fm fk+1 .
Proof . Let m Z2 . We will show that
fm+k = fm1 fk + fm fk+1
f
Math 215: Homework 13a Solutions April 26, 2013
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Proposition 9.18: For all m, n Z,
e(m n) = e(m) e(n).
Proof. Let m Z. We rst show that for all n Z0 that e(m n) = e(m) e(n) by induction
on n. First, when n = 0,
e(m n) = e(m 0) = e(0) = 0 = e
Math 215: Homework 14 Solutions
May 7, 2013
If A and B are sets, we say they have the same cardinality if there exists a bijection f : A
B, in which case we write |A| = | B|. We say a set A is countable if there exists a bijection
f : N A.
Proposition HW
Math 215: Homework 11
April 19, 2013
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This looks long, but most of the proofs are very short!
Proposition 8.A: The number 0 R does not have a multiplicative inverse.
Proof. Suppose to the contrary that there exists x R such that x 0 = 1. Pro
Math 215: Homework 11 Solutions April 12, 2013
Your name here
Project 6.27: Finish your work on this project. Specically, state and prove a proposition
that characterizes the set of n 2 such that Zn satises the cancellation property (Axiom
1.5). You can o
Math 215: Homework 10
April 3, 2013
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Proposition HW10.1: Let A be a set, and let be an equivalence relation on A. Then the
equivalence classes of form a partition of A.
Proof . We need to prove the following:
1. For all a A, there exists x A
Math 215: Homework 1a Solutions January 25, 2013
David Maxwell
Proposition 1.7: If m is an integer, then 0 + m = m and 1 m = m.
Proof . Let m Z. Then by additive commutivity
0 + m = m + 0.
But m + 0 = m by Axiom 1.2. So 0 + m = m.
On the other hand, by mu