Sean Golden
Logic and Proof
Dr. Anderson
Appendix C: pg. 525; 2, 4, 5, 7
2. Prove that for each nonnegative integer n, 2n > n.
Proof: By Mathematical Induction
Suppose that k ɛ N and that k = 0. Then 2k > k and 20 > 0 which implies that 1 > 0 and thus is true
for k=0. Moreover, if we suppose that k = 1. Then 2k > k and 21 > 1 which implies that 2 > 1 and thus is
true for k=1. Now also suppose that 2k > k for some k ≥ 0 which is the inductive hypothesis. In this case,
2k+1 > k+1. Since 2k+1 = 2*2k and 2k>k, then 2*2k > 2(k) by the inductive hypothesis. Furthermore, since
2(k) = k + k and k + k > k + 1 when k > 1, we can conclude that 2k+1 > k+1 for every k>1 and thus that 2k >
k.
4. Let r be a real number, r ≠ 1. Prove that for every integer n ≥ 1,
1 + r + r2 + r3 + … + rn-1 =
𝑟 𝑛 −1
𝑟−1
Proof: By Mathematical Induction
Suppose that k ɛ N and that k = 1. Then r1-1 =
rk-1 =
𝑟 𝑘 −1
𝑟−1
𝑟 1 −1
𝑟−1
and therefore 1 = 1. Thus, 1 + r + r2 + r3 + … +
holds for when k = 1. Now also suppose that 1 + r + r2 + r3 + … + rk-1 =
is the inductive hypothesis. So, 1 + r + r2 + r3 + … + rk-1 + rk =
𝑟 𝑘 −1
𝑟−1
𝑟 𝑛 −1
𝑟−1
+ rk =
𝑟 𝑘 −1
𝑟−1
=
𝑟 𝑘 −1+𝑟 𝑘 (𝑟−1)
𝑟−1
=
𝑟 𝑘 −1+𝑟 𝑘+1 −𝑟 𝑘
𝑟−1
=
−1+𝑟 𝑘+1
𝑟−1
=
𝑟 (𝑘+1) −1
,
𝑟−1
+
𝑟 𝑘 −1
𝑟−1
𝑟 𝑘 −1
𝑟−1
for some k ≥ 1 which
+ rk by the inductive hypothesis. Thus,
𝑟 𝑘 (𝑟−1)
𝑟−1
where t = k+1 is an integer. Therefore, we can conclude that 1 + r + r2 + r3 + … + rn-1 =
for every integer n ≥ 1.
5. Prove that 4 is a factor of 7n – 3n for every positive integer n.
Proof: By Mathematical Induction
Suppose that k ɛ N and that k = 1. Then 71 – 31 = 7 – 3 = 4 * 1 and therefore 4 is a factor of 7k –
3k when k = 1. Now also suppose that 4 is a factor of 7k – 3k for some k ≥ 1 and so 7k – 3k = 4 * t where k
≥ 1 and t ≥ 1 which is the inductive hypothesis. So, 7k+1 – 3k+1 = 4 * t by the inductive hypothesis. Thus,
7k+1 – 3k+1 = 7k+1 – 7*3k + 7*3k – 3k+1
= 7*7k – 7*3k + 7*3k – 3*3k
= 7(7k – 3k) + (7 – 3)3k
= 7(4 * t) + (7-3)3k , by inductive hypothesis
= 4(7t) + 4(3k)
= 4(7t + 3k)
= 4s, where s = 7t + 3k is an integer. Therefore, we can conclude that 7k – 3k = 4 * t where k≥1
and t≥1 and therefore that 4 is a factor of 7n – 3n for every positive integer n.
7. Prove that 3 is a factor of 22n + 1 + 1 for every positive integer n.
Proof: By Mathematical Induction
Suppose that k ɛ N and that k = 1. Then 22(1) + 1 + 1 = 23 + 1 = 8 + 1 = 3 * 3 and therefore 3 is a
factor of 22k + 1 + 1 when k = 1. Now also suppose that 3 is a factor of 22n + 1 + 1 for some k ≥ 1 and so 22n+1
+ 1 = 3 * t where k ≥ 1 and t ≥ 1 which is the inductive hypothesis. So, 22(k+1)+1 + 1 = 3 * t by the inductive
hypothesis. Thus,
22(k+1)+1 + 1 = 22k+2+1 + 1
= 22k+3 + 1
= 22 * 22k+1 + 1
= 4 * 22k+1 + 1
= 4(22k+1 + 1) – 4 +1
= 4(22k+1 + 1) – 3
= 4(3t) – 3, by the inductive hypothesis
= 3(4t) -3
= 3(4t – 1)
= 3s, where s=4t-1 is an integer. Therefore, we can conclude that 22n + 1 + 1 = 3 * t where k≥1 and
t≥1 and therefore that 3 is a factor of 22n + 1 + 1 for ever positive integer n.