Solutions for exercises in chapter 1 E1.1 Prove by induction that for ...
Solutions for exercises in chapter 1 E1.1 Prove by induction that for ...
Solutions for exercises in chapter 1 E1.1 Prove by induction that for ...
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<strong>Solutions</strong> <strong>for</strong> <strong>exercises</strong> <strong>in</strong> <strong>chapter</strong> 1<br />
<strong>E1.1</strong> <strong>Prove</strong> <strong>by</strong> <strong>in</strong>duction <strong>that</strong> <strong>for</strong> every n ∈ ω,<br />
1 3 + 2 3 + 3 3 + · · · + n 3 = n 2 (n + 1) 2 /4.<br />
Both sides are 0 <strong>for</strong> n = 0. Now assume <strong>that</strong> the equation is true <strong>for</strong> n. Then<br />
1 3 + 2 3 + 3 3 + · · · + n 3 + (n + 1) 3 = n2 (n + 1) 2<br />
4<br />
+ (n + 1) 3<br />
= n2 (n + 1) 2 + 4(n + 1) 3<br />
4<br />
= (n + 1)2 [n 2 + 4(n + 1)]<br />
4<br />
= (n + 1)2 [n 2 + 4n + 4]<br />
4<br />
= (n + 1)2 (n + 2) 2<br />
,<br />
4<br />
which gives our equation with n replaced <strong>by</strong> n+1. So the equation holds <strong>by</strong> the <strong>in</strong>duction<br />
pr<strong>in</strong>ciple.<br />
E1.2 <strong>Prove</strong> <strong>that</strong> <strong>for</strong> any positive <strong>in</strong>teger n,<br />
1<br />
For n = 1 the left side is<br />
equation holds <strong>for</strong> n. Then<br />
1 1<br />
+ + · · · +<br />
1 · 2 2 · 3<br />
1 1<br />
+ + · · · +<br />
1 · 2 2 · 3<br />
1·2<br />
1<br />
n(n + 1)<br />
= n<br />
n + 1 .<br />
1 = 2 , and <strong>that</strong> is the right side too. Now assume <strong>that</strong> the<br />
1<br />
n(n + 1) +<br />
1<br />
(n + 1)(n + 2)<br />
n<br />
=<br />
n + 1 +<br />
1<br />
(n + 1)(n + 2)<br />
= n(n + 2) + 1<br />
(n + 1)(n + 2)<br />
= n2 + 2n + 1<br />
(n + 1)(n + 2)<br />
(n + 1)<br />
=<br />
2<br />
(n + 1)(n + 2)<br />
= n + 1<br />
n + 2 ,<br />
which gives the equation with n replaced <strong>by</strong> n + 1. So the equation holds <strong>for</strong> all n <strong>by</strong> the<br />
<strong>in</strong>duction pr<strong>in</strong>ciple.<br />
E1.3 <strong>Prove</strong> <strong>that</strong> <strong>for</strong> any positive <strong>in</strong>teger n,<br />
1 · 2 + 2 · 3 + · · · + n · (n + 1) =<br />
1<br />
n(n + 1)(n + 2)<br />
.<br />
3
For n = 1, the left side is 1 · 2 = 2, while the right side is 1·2·3<br />
3<br />
<strong>that</strong> the equation holds <strong>for</strong> n. Then<br />
= 6<br />
3<br />
= 2 also. Now assume<br />
n(n + 1)(n + 2)<br />
1 · 2 + 2 · 3 + · · · + n · (n + 1) + (n + 1) · (n + 2) = + (n + 1)(n + 2)<br />
3<br />
= n(n + 1)(n + 2) + 3(n + 1)(n + 2)<br />
3<br />
= (n + 1)(n + 2)(n + 3)<br />
,<br />
3<br />
which gives the equation with n replaced <strong>by</strong> n + 1. So the equation holds <strong>for</strong> all positive<br />
<strong>in</strong>tegers n.<br />
E1.4 Recall <strong>that</strong> <strong>for</strong> m a positive <strong>in</strong>teger, m! = m · (m − 1) · . . . · 1, with 0! = 1; and <strong>for</strong><br />
0 ≤ i ≤ m, � �<br />
m<br />
=<br />
i<br />
<strong>Prove</strong> <strong>that</strong> if 1 ≤ i ≤ m, then<br />
(Induction is not needed.)<br />
� � � �<br />
m m<br />
+ =<br />
i i − 1<br />
m!<br />
i!(m − i)! .<br />
� � � �<br />
m m<br />
+ =<br />
i i − 1<br />
= m!(m − i + 1)<br />
m!<br />
i!(m − i)! +<br />
� m + 1<br />
i!(m − i + 1)! +<br />
= m!(m − i + 1 + i)<br />
i!(m − i + 1)!<br />
= (m + 1)!<br />
i!(m + 1 − i)!<br />
� �<br />
m + 1<br />
= .<br />
i<br />
i<br />
�<br />
.<br />
m!<br />
(i − 1)!(m − i + 1)!<br />
m!i<br />
i!(m − i + 1)!<br />
E1.5 <strong>Prove</strong> <strong>by</strong> <strong>in</strong>duction the b<strong>in</strong>omial theorem, <strong>that</strong> <strong>for</strong> any n ∈ ω and any real numbers<br />
a, b,<br />
(a + b) n n�<br />
=<br />
i=0<br />
For n = 0 the left side is 1, while the right side is<br />
0�<br />
i=0<br />
� �<br />
0<br />
a<br />
i<br />
i b 0−i =<br />
2<br />
� �<br />
n<br />
a<br />
i<br />
i b n−i ,<br />
� �<br />
0<br />
a<br />
0<br />
0 b 0 = 1.
So the equation holds <strong>for</strong> n = 1. Now assume <strong>that</strong> it holds <strong>for</strong> n. Then<br />
(a + b) n+1 = (a + b)(a + b) n<br />
= (a + b)<br />
=<br />
(a + b) n+1 n+1 �<br />
= ci +<br />
i=1<br />
n�<br />
i=0<br />
� n<br />
i<br />
� n<br />
i<br />
n�<br />
i=0<br />
� n<br />
i<br />
�<br />
a i b n−i<br />
�<br />
a i+1 b n−i +<br />
�<br />
a i b n−i+1 .<br />
n�<br />
�<br />
n<br />
i<br />
�<br />
n+1 0 a b 〉 start<strong>in</strong>g with 1; let c1 =<br />
i=0<br />
i=0<br />
Now we number the sequence 〈 � � � � � n 1 n n 2 n−1 n<br />
� �<br />
a b , a b , . . ., 0 1<br />
n<br />
n 1 n<br />
0 a b , c2 = � � n 2 n−1<br />
1 a b , etc; so ci = � � n i n−i+1<br />
i−1 a b . Hence, cont<strong>in</strong>u<strong>in</strong>g the above,<br />
n�<br />
�<br />
a i b n−i+1<br />
n+1 �<br />
� �<br />
n<br />
=<br />
i − 1<br />
i=1<br />
n�<br />
� �<br />
n<br />
=<br />
i − 1<br />
i=1<br />
= a n+1 + b n+1 +<br />
= a n+1 + b n+1 +<br />
n+1 �<br />
�<br />
n + 1<br />
=<br />
i<br />
i=0<br />
which f<strong>in</strong>ishes the <strong>in</strong>ductive proof.<br />
a i b n−i+1 +<br />
n�<br />
i=0<br />
� �<br />
n<br />
a<br />
i<br />
i b n−i+1<br />
a i b n−i+1 + a n+1 + b n+1 +<br />
n�<br />
�� �<br />
n<br />
+<br />
i − 1<br />
n�<br />
�<br />
n + 1<br />
i=1<br />
i=1<br />
�<br />
a i b n+1−i ,<br />
i<br />
�<br />
a i b n−i+1<br />
n�<br />
i=1<br />
� ��<br />
n<br />
a<br />
i<br />
i b n−i+1<br />
� �<br />
n<br />
a<br />
i<br />
i b n−i+1<br />
E1.6 <strong>Prove</strong> the follow<strong>in</strong>g additional properties of the Ackermann function (some can be<br />
proved without <strong>in</strong>duction):<br />
(i) A(m, n) < A(m, n + 1).<br />
(ii) If n < p, then A(m, n) < A(m, p).<br />
(iii) A(1, n) = n + 2.<br />
(iv) A(m, n + 1) ≤ A(m + 1, n). (H<strong>in</strong>t: <strong>in</strong>duction on n.)<br />
(v) A(m, n) ≤ A(m + 1, n).<br />
(vi) A(2, n) = 2n + 3.<br />
(i): We consider two cases.<br />
Case 1. m = 0. Then A(0, n) = n + 1 < n + 2 = A(0, n + 1).<br />
Case 2. m > 0. Write m = p + 1. Then<br />
A(m, n + 1) = A(p + 1, n + 1) = A(p, A(p + 1, n))<br />
> A(p + 1, n) (us<strong>in</strong>g the statement proved <strong>in</strong> the text)<br />
= A(m, n).<br />
3
(ii): Fix m and n. We prove <strong>by</strong> <strong>in</strong>duction on p <strong>that</strong> <strong>for</strong> all p > n we have A(m, n) < A(m, p).<br />
This is true <strong>for</strong> p = n + 1 <strong>by</strong> (i). Now assume it <strong>for</strong> p, where p > n. Then<br />
A(m, n) < A(m, p) (<strong>by</strong> assumption) < A(m, p + 1) (<strong>by</strong> (i))<br />
(iii): Induction on n. For n = 0, A(1, 0) = A(0, 1) = 2. Assume <strong>that</strong> A(1, n) = n+2. Then<br />
A(1, n + 1) = A(0, A(1, n)) = A(1, n) + 1 = n + 2 + 1 = n + 1 + 2, f<strong>in</strong>ish<strong>in</strong>g the <strong>in</strong>duction.<br />
(iv): Induction on n. For n = 0, A(m + 1, 0) = A(m, 1), as desired. Now assume <strong>that</strong><br />
A(m, n + 1) ≤ A(m + 1, n). Then A(m + 1, n) ≥ A(m, n + 1) ≥ n + 2 <strong>by</strong> the fact proved<br />
<strong>in</strong> the notes, and so <strong>by</strong> (ii) we get<br />
A(m + 1, n + 1) = A(m, A(m + 1, n)) ≥ A(m, n + 2),<br />
f<strong>in</strong>ish<strong>in</strong>g the <strong>in</strong>duction.<br />
(v): <strong>by</strong> (i) and (iv).<br />
(vi): <strong>by</strong> <strong>in</strong>duction on n. n = 0: A(2, 0) = A(1, 1) = 3, as desired. Now assume <strong>that</strong><br />
A(2, n) = 2n+3. Then A(2, n+1) = A(1, A(2, n)) = A(2, n)+2 = 2n+3+2 = 2(n+1)+3,<br />
f<strong>in</strong>ish<strong>in</strong>g the <strong>in</strong>ductive proof.<br />
E1.7 The Fibonacci sequence is def<strong>in</strong>ed <strong>by</strong> f1 = f2 = 1 and fn = fn−1 + fn−2 <strong>for</strong> all<br />
n ≥ 3. <strong>Prove</strong> <strong>that</strong> <strong>for</strong> any positive <strong>in</strong>teger n, f4n is divisible <strong>by</strong> 3.<br />
Induction on n. The case n = 1:<br />
f4 = f3 + f2 = f2 + f1 + f2 = 3,<br />
as desired. Now suppose <strong>that</strong> n is a positive <strong>in</strong>teger and f4n is divisible <strong>by</strong> 3. Then<br />
f 4(n+1) = f4n+4 = f4n+3 + f4n+2 = f4n+2 + f4n+1 + f4n+2<br />
= 2f4n+2 + f4n+1 = 2(f4n+1 + f4n) + f4n+1 = 3f4n+1 + 2f4n;<br />
s<strong>in</strong>ce f4n is divisible <strong>by</strong> 3, so is f 4(n+1), f<strong>in</strong>ish<strong>in</strong>g the <strong>in</strong>ductive proof.<br />
E1.8 For the Fibonacci sequence, prove <strong>that</strong> f1+f2+· · ·+fn = fn+2 −1 <strong>for</strong> every positive<br />
<strong>in</strong>teger n.<br />
Induction on n. The case n = 1: f3 − 1 = f2 + f1 − 1 = 1 = f1. Now assume <strong>that</strong><br />
f1 + f2 + · · · + fn = fn+2 − 1. Then<br />
complet<strong>in</strong>g the <strong>in</strong>ductive proof.<br />
f1 + f2 + · · · + fn + fn+1 = fn+2 − 1 + fn+1 = fn+3 − 1,<br />
E1.9 For the Fibonacci sequence, prove <strong>that</strong> f2 + f4 + · · · + f2n = f2n+1 − 1 <strong>for</strong> every<br />
positive <strong>in</strong>teger n.<br />
Induction on n: For n = 1, f3 −1 = f2 +f1 − 1 = 1 = f2. Now assume <strong>that</strong> n is a positive<br />
<strong>in</strong>teger and f2 + f4 + · · · + f2n = f2n+1 − 1. Then<br />
f2 + f4 + · · · + f2n + f 2(n+1) = f2n+1 − 1 + f2n+2 = f2n+3 − 1 = f 2(n+1)+1 − 1,<br />
4
complet<strong>in</strong>g the <strong>in</strong>ductive proof.<br />
<strong>E1.1</strong>0 For the Fibonacci sequence, prove <strong>that</strong> f 2 1 + f2 2 + · · · + f2 n = fnfn+1 <strong>for</strong> every<br />
positive <strong>in</strong>teger n.<br />
Proof. Induction on n. For n = 1, f1f2 = 1 = f2 1 . Now assume <strong>that</strong> n is a positive<br />
<strong>in</strong>teger and f2 1 + f2 2 + · · · + f2 n = fnfn+1. Then<br />
f 2 1 + f 2 2 + · · · + f 2 n + f 2 n+1 = fnfn+1 + f 2 n+1 = (fn + fn+1)fn+1 = fn+2fn+1,<br />
complet<strong>in</strong>g the <strong>in</strong>ductive proof.<br />
<strong>E1.1</strong>1 For 〈1, 3, 2〉 and 〈0, 4, 5〉, describe the doma<strong>in</strong> of the function 〈1, 3, 2〉 ⌢ 〈0, 4, 5〉, and<br />
give its values <strong>for</strong> each element of the doma<strong>in</strong>.<br />
This is the function h with doma<strong>in</strong> 6 such <strong>that</strong> h(0) = 1, h(1) = 3, h(2) = 2, h(3) = 0,<br />
h(4) = 4, and h(5) = 5.<br />
E11.12 Let ϕ = 〈2, 3, 4〉. Express 〈1〉 ⌢ ϕ ⌢ 〈5〉 as a s<strong>in</strong>gle sequence.<br />
〈1〉 ⌢ ϕ ⌢ 〈5〉 = 〈1, 2, 3, 4, 5〉.<br />
E11.13 Use the recursion theorem (Theorem 1.11) to show <strong>that</strong> <strong>for</strong> each m ∈ ω there is<br />
a function gm with the follow<strong>in</strong>g properties:<br />
(i) g1 is the function mapp<strong>in</strong>g R <strong>in</strong>to R such <strong>that</strong> g1(a) = a <strong>for</strong> all a ∈ R.<br />
(ii) For each positive <strong>in</strong>teger m, gm+1 is the function mapp<strong>in</strong>g m+1 R <strong>in</strong>to R such <strong>that</strong><br />
<strong>for</strong> any 〈a0, . . ., am ∈ R〉,<br />
gm+1(a0, . . ., am) = gm(a0, . . ., am−1) + am.<br />
[ m X is the set of all m-tuples of members of X, <strong>for</strong> any m ∈ ω and any set X. The value<br />
gm(a0, . . ., am−1) is denoted <strong>by</strong> �<br />
i
For any positive <strong>in</strong>teger m let g ′ m be the function whose doma<strong>in</strong> is the set of all m-tuples of<br />
real numbers such <strong>that</strong> g ′ m (〈a0, . . ., am−1〉) = g(〈a0, . . ., am−1〉). We claim <strong>that</strong> g ′ satisfies<br />
the conditions of the exercise.<br />
<strong>for</strong> m ≥ 2 and a = 〈a0, . . ., am〉,<br />
g ′ 1(〈a0〉) = g(〈a0〉)<br />
= f(〈a0〉, g〈pred(A, 〈a0〉, R))<br />
= a0;<br />
g ′ m+1 (a) = g(a)<br />
= f(a, g ↾ prec(A, a, R))<br />
= (g ↾ prec(A, a, R))(〈a0, . . ., am−1〉) + am<br />
= g((〈a0, . . ., am−1〉) + am<br />
= g ′ m(〈a0, . . ., am−1〉) + am.<br />
6