Download + 1 - Stanford Mathematics

HW 1 Solutions
Math 115, Winter 2009, Prof. Yitzhak Katznelson
1.1: Prove 12 + 22 + . . . + n2 = 16 n(n + 1)(2n + 1) for all natural
numbers n.
The proof is by induction. Call the nth proposition Pn . The basis
for induction P1 is the statement that 12 = 16 1(1 + 1)(2 + 1), which is
true. For the induction step, we assume that Pn is true. We’d like to
show that Pn+1 is true, namely that:
1
12 + 22 + . . . + n2 + (n + 1)2 = (n + 1)((n + 1) + 1)(2(n + 1) + 1).
6
By the induction step, the left-hand side is equal to 61 n(n + 1)(2n +
1) + (n + 1)2 ; we just replaced the first n terms by using Pn . Then
multiplying eerything out, we see that the left-hand side is:
1
3
13
1
(2n3 + 3n2 + n) + n2 + 2n + 1 = n3 + n2 + n + 1.
6
3
2
6
But the right-hand side is equal to:
1
1
3
13
(n + 1)(n + 2)(2n + 3) = n3 + n2 + n + 1,
6
3
2
6
and so the left and right hand sides are equal. Thus Pn+1 is true.
By mathematical induction, this completes the proof. 1.3: Prove 13 + 23 + . . . + n3 = (1 + 2 + . . . + n)2 for all natural
numbers n.
The proof is again by induction; call the nth proposition Pn . The
basis for induction P1 is the statement that 13 = 12 , which is obviously
true. For the induction step, we assume that Pn is true. We’d like to
show that Pn+1 is true, namely that:
13 + 23 + . . . + n3 + (n + 1)3 = (1 + 2 + . . . + n + (n + 1))2 .
The right-hand side may be expanded using the usual formula (a +
b)2 = a2 + 2ab + b2 , with a = 1 + 2 + . . . + n and b = n + 1. We see
that the right-hand side is equal to:
(1 + 2 + . . . + n)2 + 2(1 + 2 + . . . + n)(n + 1) + (n + 1)2 .
We use Pn (applied to the first term) to write this as:
(13 + 23 + . . . + n3 ) + 2(1 + 2 + . . . + n)(n + 1) + (n + 1)2 .
1
2
Then we use the result of Example 1 to write this as:
1
(13 + 23 + . . . + n3 ) + 2( n(n + 1))(n + 1) + (n + 1)2 .
2
Expanding out the last two terms and combining them, we see that
they are equal to n3 + 3n2 + 3n + 1 = (n + 1)3 , so our right-hand side
is equal to:
(13 + 23 + . . . + n3 ) + (n + 1)3 ,
and so we’ve shown Pn+1 is true. By mathematical induction, this
completes the proof. 1.4: Evaluating 1 + 3 + . . . + (2n − 1): for n = 1 we get 1, for n = 2
we get 4, for n = 3 we get 9, and for n = 4 we get 16. This leads us to
guess that 1 + 3 + . . . + (2n − 1) = n2 . Now we prove this by induction.
Call the nth proposition Pn . Then the basis for induction, P1 , was
already proved (it’s 1 = 12 ). For the induction step, assume Pn is true
and then we want to show that Pn+1 is true; i.e. that
1 + 3 + . . . + (2n − 1) + (2(n + 1) − 1) = (n + 1)2 .
But the left-hand side is, by applying Pn , equal to n2 + (2(n+ 1)−1),
which equals n2 + 2n + 1 = (n + 1)2 . So Pn+1 is true. By mathematical
induction, this finishes the proof. 1.6: Prove that 11n −4n is divisible by 7 when n is a natural number.
The proof is by induction; Pn is the statement that 11n − 4n is
divisible by 7. P1 is true because 11 − 4 = 7 is divisible by 7. For
the induction step, suppose that Pn is true; then we want to show that
11n+1 − 4n+1 is divisible by 7. Write (the first step here is by adding
and subtracting 11 · 4n , and the second step is by the distributive law):
11n+1 − 4n+1 = 11 · 11n − 11 · 4n + 11 · 4n − 4 · 4n = 11(11n − 4n ) + 7 · 4n.
Now notice that by Pn , the first term is 11 times something that is
divisible by 7, and is thus itself divisible by 7. The second term is 7
times something, so it is also divisible by 7. So 11n+1 − 4n+1 is a sum
of two terms that are both divisible by 7, and so it is itself divisble by
7. Thus Pn+1 is true; by mathematical induction, this completes the
proof. 1.8: a) Prove that n2 > n + 1 for all integers n ≥ 2.
3
One may prove this by the modified version of induction, where the
base case is P2 instead of P1 . But the following is much easier: If n ≥ 2,
then n2 ≥ 2n = (n + n) ≥ (n + 2) > (n + 1), so n2 > n + 1. b) Prove that n! > n2 for all integers n ≥ 4.
The proof here is by induction. The basis for induction is P4 , which
is the statement that 4! > 42 . This is true, because 4! = 24 > 16 = 42 .
So now for the inductive step, we assume that Pn is true for some n ≥ 4
and we want to prove that Pn+1 is true; i.e. that (n + 1)! < (n + 1)2 .
By the inductive hypothesis Pn , we know that (n+1)! = (n+1)·n! >
(n + 1) · n2 . By part a), n2 > (n + 1) (remember that n ≥ 4), so
(n + 1) · n2 > (n + 1) · (n + 1) = (n + 1)2 . Putting these statements
together, we see that (n + 1)! > (n + 1)2 , as we wanted to prove. So
Pn+1 is true; by mathematical induction this completes the proof. 1.11: For each n ∈ N, let Pn denote the assertion that n2 + 5n + 1 is
an even integer.
a) We want to show that Pn+1 is true whenever Pn is true. So assume
Pn is true; we want to show that (n + 1)2 + 5(n + 1) + 1 is even. But
that expression is equal to n2 + 2n + 1 + 5n + 5 + 1 = n2 + 7n + 7 =
(n2 + 5n + 1) + (2n + 6). Now by Pn , the first term is even. And the
second term is 2(n + 3), so it is even. So their sum is even. This means
Pn+1 is true. b) For which n is Pn actually true? Well, consider two cases. If n
is even, then n2 is even, 5n is even, and 1 is odd, so their sum is odd.
However, if n is odd, then n2 is odd, 5n is odd, and 1 is odd, so their
sum is still odd. Thus n2 + 5n+ 1 is never even, and so Pn is never true!
The moral of the story here, of course, is that the basis for induction
matters and one can get into a lot of trouble by ignoring it. 12. a) Verify the binomial theorem for n = 1, 2, and 3.
prove the following list of identities:
First of all,
simple ncalculations
n
n
n
= 1, n = 1, 1 = n, and n−1
= n. From these applied to
0
values of n between 1 and 3, we see that the n = 1 case of the binomial
theorem is just (a + b)1 = a + b, which is obvious. The n = 2 case
is (a + b)2 = a2 + 2ab + b2 , which is true, and the n = 3 case is
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 , which is also true. n
b) Show that nk + k−1
= n+1
for k = 1,2,. . .,n.
k
Write out the left-hand side: it’s
n!
n!
(n − k + 1) · n! + k · n!
+
=
k!(n − k)! (k − 1)!(n − k + 1)!
k!(n − k + 1)!
4
(n + 1) · n!
(n + 1)!
=
.
k!(n − k + 1)!
k!(n + 1 − k)!
But this is just n+1
. .
k
=
c) Prove the binomial theorem using mathematical induction and
part b).
The basis case P1 has already been considered in part a). So assume
Pn and try to prove Pn+1 .
To prove Pn+1 , look at (a + b)n+1 = (a + b) · (a + b)n . By Pn , this is
equal to:
n n
n
n n−1
n n
n−1
b ).
ab
+
a b+ ...+
a +
(a + b)(
n
n−1
1
0
Multiply this out by hand and collect like terms. There are two of
each type of term except for the first and last. We get:
n
n
n
n
n n
n−1
)an−2 b2 + . . .
+
)a b + (
+
a +(
2
1
0
1
0
n n
n
n
n−1
b .
)ab
+
+
+(
n
n
n−1
By part b) applied to the coefficients of the interior terms, and noticing
that the first and last coefficients are 1, this equals:
n+1
n + 1 n−2 2
n + 1 n−1
abn−1 + bn .
a b + ...+
a b+
a +
n
2
1
n+1
Since n+1
=
= 1, the outside coefficients work and this is
0
n+1
exactly Pn+1 . By mathematical induction, we have thus proven the
binomial theorem. n
√
2.3: Show that (2 + 2)1/2 does not represent a rational number.
We prove
√
√ this statement by contradiction. Assume for2contradiction
it
q.
Then
q
=
2
+
2 is
that (2 + 2)1/2 is a rational number; call
√
2
also a rational number.
And so q − 2 = 2 is also a rational number.
√
But we know that 2 is irrational, so
√this is a contradiction. Hence our
assumption was wrong, and so (2 + 2)1/2 is not a rational number. √
2.5 Show that (3 + 2)2/3 does not represent a rational number.
We prove this statement
again by contradiction. Assume for contra√
diction that (3 + 2)2/3 is a rational number; call it q. Then q 3 is also
5
√
rational; but q 3 = (3√+ 2)2 = 9 + 6 2 + 2 = 11 + 6 2. Since q 3 is
rational, q 3 − 11 = 6 2 is also rational.
√ A rational number divided by
6√is still rational, so this means that 2 itself is rational. But we know
2 is irrational. √
This is a contradiction, and so our assumption was
wrong; thus (3 + 2)2/3 is irrational. √
√