Math 104, Solution to Homework 1 Instructor: Guoliang Wu June 19, 2009
Ross, K. A., Elementary Analysis: The theory of calculus : 1.2 Prove Prove 3 + 11 + + (8n (8n 5) = 4n 4n2 n for all natural numbers n. Solution: (1) When n = 1, the assertion holds since 3 = 4 12 1.
···
−
(2) Suppose 3 + 11 + that
3 + 11 +
−
· −
(8 n − 5) = 4n 4n2 − n, we want to show · · · + (8n
(8n − 5) + [8(n [8(n + 1) − 5] = 4(n 4(n + 1)2 − (n + 1). 1). · · · + (8n
The left hand side equals (by induction assumption)
4n2
[8(n + 1) − 5] = 4n 4n2 + 7n 7n + 3 = 4(n 4(n + 1)2 − (n + 1). 1). − n + [8(n
Thus, by induction, we have proved the claim. 1.3 Prove Prove 13 +2 3 + + n3 = (1+2+ + n)2 for all natural numbers n. Solution: (1) When n = 1, the assertion holds since 13 = 1 = 12 .
···
···
(2) Suppose 13 + 23 + that
13 + 23 +
· · · + n3 = (1+ 2+ · · · + n)2, we want to show
(n + 1)3 = [1 + 2 + · · · + n + (n (n + 1)]2 . · · · + n3 + (n
The left hand side equals (by induction assumption)
(1 + 2 +
(n + 1)3 , · · · + n)2 + (n
and the right hand side equals (using the formula for (a + b)2 =
1
...)
· · · + n)2 + 2(1 + 2 + · · · + n)(n + 1) + (n + 1)2 n(n + 1) =(1 + 2 + · · · + n)2 + 2 (n + 1) + (n + 1)2 (Here we use the formula 2 1 + 2 + · ·· + n = n(n+1) , which itself can be prove by induction.) 2 =(1 + 2 + · · · + n)2 + n(n + 1)2 + (n + 1)2 =(1 + 2 + · · · + n)2 + (n + 1)3 . (1 + 2 +
Thus, we obtain the desired equality. By induction, we have proved the claim. 1.6 Prove that 11n
− 4n is divisible by 7 when n is a natural number. (1) When n = 1, the assertion holds since 11 − 4 = 7 is
Proof. divisible by 7.
(2) Suppose that 11n 4n is divisible by 7, we want to show that 11n+1 4n+1 is also divisible by 7.
−
−
11n+1
− 4n+1 = 11(11n) − 4(4n) = 11(11n − 4n) + 11(4n) − 4(4n) = 11(11n − 4n ) + 7(4n ). The first term is a divisible by 7 since 11n −4n is, and the second term 7(4n ) is clearly a multiple of 7. Therefore, 11n+1 − 4n+1 is also divisible by 7.
By induction, we proved the claim. 1.7 Prove that 7n n.
− 6n − 1 is divisible by 36 for all positive integers
Proof. (1) When n = 1, the assertion holds since 71 is divisible by 7.
(2) Suppose that 7n that 7n+1 6(n + 1)
−6−1 = 0
− 6n − 1 is divisible by 36, we want to show − − 1 is divisible by 36. 7n+1 − 6(n + 1) − 1 = 7(7n ) − 6(n + 1) − 1 = 7(7n − 6n − 1) + 7(6n + 1) − 6(n + 1) − 1 = 7(7n − 6n − 1) + 36n. 2
The first term is a divisible by 36 since 7n 6n 1 is, and the second term 36n is clearly a multiple of 36. Therefore, 7n+1 6(n + 1) 1 is divisible by 36.
−
−
−
−
By induction, we proved the claim. 1.9 (a) Decide for which integers the inequality 2n > n2 is true. (b) Prove you claim in (a) by mathematical induction. Solution: (a) We check that when n = 2, equality n = 3, 23 = 8 < 32 = 9. When n = 2, equality holds
holds. For again. For n 2 5 2 6 n = 5, 6, , 2 > n holds since 2 = 32 > 5 = 25, 2 = 64 > 62 = 36, etc. We may guess that 2n > n 2 is true for n 5.
· ··
≥
(b) We use mathematical induction (or rather, a slight variation. See Exercise 1.8 in the textbook.) to prove our assertion in (a). (1) When n = 5, we have checked that 2n > n2 is true. (2) Suppose 2n > n2 for some n 5, we need to show that n+1 2 2 > (n + 1) is also true. Observe that
≥
2n+1 = 2(2n ) > 2n2 = n2 + n2 = n2 + 2n + 1 + (n2 = (n + 1)2 + (n
− 2n − 1)
− 1)2 − 2 ≥ (n + 1)2 + 42 − 2 > (n + 1)2.
√2)1/2 does not represent a rational number. √ √ Denote by a = (2 + 2)1/2 . Then a2 = 2 + 2 and
2.3 Show that (2 + Solution: (a2 2)2 =
−
2. Thus, a is a root of the equation a4
− 4a2 + 2 = 0.
By Rational Zeros Theorem, the only possible rational roots of this equation are 1, 2. We check that none of them is a root. Therefore, a = (2 + 2)1/2 does not represent a rational number.
± ±√
√
2.5 Show that (3 + 2)2/3 does not represent a rational number. 2/3 Solution: Denote by b = (3 + 2) . Then b3 = (3 + 2)2 = 11 + 6 2 and (b3 11)2 = 72. Thus, b is a root of the equation
√
√
−
b6
− 22b3 + 49 = 0. 3
√
By Rational Zeros Theorem, the only possible rational roots of this equation are 1, 7, 49. We check that none of them is a root. Therefore, b = (3 + 2)2/3 does not represent a rational number.
± ± ±√
Extra. If r Q, r = 0, and x is irrational, prove that r + x and rx are irrational.
∈
We can prove the claim by contradiction. First, we know r = 0 is a rational number, which can be written as m n where m, n Z and m, n = 0. Solution:
∈
1) Suppose r + x is also rational, then it can be written as r + x = pq , with p, q integers and q = 0. Then
x = (r + x)
− qm − r = pq − mn = pn qn
is rational since pn qm and qn are both integers and qn = 0. This is a contradiction to the assumption that x is irrational, and hence r + x cannot be rational.
−
2) Similarly, suppose rx is rational, then we can write it as rx = pq with p , q integers and q = 0. Then
p m p n x = (rx)/r = / = q
n
qm
is rational since p n and qm are both integers and qm = 0. This is again a contradiction and hence rx cannot be rational.
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