Homework 4 Solutions:
Thm. 2.20. There do not exists natural numbers m and n such that 24m3 = n3 .
Proof: Let m, n ∈ N.
tm
Using the Fundamental Theorem of Arithmetic, express m = pr11 · · · prnn and n = q1t1 · · · qm
with pi , qj prime and ri , tj ∈ N for any i and j.
3tm
tm 3
3rn
1
.
) = q13t1 · · · qm
and n3 = (q1t1 · · · qm
Then m3 = (pr11 · · · prnn )3 = p3r
1 · · · pn
Also, 24 = 23 ∗ 3.
Case 1: pi 6= 3 for any i.
Then the prime factorization of 24m3 contains 31 .
Since every exponent in the prime factorization of n3 is a multiple of 3, we see that 24m3
and n3 have different prime factorizations.
Case 2: pi = 3 for some i.
Then the prime factorization of 24m3 contains 33ri +1 , and 3ri + 1 is not a multiple of 3.
Since every exponent in the prime factorization of n3 is a multiple of 3, we see that 24m3
and n3 have different prime factorizations.
Thus in either case, 24m3 and n3 have different prime factorizations, so 24m3 6= n3 .
Ex. 2.22. Show that
√
12 is irrational.
Proof: For the sake of contradiction, assume that
a, b ∈ Z and b 6= 0.
√
Then 12b = a =⇒ 12b2 = a2 .
√
12 is rational, i.e.
√
12 =
a
b
with
However, 12 = 22 ∗ 3, so following the logic of Thm. 2.20, the prime factorization of 12b2
has an odd number of 3’s, but the prime factorization of a2 has an even number of (or zero)
3’s.
Thus 12b2 6= a2 , and we have our contradiction.
√
∴ 12 is irrational.
Thm. 2.26 Let p be a prime and let a be an integer. Then p does not divide a if and
only if gcd(a, p) = 1.
Proof: By definition, gcd(a, p) divides both a and p.
Since p is prime, only 1 and p divide p, so either gcd(a, p) = 1 or gcd(a, p) = p.
(=⇒) Assume p - a.
If gcd(a, p) = p, then p | a, which is a contradiction.
So gcd(a, p) = 1.
1
(⇐=) Assume gcd(a, p) = 1.
If p | a, then p divides both p and a, which implies gcd(a, p) = p.
Thus p - a.
.
Thm. 2.32. For all natural numbers n, we have gcd(n, n + 1) = 1.
Proof: For the sake of contradiction, assume gcd(n, n + 1) = d where d > 1.
Then d divides n and n + 1, so n = d ∗ x and n + 1 = d ∗ y for some x, y ∈ Z.
Then (n + 1) − n = 1 = dy − dx = d(y − x) =⇒ d | 1, which is a contradiction.
∴ gcd(n, n + 1) = 1.
Thm. A.19. For every natural number n,
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)
.
6
Proof:
Base case, n = 1: 12 = 1 =
1(2)(3)
6
=
1(1+1)(2∗1+1)
.
6
Induction step, assume Pk is true, i.e. 12 + 22 + · · · + k 2 =
k(k+1)(2k+1)
.
6
Then 12 + 22 + · · · + k 2 + (k + 1)2
=
=
=
=
=
=
k(k+1)(2k+1)
+ (k + 1)2
6
+ 6(k+1)
)
(k + 1)( k(2k+1)
6
6
2
(k + 1) 2k +k+6k+6
6
2k2 +7k+6
(k + 1)
6
(k+2)(2k+3)
(k + 1)
6
(k+1)((k+1)+1)(2(k+1)+1)
, and
6
so Pk+1 is true.
Thm. A.21. For every natural number n,
13 + 23 + · · · + n3 = (1 + 2 + ... + n)2 .
Proof:
Base case, n = 1: 13 = 1 = (1)2 .
Induction step: Assume Pk is true, i.e.
13 + 23 + · · · + k 3 = (1 + 2 + ... + k)2 .
Then 13 + 23 + · · · + k 3 + (k + 1)3
2
= (1 + 2 + ... + k)2 + (k + 1)3
= ( k(k+1)
)2 + (k + 1)3 (by Thm. A.10.)
2
2
4(k+1)
)
4
2
(k + 1)2 ( k +4k+4
)
4
2
(k + 1)2 ( (k+2)
)
22
( (k+1)(k+2)
)2
2
= (k + 1)2 ( k4 +
=
=
=
= (1 + 2 + · · · + (k + 1))2 , so Pk+1 is true.
Consider the following game involving two plyers, whom we will call player 1 and player
2. Two piles each containing the same number of rocks sit between the players. At each
turn a player may remove any number of rocks (other than zero) from one of the piles. The
player to remove the last rock wins. Player 1 always goes first.
Thm. A.29 For any natural number n of rocks in each pile to begin, player 2 has a
winning strategy.
Proof (by strong induction):
Base case, n = 1: Here both piles have 1 rock. Player 1 must choose 1 rock which
eliminates 1 pile. Then player 2 chooses the other rock, which removes all rocks from the
game, so player 2 wins.
Induction step: Assume that for any k ≤ n, Pk is true, i.e. if each pile consists of k ≤ n
rocks on player 1’s turn, then player 2 has a winning strategy.
Now assume that the game starts with n + 1 rocks in each piles.
Player 1 must choose some amount of rocks from one of the pile, say he chooses j rocks
from pile x, where j ≥ 1 and x is either 1 or 2.
Now pile x has n + 1 − j rocks and the other pile, call it pile y, still has n + 1 rocks.
Let player 2’s strategy be taking j rocks from pile y.
After player 2’s turn, it will be player 1’s turn and both piles will have n + 1 − j ≤ n
rocks.
Thus by the induction hypothesis, from this point on player 2 has a winning strategy.
∴ Pk+1 is true.
Thm. A.31. Every natural number greater than 7 can be written as a sum of 3’s and
5’s.
Proof (by strong induction):
Base cases (we will show this is true for n = 8, 9, and 10, the reason for this will become
clear in the induction step).
3
n = 8 : 8 = 3 + 5.
n = 9 : 9 = 3 + 3 + 3.
n = 10 : 10 = 5 + 5.
Induction step: Now let n + 1 > 10 and assume Pk is true for all 7 < k ≤ n, i.e. any
natural number between 8 and n inclusive can be expressed as a sum of 3’s and 5’s.
Then n + 1 = (n − 2) + 3, where 7 < n − 2 ≤ n.
By the induction hypothesis, n − 2 can be expressed as a sum of 3’s and 5’s.
Say there are x 3’s and y 5’s in the expression for n − 2 for some positive integers x and
y.
Then (n − 2) + 3 = n + 1 can be expressed as the sum of (x + 1) 3’s and y 5’s.
∴ Pk+1 is true.
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