Download MATH10040: Numbers and Functions Homework 1: Solutions

MATH10040: Numbers and Functions
Homework 1: Solutions
1. P and Q are statements. Is the following statement true or false?
If P does not imply Q then Q implies P.
Explain your reasoning.
Solution: The statement is true:
If P does not imply Q then the statement P implies Q is false. This
can only happen if P is true and Q is false. Thus Q is false and hence
Q implies P is true.
2. Two hundred soldiers stand in a rectangular array with 10 rows and
20 columns.
The tallest soldier in each row is selected. Of these 10 soldiers, S is
the smallest.
The smallest soldier in each column is selected. Of these 20 soldiers,
T is the tallest.
What can be said about the relative size of S and T ? Prove your
assertion.
Solution:
Proposition: T ≤ S always.
Proof S is in some row; let’s call it row i. T is in some column; let’s
call it column j.
Let R be the soldier in row i and column j. We compare R to S and
T:
S is the tallest soldier in row i, and R is in row i. So R ≤ S.
T is the smallest soldier in column j and R is in column j. So T ≤ R.
Thus T ≤ R and R ≤ S. So T ≤ S.
3. Recall that a number is rational if it can be written as a/b where a
and b are integers (and b 6= 0 of course).
(a) Show that the product of two rational numbers is rational.
(b) Show that the sum or difference of two rational numbers is rational.
(c) Suppose that a 6= 0 is rational and b is irrational. Show that
a + b and ab are irrational.
Solution:
(a) Let p, q be rational numbers. So there are integers a, b, c, d such
that p = a/b and q = c/d. Therefore
pq =
a c
ac
· =
b d
bd
which is a rational number since ac, bd ∈ Z.
(b) Again, let p = a/b and q = c/d where a, b, c, d ∈ Z (and b, d 6= 0).
Then
a c
ad bc
ad ± bc
p±q = ± =
±
=
b d
bd bd
bd
is rational since bd, ad ± bc ∈ Z.
(c) Let c = a + b, d = ab.
Suppose, FTSOC, that c is rational. Since a is rational, it follows
from (b) that c − a = b is rational: a contradiction. So c is
irrational.
Similarly, suppose FTSOC that d is rational. Now a 6= 0 is
rational, hence 1/a is rational. By (b), it follows that (1/a)·d = b
is rational: again, a contradiction.
√
4. We will show later that 2 is an irrational number. Give an example
of two distinct positive irrational numbers a and b which have the
property that a + b is rational. Prove your assertions.
√
√
Solution: Take a = 2 and b = 2 − 2.
√
√
First we show that b is irrational:
2 is irrational, hence so is − 2.
√
Since 2 is rational, b = 2 + (− 2) is irrational by part (c) of the last
exercise.
√
Note that b > 0 since√2 > √
2. Also b 6= a. √
√
(Proof: b = a =⇒ 2 + 2 = 2 =⇒ 2 2 = 2 =⇒ 2 = 1, a
contradiction.)
So a, b are distinct postive irrational numbers whose sum is 2.
5. Prove by induction on n that for all natural numbers n ≥ 1
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)
.
6
Solution: For n = 1 the statement reads
12 =
and so is true.
6
1 · 2 · (2 + 1)
= =1
6
6
Suppose we have proved the statement for some n. Then
n(n + 1)(2n + 1)
12 + · · · + n2 + (n + 1)2 =
+ (n + 1)2 (by our inductive hyp.)
6
n(2n + 1)
2n2 + 7n + 6
= (n + 1) ·
+ n + 1 = (n + 1) ·
6
6
(2n + 3)(n + 2)
= (n + 1) ·
6
which is precisely the statement for n + 1.
So we’re done, by induction.
6. Suppose that x is a number other than 1. Prove by induction on n
that for all natural numbers n ≥ 1
1 + x + x2 + · · · + xn =
xn+1 − 1
.
x−1
Solution: For n = 1, the statement reads
1+x=
x2 − 1
=1+x
x−1
which is clearly true.
Suppose that we have proved the statement for some n.
Then
xn+1 − 1
+ xn+1 (by our inductive hyp.)
x−1
xn+1 − 1 + xn+1 (x − 1)
xn+1 − 1 + xn+2 − xn+1
=
=
x−1
x−1
xn+2 − 1
=
x−1
1 + x + · · · + xn + xn+1 =
which is the statement for n + 1.
7. Prove by induction on n that
n+1
1
1
1
1−
1−
··· 1 − 2 =
4
9
n
2n
for all integers n ≥ 2.
Solution: When n = 2 the statement is
1−
1
3
=
4
4
which is visibly true.
Suppose the statement is true for n. Then
1
1
1
n+1
1
1−
··· 1 − 2 · 1 −
=
· 1−
(by our inductive hyp.)
4
n
(n + 1)2
2n
(n + 1)2
n + 1 (n + 1)2 − 1
(n + 1)2 − 1
=
=
2n
(n + 1)2
2n(n + 1)
2
n + 2n
n+2
=
=
2n(n + 1)
2(n + 1)
which is the statement for n + 1.
8. Let p1 = 2, p2 = 3, p3 = 5, p4 = 7, . . . be the sequence of primes.
What is p13 ? Prove that pn ≥ 2n + 15 for all n ≥ 13.
Solution: We prove the statement by induction on n ≥ 13.
p13 = 41 = 2 · 13 + 15. So the statement is true for 13.
Suppose now that the statement is true for some n ≥ 13. Then
pn+1 ≥ pn + 2 (why?) and pn ≥ 2n + 15 by our inductive hypothesis.
Thus pn+1 ≥ (2n + 15) + 2 = 2(n + 1) + 15 as required.
9. Explain carefully what is wrong with following argument:
Proposition Everybody has the same number of hairs.
Proof: We’ll prove this by induction on n, the number of people.
If n = 0 or 1, the statement is clearly true.
Assume the statement is true for any n people, and suppose there is
a group of n + 1 people. Remove one person from the group. By our
inductive hypothesis, the remaining n all have the same number of
hairs. Remove a different person. The remaining n people all have
the same number of hairs. Therefore the first person has the same
number of hairs as all the rest, and hence all n + 1 have the same
number of hairs.
Solution: The inductive step in the argument as given fails when
n = 1: In this case n + 1 = 2 and when we remove a ‘different person’
only the first person remains and there is no ‘all the rest’.
10. Prove or disprove the following statement: For all natural numbers n,
n2 − n + 41 is a prime number.
Solution: The statement is false: taking n = 41, 412 − 41 + 41 = 412
is not prime (it has factors 1, 41 and 412 ).