Download Assignment #1 Solutions

1
MATHEMATICS 271 L60 SUMMER 2003
ASSIGNMENT 1 SOLUTION
1. Let P be the statement: “If a | b and a | c then a | (b + c) and a | (b − c)”
(a) Is P true for all integers a, b and c? Prove your answer.
(b) Write the converse of P. Is the converse of P true for all integers a, b and c? Prove
your answer.
(c) Write the contrapositive of P. Is the contrapositive of P true for all integers a, b
and c? Explain.
(d) Write the negation of P. Is the negation of P true for some integers a, b and c?
Explain.
Solution:
(a) P is true for all integers a, b and c and here is a proof. Let a, b and c be integers
so that a | b and a | c. Since a | b and a | c, there are integers m and n so that b = am
and c = an, and so, b + c = am + an = a (m + n) and b − c = am − an = a (m − n).
Since b + c = a (m + n) and b − c = a (m − n), where a, m + n and m − n are integers, we
conclude that a | (b + c) and a | (b − c).¤
(b) The converse of P is “If a | (b + c) and a | (b − c) then a | b and a | c”. The converse
of P is not true for all integers a, b and c. For example, when a = 2 and b = c = 1, we
have a | (b + c) and a | (b − c) (because 2 | 2 and 2 | 0), but a - b (because 2 - 1).¤
(c) The contrapositive of P is “If a - (b + c) or a - (b − c) then a - b or a - c”. The
contrapositive of P is true for all integers a, b and c. This is because the contrapositive of
P is logically equivalent to P which is proven to be true in (a).¤
(d) The negation of P is “ a | b and a | c, but a - (b + c) or a - (b − c)”. The negation
of P is not true for some integers a, b and c, because we have shown in (a) that P is true
for all integers a, b and c.¤
2. Prove or disprove each of the following:
(a) For all real numbers x and y, if x is rational and y is irrational then x + y is
irrational.
(b) For all real numbers x and y, if x is rational and y is irrational then xy is irrational.
(c) For all real numbers x and y, if x+y is irrational then x is irrational or y is irrational.
(d) For all real numbers x and y, if xy is irrational then x is irrational or y is irrational.
Solution:
(a) This statement is true and here is a proof. Let x be a rational number and y be an
irrational number. We show that x + y is irrational by a contradiction proof. Suppose that
x + y is rational. Then, since x and x + y are rational, there are integers m, n, p, q so that
x= m
and x + y = pq where n 6= 0 6= q. It follows that y = (x + y) − x = pq − m
= pn−mq
,
n
n
qn
which implies that y is rational (for being a quotient of the integers pn − mq and qn, and
note that nq 6= 0 because n 6= 0 6= q). This contradicts the assumption that y is irrational.
Thus, x + y is irrational.¤
2
√
(b) This statement is false. For example, when x = 0 and y = 2, we see that x is
rational and y is irrational. However, xy = 0 is rational.¤
(c) This statement is true and here is a proof. Let x and y be real numbers so that
x + y is irrational. We show that x is irrational or y is irrational by a contradiction proof.
Suppose that both x and y are rational; that is, there are integers m, n, p, q so that
x= m
and y = pq where n 6= 0 6= q. It follows that x + y = pq + m
= pn+mq
, which implies
n
n
qn
that x + y is rational (for being a quotient of the integers pn + mq and qn, and note that
nq 6= 0 because n 6= 0 6= q). This contradicts the assumption that x + y is irrational. Thus,
x is irrational or y is irrational.¤
(c) This statement is true and here is a proof. Let x and y be real numbers so that
xy is irrational. We show that x is irrational or y is irrational by a contradiction proof.
Suppose that both x and y are rational; that is, there are integers m, n, p, q so that
x= m
and y = pq where n 6= 0 6= q. It follows that xy = pq m
= pm
, which implies that xy
n
n
qn
is rational (for being a quotient of the integers pm and qn, and note that nq 6= 0 because
n 6= 0 6= q). This contradicts the assumption that xy is irrational. Thus, x is irrational or
y is irrational.¤
3. In this question, k, m and n are integers. Prove or disprove each of the following:
(a) m is even and n is odd if and only if m + 2n is even and m + n is odd.
(b) If k | m and m | n then k | n.
(c) If k | mn then k | m and k | n.
(d) For all natural numbers n, n2 + n + 41 is a prime.
Solution:
(a) This statement is true and here is a proof.
(=⇒) We prove that if m is even and n is odd then m + 2n is even and m + n is odd.
Suppose that m is an even integer and n is an odd integer, that is, there are integers
x and y so that m = 2x and y = 2y + 1. Then, m + 2n = 2x + 2n = 2 (x + n) and
m + n = 2x + 2y + 1 = 2 (x + y) + 1, which means that m + 2n is even and m + n is odd.
(⇐=) We prove that if m + 2n is even and m + n is odd then m is even and n is odd.
Suppose that m and n are integers so that m + 2n is even and m + n is odd, that is,
m + 2n = 2s and m + n = 2t + 1 for some integers s and t. Then,
m = 2 (m + n) − (m + 2n) = 2 (m + n) − 2s = 2 (m + n − s) and
n = (m + 2n) − (m + n) = 2s − (2t + 1) = 2 (s − t − 1) + 1,
which imply that m is even and n is odd.¤
(b) This statement is true and here is a proof. Suppose that k, m and n are integers
so that k | m and m | n, which mean that m = ka and n = mb for some integers a and b.
Then, n = mb = (ka) b = k (ab) which means that k | n.¤
(c) This statement is false because when k = 4 and m = n = 2, we have k | mn
(because 4 | 4), but k - m (because 4 - 2).¤
(d) This statement is false because when n = 41, we have
2
n + n + 41 = 412 + 41 + 41 = 41 × (41 + 1 + 1) = 41 × 43 which is not a prime.¤
3
4. Prove or disprove each of the following:
(a) For all real numbers x, x2 − x + 1 > 0.
(b) For all real numbers x and y, if x2 − x + 1 = y 2 − y + 1 then x = y.
(c) For all posive real numbers x, there exists a positive number y so that 0 < y < x.
(d) There exists a positive number y so that for all positive real numbers x, 0 < y < x.
Solution:
(a) This statement is true and here is a proof. Let x be a real number. Then
¡ ¢
¡
¡
¢2
¢2
2
x − x + 1 = x2 − 2 12 x + 14 + 34 = x − 12 + 34 ≥ 34 > 0. This is because x − 12 ≥ 0.¤
(b) This statement is false because when x = 0 and y = 1, we have
x2 − x + 1 = 02 − 0 + 1 = 1 = 12 − 1 + 1 = y 2 − y + 1, but x = 0 6= 1 = y.¤
(c) This statement is true and here is a proof. Let x be a posive real number. We
consider y = x2 . From 0 < 12 < 1,we get (by multiplying by the positive number x), that
0 < x2 < x, that is, 0 < y < x.
(d) This statement is false because for any real number y, as shown in (c) we can find
a positive real number x so that x < y.