Download Formal Methods Key to Homework Assignment 3, Part 2

Formal Methods
Key to Homework Assignment 3, Part 2
February 12, 2007
• Prove that multiplication of rational numbers is well-defined.
Proof. Suppose m, n, p, q, r, s, t and u are integers such that n, q, s, and u are nonzero,
m/n = r/s, and p/q = t/u. We want to see that
mp
rt
=
,
nq
su
or, equivalently
mp
rt
= .
nq
su
From the definition of equality of rational numbers, we need to see that
mpsu = nqrt.
But by assumption m/n = r/s and p/q = t/u. So ms = nr and pu = qt. Substituting
into the left-hand side of the equation mpsu = nqrt, we get
nrqt = nqrt.
Reversing the steps, then, we see that
mp
rt
=
,
nq
su
and multiplication of rational numbers is well-defined.
89. Is the converse of “if n is any prime, then 2n + 1 prime” true? If your answer is yes,
prove the statement. Otherwise find a counterexample.
The converse is “if 2n + 1 is prime, then n is prime.” This is false. For example,
24 + 1 = 17, but 4 isn’t prime.
94. Prove that if a and b are rational numbers with a < b, then there exists a rational
number r such that a < r < b.
1
Proof. Define r = (a + b)/2. Then if m, n, p, and q are integers with n and q nonzero,
such that a = m/n and b = p/q, we have
m p 1
mq + np 1
mq + np
r=
+
=
=
.
n
q 2
nq
2
2nq
Since mq + np is an integer and 2nq is a nonzero integer, we see that r is a rational
number.
To check that a < r < b, we can check that the differences r − a and b − r are both
positive. We have that
r−a=
a+b
a + b − 2a
b−a
−a=
=
2
2
2
Since a < b, we know that b − a > 0. So r − a > 0 and a < r. The argument that b − r
is positive is entirely analogous:
b−r =b−
a+b
2b − a − b
b−a
=
=
,
2
2
2
which, as we’ve just seen is positive. So r is a rational number such that a < r < b.
95. Prove that if x is a positive real number, then x + 1/x ≥ 2.
Proof. This is similar to problem 80, which we proved by contradiction. So let’s try
assuming the contrary. That is, we assume that x is a positive real number such that
x + 1/x < 2. Since x is positive, we can multiply both sides of this inequality by x and
get
x2 + 1 < 2x,
or, equivalently,
x2 − 2x + 1 = (x − 1)2 < 0.
However, since x is a real number (x − 1)2 ≥ 0, and this is a contradiction. So the
assumption that x + 1/x < 2 must be false, and x + 1/x ≥ 2 for all positive real
numbers x.
√
√
√
96. (a) Find positive real numbers x and y such that x + y 6= x + y.
√
√
√
√
If x = y = 1, then x and y are positive, and x + y = 2 but x + y = 2.
√
√
Since 2 is rational and 2 is irrational, 2 6= 2.
√
√
√
(b) Prove that if x and y are positive real numbers, then x + y ≤ x + y.
First Proof. We’ve proved a number of inequalities involving real numbers by
using contradiction. So let’s try contradiction. So we assume that x and y are
positive real numbers such that
√
√
√
x + y > x + y.
Since the function f (z) = z 2 is increasing on the positive reals, we know that if
0 < a < b then 0 < a2 < b2 . So
√
√
√
( x + y)2 > ( x + y)2 .
2
Simplifying, gives
√
x + y > x + 2 xy + y,
and subtracting x + y from both sides gives
√
0 > 2 xy,
√
√
but since x and y are positive reals, xy >√0. So 2 xy > 0, which is a contra√
√
diction. So the √
assumption that x + y > x + y must be false, and we have
√
√
that x + y ≤ x + y, for all positive real numbers x and y.
Second Proof. We can reverse the steps in the argument given in the first proof
and give a direct proof of the the result. Suppose that x and y are positive real
√
√
numbers. Then xy > 0, and 2 xy > 0. Thus
√
x + y ≤ x + 2 xy + y,
or, equivalently,
√
√
√
( x + y)2 ≤ ( x + y)2 .
√
Since the function g(z) = z is increasing on the positive reals, we can take
square roots of both sides of this inequality, and get
√
√
√
x + y ≤ x + y.
Note. Since
√
xy > 0, this proof actually shows that
√
√
√
x + y < x + y.
3