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Formal Methods Key to Homework Assignment 3, Part 2
• Prove that multiplication of rational numbers is
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
From the definition of equality of rational numbers , we need to see that
mpsu = nqrt.
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.
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
Since mq + np is an integer and 2nq is a nonzero integer,
we see that r is a rational
To check that a < r < b, we can check that the differences r − a and b − r are both
positive. We have that
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:
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
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
96. (a) Find positive real numbers x and y such that .
If x = y = 1, then x and y are positive, and but .
Since 2 is rational and is irrational, .
(b) Prove that if x and y are positive real numbers, then .
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
Since the function f(z) = z2 is increasing on the positive reals, we know that if
0 < a < b then 0 < a2 < b2. So
Simplifying , gives
and subtracting x + y from both sides gives
but since x and y are positive reals,
. So ,
which is a contradiction.
So the assumption that must be false, and we have
that , 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 , and . Thus
Since the function is
increasing on the positive reals, we can take
square roots of both sides of this inequality, and get
Note. Since , this proof actually shows that