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Chapter 5 Existence and Proof by contradiction
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5.1 Counterexamples
5.2 Proof by Contradiction
5.3* A Review of Three Proof Techniques
5.4 Existence Proofs
5.5 Disproving Existence Statements
Section 5.1 Counterexamples
Notice that some quantified statements of the type xS, R(x) are
false.
We have seen that  (x  S, R(x))   x  S,  R(x),
that is, xS, R(x) is false, then there exists some element xS for
which R(x) is false. Such an element x is called a counterexample of
the statement xS, R(x). Finding a counterexample verifies that
xS, R(x) is false.
Examples
Example: Consider the statement: If x  R, then (x2-1)2>0. Show that
this is false by exhibiting a counterexample.
Solution: For x=1, (x2-1)2=(12-1)2=0. Thus x=1 is a counterexample.
If a statement P is shown to be false in some manner, then P is said to
be disproved.
Example
Disprove the statement: For every odd positive integer n, 3 | (n2-1).
Solution: Since 3 | (32-1), it follows that n=3 is a counterexample.
Counterexample for x  S, P(x)Q(x)
We see that a quantified statement of the type
x  S, R(x) is false if  x  S,  R(x) is true.
Similarly, the quantified statement
x  S, P(x)Q(x) is false if  x  S,  (P(x)Q(x)) is true,
Where  (P(x)Q(x))  P(x)  ( Q(x)).
Thus, x  S, P(x)Q(x) is false if  x  S, (P(x)  ( Q(x))).
That is, to show that x  S, P(x)Q(x) is false, we need to exhibit a
counterexample, which is then an element x  S for which P(x) is
true and Q(x) is false.
Examples
Example
Disprove the statement: Let n  Z. If n2+3n is even, then n is odd.
Solution: If n=2, then n2+3n= 22+3(2)=10 is even and 2 is even. Thus
n=2 is a counterexample.
Example
Show that the statement: Let n  Z. If 4 | (n2-1), then 4 | (n-1) is false.
Solution: Since 4 | (32-1) but 4 | (3-1), it follows that n=3 is a
counterexample.
Section 5.2 Proof by Contradiction
Suppose that we would like to show that a certain mathematical
statement R is true. We may assume R is false statement and, from
this assumption, we are able to arrive at or deduce a statement that
contradicts some assumption we made in the proof or some known
fact.
Therefore, we have established the truth of the implication
( R) C
However, because ( R) C is true and C is false, it follows that  R is
false and so R is true, as desired. This technique is called proof by
contradiction.
Often a proof by contradiction begins with
Suppose that R is false.
or
Assume, to the contrary, that R is false.
Proof by Contradiction Cont.
If R is the statement x  S, P(x)Q(x), then a proof by contradiction
of the statement consists of verifying the implication
(x  S, P(x)Q(x))  C for some contradiction C.
Note that (x  S, P(x)Q(x))   x  S, (P(x)  ( Q(x))), it follows
that a proof by contradiction of x  S, P(x)Q(x) might begin with
Assume, to the contrary, that there exists some element x  S for
which P(x) is true and Q(x) is false.
The remainder of the proof then consists of showing that this
assumption leads to a contradition.
Example
Result: There is no smallest positive real number.
Proof: Assume, to the contrary, that there is a smallest positive real
number, say r. Since 0<r/2<r, it follows that r/2 is a positive real
number that is smaller than r. This, however, is a contradiction.
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Example
Result: If a is an even integer and b is an odd integer, then 4 | (a2+2b2).
Proof: Assume, to the contrary, that there exists an even integer a and
an odd integer b such that 4|(a2+2b2). Thus a=2x, b=2y+1, and
a2+2b2=4z for some integer x, y, and z. Hence (2x)2+2(2y+1)2=4z.
Simplifying, we obtain 4x2+8y2+8y+2=4z or, equivalently,
2=4z-4x2-8y2-8y= 4(z-x2-2y2-2y).
Since z-x2-2y2-2y is an integer, 4 | 2, which is impossible.
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Example
The following results concern irrational numbers. Recall that a real
number is rational if it can be expressed as m/n for some m, n  Z,
where n0. A real number is irrational if it is not rational.
Result: The sum of a rational number and an irrational number is
irrational.
Proof: Assume, to the contrary, that there exist a rational number x and
an irrational number y whose sum is a rational number z. Thus
x+y=z, where x=a/b and z=c/d for some integers a, b, c, d  Z and b,
d 0. This implies that
y=c/d-a/b=(bc-ad)/bd.
Since bc-ad and bd are integers and bd 0 it follows that y is
rational, which is a contradiction.
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Section 5.3 A Review of Three Proof Techniques
In order to prove the truth of a statement x  S, P(x)Q(x), we have
been introduced to three proof techniques: direct proof, proof by
contrapositive, and proof by contradiction.
Prove x  S, P(x)Q(x).
• If we use a direct proof, then
Assume that P is true, and show that Q is true.
• If we use a proof by contrapositive, then
Assume that Q is false, and show that P is false.
• If we use a proof by contradiction, then
Assume that P Q is false, and obtain a contradiction.
Section 5.4 Existence Proofs
An existence theorem concerning an open sentence R(x) over a
domain S can be expressed as a quantified statement
 x  S, R(x): There exists x  S such that R(x).
Such a statement is true provided that R(x) is true for some x  S.
A proof of existence theorem is called an existence proof.
An existence proof may consist of displaying or constructing an
example of such an object or perhaps, with the aid of known results,
verifying that such objects must exist without every producing a
single example of the desired example.
Example
Result. There exists an integer whose cube equals its square.
Proof: Since 13=12=1, the integer 1 has the desired property.
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Result. There exist real numbers a and b such that (a+b)2=a2+b2.
Proof: Let a=1 and b=0. Then
(a+b)2=(1+0)2=12=12+02=a2+b2.
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Example
Result: There exist irrational numbers a and b such that ab is rational.
Proof. Consider the number 2 2 . This is either rational or irrational. We
consider two cases.
Case 1: 2 2 is rational. Then we can take a=b= 2 , and we have the
desired result.
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Case 2: 2 2 is irrational. In this case, consider ab, where a= 2 and b= 2
. Observe that
ab= ( 2 2 ) 2  2 2 2  2 2  2,
Which is rational.
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Existence of Solutions
Recall
Intermediate Value Theorem of Calculus
If f is a function that is continuous on the closed interval [a, b] and k is a
number between f(a) and f(b), then there exists a number c  (a, b)
such that f(c)=k.
Example
Result. The equation x5+2x-5=0 has a real number solution between
x=1 and x=2.
Proof. Let f(x)=x5+2x-5. Since f is a polynomial function, it is continuous
on R and so f is continuous on [1, 2]. Now f(1)=-2 and f(2)=31. Since
0 is between f(1) and f(2), it follows by the Intermediate Value
Theorem of Calculus that there is a number c between 1 and 2 such
that f(c)= c5+2c-5=0. Hence c is a solution.
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Uniqueness
An element belonging to some prescribed set A and possessing a
certain property P is unique if it is the only element of A having
property P. Typically, to prove that only one element of A having
property P, we proceed in one of two ways:
(1) We assume that a and b are elements of A possessing property P
and show that a=b. (direct proof)
(2) We assume that a and b are distinct elements of A possessing
property P and show that a=b. (proof by contradiction)
Example
Result. The equation x5+2x-5=0 has a unique real number solution
between x=1 and x=2.
Proof. Assume, to the contrary, that the equation x5+2x-5=0 has two
distinct real number solutions a and b between x=1 and x=2. We
may assume that a<b. Since 1<a<b<2, it follows that
a5+2a-5< b5+2b-5. On the other hand, a5+2a-5= b5+2b-5=0, which
produces a contradiction.
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Section 5.5 Disproving Existence Statements
To disprove a quantified statement of the type  x  S, R(x) requires a
different approach. Note
( x  S, R(x))   x  S,  R(x),
It follows that the statement  x  S, R(x) is false if R(x) is false for
every x  S.
Result: Disprove the statement: There is a real number x such that
x6+2x4+x2+2=0.
Solution: Let x  R. Since x6 ,x4, and x2 are all even powers of the real
number x, it follows that x6 0, x4 0, and x2 0. Therefore,
x6+2x4+x2+22 and so x6+2x4+x2+20. Hence the equation
x6+2x4+x2+2=0 has no real number solution.
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