Proofs.
Indirect Proofs
Math 2326
Fall 2008
9ften, it is very difficult to give a direct approach to ! x (P(x) " C(x))D
the connection between P!x) and C!x) might not be suitable to this approach.
For example, in class, we prove that the sIuare of an even number is even. It
turns out that the converse of this is also true. If n is an integer and n2 is even,
then n is even. Cuantified, ! n !n2 is even " n is even). If we were to taJe a
direct approach, we would start.
Suppose than n2 is even. Lhen there is an integer J for which n2 # 2J.
At this point, we would be stucJ. It is hard to get from n2 # 2J to n # 2N.
Any proof techniIue other than a direct proof is often called an indirect
proof. Lhe two most common indirect methods are called Proof by
Contraposition and Proof by Contradiction.
p
I
p"I
L
L
L
L
F
F
F
L
L
F
F
L
In the direct proof, the approach was to show the second
line in the Lruth table does not occur by showing that
I has to be true when p is true. In a proof by
contraposition, we show that the second line does not
occur by showing that when I is false, p is also false.
Lhat is, in a direct proof, we assume p is true and show that we have to be in the
top line of the truth table. For a proof by contraposition, we assume I is false and
show that we have to be in the bottom line. Alternatively !and betterO) a proof
by contraposition is a direct proof of the contrapositive. Lhat is,
! x (P(x) " C(x)) # ! x (¬C(x) " ¬P(x)).
We start by assuming C is false, and directly show that P is false. LetRs taJe our
example above. If n2 is even, then n is even. Lhe contrapositive of this is SWhen
n is odd, n2 is also odd.T We essentially proved this already in the last set of notes
page 2
!as a corollary to proving that the product of odd numbers is odd) but letRs do it
again. Lhe proof would go.
Let n be odd. Lhen for some integer J, n # 2J % 1. Vow n2 # !2J % 1)2
# 4J2 % 4J % 1 # 2!2J2 % 2J) % 1, so n2 is also odd. Xy contraposition,
when n2 is even, n must be even.
Here is an example from the !extra credit) from the last exam.
Suppose that {u, v, w} is a linearly independent set of vectors. Prove
that {u, w} is also a linearly independent set.
Here is the solution I gave. LhinJ of this as a p " I statement, where p is
the proposition that Zu, v, w[ is independent and I is the proposition
Zu, w[ independent. We said in class that p " I # ¬I " ¬p. I will show
that instead. Here, we can assume Zu, w[ is linearly dependent, so for
some x, y, not both \ero, xu ] yw = 0. With this, xu ] 0v ] yw = 0 is a
dependence relation for Zu, v, w[, so Zu, v, w[ is dependent as well.
As you can see, this is exactly a direct proof of the contrapositive.
Another example. Prove the following. If x % y > 100 then x > 50 or y >
50. In Iuantified form, ! x ! y !x % y > 100 " !x > 50 $ y > 50) ). Lhis can
probably be done directly, but it is not straightforward. It is almost
embarrassingly easy by contraposition, provided the details are done right. First,
the contrapositive of x % y > 100 " !x > 50 $ y > 50) is
¬!x > 50 $ y > 50) " ¬!x % y > 100).
Lo use this we apply De MorganRs law and carefully negate the ineIualities.
!x ! 50 $ y ! 50) " !x % y ! 100).
Lhe hard part is done. Vow the proof. Assuming x ! 50 and y ! 50, we have
x % y ! 50 % 50 # 100. Lhat is, x % y ! 100.
A complication. bonsider the problem. Prove that the sum of a rational
number and an irrational number is irrational. Here, irrational means not rational.
In Iuantified form, ! x ! y !!x is rational $ y is irrational) " !x % y is irrational)
Lhe contrapositive is not much better. If x % y is rational, then x is irrational or y
page 3
is rational. Lhere is a tricJ for this Jind of problem. Some hypotheses can be
SsucJedT into the universe of discourse. I claim
! x ! y !!x is rational $ y is irrational) " !x % y is irrational)
is logically eIuivalent to
! r ! y !y is irrational " r % y is irrational),
where the universe of discourse for r is the set of rational numbers and the
universe for y is the set of all real numbers. If you believe this, the contrapositive
claim is ! r ! y !r % y is rational " y is rational). Here is how the proof would
go.
m
Since r is rational, for some integers m, n with n " 0, r # . Since r % y
n
p
is rational, there are integers p, I with I " 0 and r % y # . Vow
I
p m pn % Im
y # !r % y) % r # %
#
, which is a Iuotient of two integers
I
n
In
with non\ero denominator so y is rational.
In general,
! x ! y !!P!x) $ C!y) $ R!x, y)) " S!x, y))
is logically eIuivalent to
! !x with P!x) true) ! !y with C!y) true) !R!x, y)) " S!x, y)).
Another example.
If x and y are positive and xy > 100 then x > 10 or y > 10.
Cuantified. ! x ! y ! !x > 0 $ y > 0 $ xy > 100) " !x > 10 & y > 10) )
Modified Iuantified version. ! x > 0 ! y > 0 !xy > 100 " !x > 10 & y > 10)),
contrapositive. ! x > 0 ! y > 0 !!x ! 10 $ y ! 10) " xy ! 100).
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In that last form, the result is easy to prove. It goes liJe this.
Let x and y be positive real numbers with x ! 10 and y ! 10. Lhen
xy ! 10&10 so xy ! 100, as desired.
Lechnically, something liJe this should be written.
Problem. Prove that if x and y are positive and xy > 100, then x > 10 or y > 10.
Solution. Suppose x and y are positive, x ! 10 and y ! 10. Lhen xy ! 10&10, or
xy ! 100. Xy way of contraposition, the original result is proved.
Finally, for this problem, note that the positivity of x and y is needed. bertainly
if x # %3 and y # %50, xy # 150 > 100, but x > 10 is false and y > 10 is false.
Lhe reason. multiplying by a negative number changes the direction of an
ineIuality. Lo Jeep this from happening, we needed x and y to be positive.
Lhere is another indirect approach, called Proof by Contradiction. Lhe
techniIue applies more generally than to Nust ! x (P(x) " C(x)). It can also be
used occasionally for the simpler form ! x P(x). Lhe idea is this. When is
a " F a true statement? Since L " F is false, we need a to be false. Said
differently, if we Jnow ¬p " F is a true statement, then we Jnow ¬p is false so p
must be true. Lhe techniIue gets its name from that fact that a proposition which
is always false is called a contradiction.
Restated, to prove ! x P(x), show that ¬P!x) implies a contradiction. For
the more complicated ! x (P(x) " C(x)), recall that
¬(P(x) " C(x)) # P!x) $ ¬C!x).
How do we get the contradiction? Usually it is of the form
!something) $ !its negation).
Since a $ ¬a # F, this does it.
As an example, taJe our previous example yet again.
If x and y are positive and xy > 100 then x > 10 or y > 10.
Cuantified. ! x ! y ! !x > 0 $ y > 0 $ xy > 100) " !x > 10 & y > 10) ).
page 5
Lhis time, we will V9L modify the universe of discourse. Instead of using
contraposition, we give a proof by contradiction. Lhe negation of
!x > 0 $ y > 0 $ xy > 100) " !x > 10 & y > 10)
is
!x > 0 $ y > 0 $ xy > 100) $ !x ! 10 $ y ! 10).
We will show this is false !a contradiction). We have done all the worJ already.
If x > 0 and y > 0 and x ! 10 and y ! 10, then xy ! 10&10, or xy ! 100. Lhis
means that xy > 100 AND xy ! 100, and this is our contradiction.
Any proof by contraposition can be done by contradiction, but it is often
considered bad form to do so. Lhe exception being cases where a proper proof
by contraposition worJs best by changing the universe of discourse !as in the last
two examples.) Lhe reason contraposition is often preferred is that a proof by
contradiction will be almost identical to the contraposition proof, but longer.
For example, we previously proved that if n2 is even, then n is even. Lhe
proof went.
If n is odd, then for some integer J, n # 2J % 1. Vow n2 # !2J % 1)2
# 4J2 % 4J % 1 # 2!2J2 % 2J) % 1, so n2 is also odd. Xy contraposition,
when n2 is even, n must be even.
Here is how the proof would go if we used a proof by contradiction instead.
Suppose n2 is even but n is odd. Since n is odd, n # 2J % 1 for some
integer J. Vow n2 # !2J % 1)2 # 4J2 % 4J % 1 # 2!2J2 % 2J) % 1, so n2 is
also odd, contradicting the assumption that n is even.
Here, we did exactly the same calculation, and showed that n2 must be odd. In
the contraposition proof, we were essentially done. In the contradiction proof, we
had to say this contradicted the original assumption that n2 was even.
In general, when proving p " I by contraposition, you prove ¬I " ¬p.
With a proof by contradiction, you prove p $ ¬I " F. If the false statement is of
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the form p $ ¬p, the contraposition proof is to be desired.
Here is the classical proof by contradiction. Lhis is one case where there is
no choice%%contrapostion Nust does not worJ.
2 is irrational.
Lhis might not looJ liJe a proposition of the form ! x (P(x) " C(x)) or even one
of the form ! x P(x), but appearances are deceiving. Remember the definition of
m
rational. x is rational if there are integers m, n with n " 0 and x # . Lo be
n
irrational means there are no such integers. We might put this in Iuantified form
(
m+
! m ! (n " 0) )) 2 " ,, .
n ,'*
')'
Lhe universe of discourse for the denominator really should exclude 0 as above.
Unfortunately, it is hard to prove 2 is irrational from here. Rather than using
the usual definition of rational, it helps greatly to use a modified version.
x is rational if.
m
there are integers m, n with n " 0, x # ,
n
AND at least one of m and n is odd.
m
Lhe reason this is true is that once we have m and n with x # , we can divide
n
66
out common factors of 2. Lhat is, 66f means
. Lhis is obviously rational
100
33
!Iuotient of two integers) and we can divide out a 2 to get 66f #
. With these
50
preliminaries, here goes the proof.
Suppose 2 is rational !that is, assume ¬p). Lhen there are integers m
m
and n with n " 0 and at least one of m and n is odd with 2 # . Multiply by
n
2
2
2
n. n 2 # m. Vow sIuare. 2n # m . We see that m is twice an integer so m2
is even. We have shown previously that if the sIuare of an integer is even, then
the integer is even. Lhat is, since m2 is even, we Jnow that m is even. Lhus,
there is an integer J so that m # 2J. Vow 2n2 # m2 # !2J)2 # 4J2. Dividing by
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2, n2 # 2J2. Lhis means that n2 is also even, and hence, that n is even. Xut then
m and n are both even contradicting the assumption that at least one of them is
odd.
Many people find the above proof unsatisfying and even unconvincing. It
certainly is not the original GreeJ proof. Lheir proof was closer to this. If
m
2 # , then as we saw above, both m and n are even. Divide out a 2 to get
n
!mh2) m1
2 #
. Xy exactly the same reasoning, m1 and n1 are also both
#
n1
!nh2)
m2
even. We can cancel out another factor of 2 to get 2 #
. Again, we are
n2
forced to conclude that m2 and n2 are even. blearly we can factor out 2Rs forever,
but this is absurd !ie a contradiction).
A final example. Vo positive sIuares !of integers) can differ by 2. Lhat is,
it is never the case that m2 % n2 # 2.
Here is the proof. Assume the result is wrong, that there are positive
integers m and n with m2 % n2 # 2. If we factor this, !m % n)!m % n) # 2. If the
product of two positive integers is 2, then the larger is 2 and the smaller is 1, so
m % n # 2 and m % n # 1. Adding, 2m # 3, so m # 32 , contradicting the
assumption that m is a positive integer.
Summary of proof techniIues. Lo prove ! x (P(x) " C(x))
Directly. Assume P!x) is true for a generic x, try to prove C!x) is also true
by bontraposition. Assume C!x) is false for x, prove that P!x) is also false
by bontradiction. Assume P!x) is true and C!x) is false, obtain a contradiction