Hands-On Exercise 3.3.2 Letxand y be two real numbers. Prove that ifx6= 0 andy6= 0, thenxy6= 0.
4
Another indirect proof is theproof by contradiction. To prove thatp⇒q, we proceed as follows:
1. Supposep⇒qis false; that is, assume thatpis true andqis false. Proof by contradiction. 2. Argue until we obtain a contradiction, which could be any result that we know is false.
How does this prove thatp⇒q? Assuming that the logic used in every step in the argument is correct, yet we still end up with a contradiction, then the only possible flaw must come from the supposition thatp⇒q is false. Consequently,p⇒q must be true.
This is what a typical proof by contradiction may look like:
Proof: Supposep⇒qis false. Thenpis true andqis false. Then . . .
.. .
. . . , which is a contradiction. Therefore,p⇒qmust be true.
There is a more general form for proving a statementr, which needs not be an implication. To prove the propositionrby contradiction, we follow these steps:
1. Supposeris false. Another form of proof
by contradiction. 2. Argue until we obtain a contradiction.
Proof: Supposeris false. Then . . . ..
.
. . . , which is a contradiction. Therefore,r must be true.
Example 3.3.4 Show that ifx3−7x2+x−7 = 0, thenx= 7.
Solution: Assume x3−7x2+x−7 = 0, we want to show that x= 7. Suppose x6= 7, then
x−76= 0, and
0 =x3−7x2+x−7 =x2(x−7) + (x−7) = (x2+ 1)(x−7)
would have implied thatx2+ 1 = 0, which is impossible. Therefore, we must havex= 7.
N Example 3.3.5 Show that if P is a point not on a line L, then there exists exactly one perpendicular line fromP ontoL.
Solution: Suppose we can find more than one perpendicular line fromP ontoL. Pick any two of them, and denote their intersections with L as Q and R. Then we have a triangle P QR, where the anglesP QRandP RQare both 90◦. This implies that the sum of the interior angles
54 Chapter 3 Proof Techniques
of the triangleP QRexceeds 180◦, which is impossible. Hence, there is only one perpendicular
line fromP ontoL. N
Example 3.3.6 Show that ifx2<5, then|x|<√5.
Solution: Assume x2<5, we want to show that|x|<√5. Suppose, on the contrary, we have |x| ≥√5. Then either x≥√5, orx≤ −√5. If x≥√5, then x2 ≥5. If x≤ −√5, we again
havex2≥5. In either case, we have a contradiction. Hence|x|<√5. N Hands-On Exercise 3.3.3 Prove that ifx2≥49, then|x| ≥7.
4
Example 3.3.7 Prove that the logical formula [(p⇒q)∧p]⇒q is a tautology.
Solution: Suppose [(p⇒q)∧p]⇒qis false for some statements pandq. Then we find
• (p⇒q)∧pis true, and
• q is false.
For the conjunction (p⇒q)∧pto be true, we need
• p⇒qto be true, and
• pto be true.
Havingptrue andqfalse would makep⇒qfalse. This directly contradicts what we have found. Therefore, the logical formula [(p⇒q)∧p]⇒qis always true, hence it is a tautology. N Example 3.3.8 Prove, by contradiction, that ifxis rational andy is irrational, thenx+y is irrational.
Solution: Letxbe a rational number andy an irrational number. We want to show thatx+y is irrational. Suppose, on the contrary, thatx+y is rational. Then
x+y=m n
for some integersmandn, wheren6= 0. Sincexis rational, we also have x= p
q for some integerspandq, where q6= 0. It follows that
m n =x+y= p q +y. Hence, y= m n − p q = mq−np nq ,
3.3 Indirect Proofs 55
wheremq−npandnqare both integers, withnq6= 0. This makesyrational, which contradicts the assumption thaty is irrational. Thus,x+y cannot be rational, it must be irrational. N
Hands-On Exercise 3.3.4 Prove that
√
x+y6=√x+√y for any positive real numbersxandy.
Hint. The words “for any” suggest this is a universal quantification. Be sure you negate the problem statement properly.
4
Example 3.3.9 Prove that√2 is irrational.
Solution: Suppose, on the contrary,√2 is rational. Then we can write
√
2 = m n
for some positive integersmandnsuch thatmandndo not share any common divisor except 1 (hence m
n is in its simplest term). Squaring both sides and cross-multiplying yields 2n2=m2.
Thus, 2 dividesm2. Consequently, 2 must also dividem. Then we can writem= 2sfor some
integers. The equation above becomes
2n2=m2= (2s)2= 4s2. Hence,
n2= 2s2,
which implies that 2 dividesn2; thus, 2 also dividesn. We have proved that both mandnare divisible by 2. This contradicts the assumption thatmandndo not share any common divisor.
Therefore,√2 must be irrational. N
Hands-On Exercise 3.3.5 Prove that√3 is irrational.
56 Chapter 3 Proof Techniques
Very often, a proof by contradiction can be rephrased into a proof by contrapositive or even a direct proof, both of which are easier to follow. If this is the case, rewrite the proof.
Example 3.3.10 Show that x2+ 4x+ 6 = 0 has no real solution. In symbols, show that
@x∈R(x2+ 4x+ 6 = 0).
Solution: Consider the following proof by contradiction:
Suppose there exists a real numberxsuch thatx2+ 4x+ 6 = 0. Using calculus,
it can be shown that the functionf(x) =x2+ 4x+ 6 has an absolute minimum
atx=−2. Thus,f(x)≥f(−2) = 2 for anyx. This contradicts the assumption that there exists anxsuch thatx2+ 4x+ 6 = 0. Thus,x2+ 4x+ 6 = 0 has no real solution.
A close inspection reveals that we do not really need a proof by contradiction. The crux of the proof is the fact thatx2+ 4x+ 6≥2 for allx. This already shows thatx2+ 4x+ 6 could never
be zero. It is easier to use a direct proof, as follows.
Using calculus, we find that the functionf(x) =x2+ 4x+ 6 has an absolute
minimum atx=−2. Therefore, for any x, we always havef(x)≥f(−2) = 2. Hence, there does not exist anyxsuch thatx2+ 4x+ 6 = 0.
Do you agree that the second proof (the direct proof) is more elegant? N Recall that a biconditional statement p⇔q consists of two implications p⇒qand q⇒p. Hence, to provep⇔q, we need to establish these two “directions” separately.
Example 3.3.11 Letnbe an integer. Prove thatn2is even if and only if nis even.
Solution: (⇒) We first prove that if n2 is even, then n must be even. We shall prove its
contrapositive: ifnis odd, thenn2 is odd. If nis odd, then we can write n= 2t+ 1 for some
integert. Then
n2= (2t+ 1) = 4t2+ 4t+ 1 = 2(2t2+ 2t) + 1, where 2t2+ 2tis an integer. Thus,n2 is odd.
(⇐) Next, we prove that if nis even, then n2 is even. Ifnis even, we can write n= 2t for some integert. Then
n2= (2t)2= 4t2= 2·2t2,
where 2t2 is an integer. Hence,n2 is even, which completes the proof.
N Hands-On Exercise 3.3.6 Letnbe an integer. Prove thatnis odd if and only ifn2 is odd.
3.3 Indirect Proofs 57
Summary and Review
• We can use indirect proofs to prove an implication.
• There are two kinds of indirect proofs: proof by contrapositive and proof by contradiction.
• In a proof by contrapositive, we actually use a direct proof to prove the contrapositive of the original implication.
• In a proof by contradiction, we start with the supposition that the implication is false, and use this assumption to derive a contradiction. This would prove that the implication must be true.
• A proof by contradiction can also be used to prove a statement that is not of the form of an implication. We start with the supposition that the statement is false, and use this assumption to derive a contradiction. This would prove that the statement must be true.
• Sometimes a proof by contradiction can be rewritten as a proof by contrapositive or even a direct proof. If this is true, rewrite the proof.
Exercises 3.3
1. Letnbe an integer. Prove that ifn2is even, then nmust be even. Use
(a) A proof by contrapositive. (b) A proof by contradiction.
Remark: The two proofs are very similar, but the wording is slightly different, so be sure you present your proofs clearly.
2. Letnbe an integer. Show that ifn2 is a multiple of 3, thennmust also be a multiple of
3. Use
(a) A proof by contrapositive. (b) A proof by contradiction.
3. Letnbe an integer. Prove that ifnis even, thenn2= 4sfor some integers.
4. Letmandnbe integers. Show that mn= 1 implies thatm= 1 orm=−1.
5. Letxbe a real number. Prove by contrapositive: ifxis irrational, then √xis irrational. Apply this result to show that √4
2 is irrational, using the assumption that√2 is irrational. 6. Letxandybe real numbers such thatx6= 0. Prove that ifxis rational, andyis irrational,
thenxy is irrational. 7. Prove that√5 is irrational. 8. Prove that √32 is irrational.
9. Letaandbbe real numbers. Show that if a6=b, thena2+b26= 2ab. 10. Use contradiction to prove that, for all integersk≥1,
2√k+ 1 +√ 1
k+ 1 ≥2
√
k+ 2.
11. Letmandnbe integers. Show that mnis even if and only ifmis even ornis even. 12. Letxandy be real numbers. Show thatx2+y2= 0 if and only if x= 0 and y= 0.
13. Prove that, ifxis a real number such that 0< x <1, thenx(1−x)≤ 1 4.
14. Let m and n be positive integers such that 3 divides mn. Show that 3 divides m, or 3 divides n.
58 Chapter 3 Proof Techniques
15. Prove that the logical formula
(p⇒q)∨(p⇒q) is a tautology.
16. Prove that the logical formula
[(p⇒q)∧(p⇒q)]⇒p is a tautology.