Proofs and Proving
Math 1200
• Direct proofs
(a): If x and y are rational numbers, then x + y is rational.
Proof: Since x is rational there exist integers a and b with b 6= 0 such that
x = ab . Similarly, since y is rational, there exist integers c and d with d 6= 0 such
that y = dc . Then x + y = ad+bc
which is rational as ad + bc is an integer and
bd
bd is an integer with bd 6= 0.
Is it the case that if x and y are irrational, then x + y is irrational? The
answer is no, which is understood to mean that if xand y are irrational, then
x + y is not necessarily irrational.
This proof is by counterexample.
In particular .10100100010000100000 . . . and .01011011101111011111 . . . are both
irrational but their sum, .11111111111111111111 . . . is the rational number 91 .
Note that to prove a statement is false it is enough to prove that it is false in
just one instance. To prove that a statement is true requires proof that it is
always true.
(b): If m and n are each divisible by 3, then so is m + n.
Proof: Since m is divisible by 3, m = 3k for some integer k. Since n is divisible by 3, n = 3l for some integer l. Then m + n = 3k + 3l = 3(k + l). As
k + l is an integer, m + n is divisible by 3.
Is it the case that if m and n are not divisible by 3, then m + n is not divisible by 3? Again, as it turns out, the answer is no. Neither 1 nor 2 are
divisible by 3 but their sum, 3, is.
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• Indirect Proofs including proof by contrapositive, proof by contradiction
(a): If n2 is odd then n is odd.
Proof: This is a proof by contrapositive. The contrapositive of the statement
is, “If n is even, i.e., not odd, then n2 is even.” If n is even, then n = 2k for
some integer k. Then n2 = (2k)2 = 4k 2 = 2(k 2 ). As n2 is divisible by 2, n2 is
even.
(b): If n2 is even then n is even.
Proof: The proof is similar to the previous one. The contrapositive of the
statement is, “If n is odd, then n2 is odd.” Since n is odd, n = 2k + 1 for some
integer k. Then n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 which is an
odd number.
√
(c): 2 is irrational.
Proof: This is a √
proof by contradiction. The statement is that it is impossible to represent 2 as a quotient of integers. We will assume otherwise, i.e.,
that it is possible and obtain a contradiction. Every rational number can be
represented
√ as a fraction in lowest terms. We will prove that no fraction representing 2 can be written in lowest terms. That’s √the contradiction. We
interpret it as telling us that no fraction can represent 2.
√
Assume there exist integers a and b such that 2 = ab . Assume further that a
and b are chosen so that they have no common factors, i.e., that the fraction
a
a2
2
2
2
b is in lowest terms. Then 2 = b2 from which a = 2b . But then a is even.
2
2
Then a is even. Write a = 2c for some integer c. Then a = (2c) = 4c2 = 2b2
from which b2 = 2c2 . But then, b is even. The fraction ab is not in lowest terms.
That’s the contradiction.
• Proofs by cases
(a): At a party I attended, some people shook hands. At the end, two people were
surprised to find that they had both shaken the same number of hands.
Solution: Let n be the number of people at the party. Label n + 1 boxes
0, 1, . . . , n − 1 and put a person’s name in the box labeled k provided they
shook k people’s hands. There are n names and n boxes so the pigeonhole principle cannot be directly applied. Consider two cases. The first is that there is
someone’s name in the box labeled 0. The second is that the box labeled 0 is
empty.
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In the first case, the box labeled n − 1 is empty since nobody shakes everyone’s hands. The n names are distributed among the boxes 0, 1, . . . , n − 2 and
as there are in all n − 1 boxes, one must contain at least two names.
In the second case, the box labeled 0 is empty so the n names are distributed
among the n − 1 boxes labeled 1, 2, . . . , n − 1. One box must contain at least
two names.
(b): For all integers n, n3 − n is divisible by 3.
There are three cases here. The first is n is divisible by 3. The second is
that upon division of n by 3, the remainder is 1, i.e., n = 3k + 1 for some integer
k. The third is that upon division of n by 3, the remainder is 2, i.e., n = 3k + 2
for some integer k.
In the first case, since n is divisible by 3, n = 3k for some k. Then n3 − n =
27k 3 − 3k = 3(9k 3 − k).
In the second case n3 − n = (3k + 1)3 − (3k + 1) = 27k 3 + 27k 2 + 6k =
3(9k 3 + 9k 2 + 2k).
In the third case n3 − n = (3k + 2)3 − (3k + 2) = 27k 3 + 54k 2 + 33k + 6 =
3(9k 3 + 18k 2 + 11k + 2).
• Exercise: Each of the following is intended as an argument to prove the result that,
for any positive integer n, if n2 is a multiple of 3, then n is a multiple of 3.
(a): Assume that n2 is an odd positive integer that is divisible by 3. That is n2 =
(3n + 1)2 = 9n2 + 6n + 1 = 3n(n + 2) + 1. Therefore, n2 is divisible by 3. Assume
that n2 is even and a multiple of 3. That is, n2 = (3n)2 = 9n2 = 3n(3n).
Therefore, n2 is a multiple of 3. If we factor n2 = 9n2 , we get 3n(3n); which
means that n is a multiple of 3.
(b): Suppose to the contrary that n is not a multiple of 3. We will let 3k be an
integer that is a multiple of 3, so that 3k + 1 and 3k + 2 are integers that are
not multiples of 3. Now n2 = (3k + 1)2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1. Since
3(3k 2 + 2k) is a multiple of 3, 3(3k 2 + 2k) + 1 is not. Now we will do the other
possibility, 3k + 2. So, n2 = (3k + 2)2 = 9k 2 + 12k + 4 = 3(3k 2 + 4k + 1) + 1 is
not a multiple of 3. Because n2 is not a multiple of 3, we have a contradiction.
(c): Let n be an integer such that n2 = 3x where x is an integer. Then 3 is a divisor
of n2 . Since n2 = 3x, nn = 3x. Thus, 3 is a divisor of n. Therefore if n2 is a
multiple of 3, then n is a multiple of 3.
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(d): Let n be a positive integer such that n2 is a multiple of 3. Then n = 3m where
m is some positive integer. So n2 = (3m)2 = 9m2 = 3(3m2 ). This breaks down
into 3m times 3m which shows that m is a multiple of 3.
Read these arguments critically. Here are some things to think about. Does the
argument make sense? If the argument doesn’t make sense, can you edit it (making
notational or other small changes) so that it does make sense? Does the argument
(or your minor revision) prove the result? If not, does it prove something else?
Carefully write in your own words a proof of the given result.
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