Direct Proof
Ex. Give a direct proof of the following:
For every integer n, if n is odd then n2 is odd.
Proof:
Let n be an odd integer.
Then n = 2k + 1 for some integer k.
Then n2 = (2k + 1)2
= 4k2 + 4k + 1
= 2(2k2 + 2k) + 1
Therefore, n2 is odd.
Ex. Give a direct proof of the following:
If r and s are rational numbers then r + s is a rational number.
Proof:
Let r and s be rational numbers.
Then r = a⁄b and s = c⁄d for some integers a, b, c, d with b ≠ 0 and d ≠ 0.
We then have
r  s  ba  dc
ad
cb
 bd
 db
 adbd cb
Note that ad+cb is an integer and bd is a nonzero integer.
Therefore r + s is a rational number.
Indirect Proof (proof by contraposition)
Ex. Give an indirect proof of the following:
For every integer n, if n2 is odd then n is odd.
Proof:
We will show ( n2 is odd n is odd ) indirectly by showing ( n is not odd
We want to prove that if n is even, then n2 is even.
Let n be an even integer.
Then n = 2k for some integer k.
Then n2 = 4k2
= 2(2k2)
2
Thus n is an even integer.
n2 is not odd ).
Proof by Contradiction
Ex. Give a proof by contradiction of the following:
2 is irrational.
Proof:
Suppose that 2 is not irrational (suppose it is rational).
Then 2  ba for some integers a and b.
Without loss of generality we can assume that ba is in reduced form, that is we can assume that a and b
share no common divisor.
Square both sides to obtain 2 
a2
b2
.
Then 2b2 = a2.
We now can see that 2 divides a2.
This implies that 2 divides a.
This implies that 4 divides a2.
So, we can write a2 as 4q for some integer q.
Our equation 2b2 = a2 now becomes 2b2 = 4q.
Thus b2 = 2q.
So, 2 divides b2.
Thus 2 divides b.
This is a contradiction. We assumed that 2  ba where a and b share no common divisor, yet we have
arrived at the fact that a and b must both be divisible by 2.
Therefore our assumption that
is irrational.
2
a
b
cannot be correct.
2 cannot be a rational number. Hence
2
Proving “if and only if” statements
Ex. Prove that r is a rational number if and only if 2r is a rational number.
Proof:
( ) First we shall show that if r is a rational number then 2r is a rational number.
Let r be a rational number.
Write r as a⁄b with a and b integers, b ≠ 0.
Then 2r = 2a⁄b . Since 2a and b are integers and b ≠ 0, we see that 2r is a rational number.
( ) Next we shall show that if 2r is a rational number then r is a rational number.
Let 2r be a rational number.
Write 2r as a⁄b with a and b integers, b ≠ 0.
Then r = a⁄2b . Since a and 2b are integers and 2b ≠ 0, we see that r is a rational number.
We have now proved that r is rational iff 2r is rational.
Proving Statements are Equivalent
Ex. Show that the following are equivalent:
p1 : n is an even integer
p2 : n + 1 is an odd integer
p3 : n2 is an even integer
Proof:
p1 p2
Suppose n is even. Then n = 2k for some integer k.
Then n + 1 = 2k + 1.
Thus n + 1 is odd.
p2
p3
Suppose n + 1 is odd. Then n + 1 = 2k + 1 for some integer k.
Then n = 2k.
Thus n2 = 4k2 = 2(2k2).
Therefore n2 is even.
p3
p1
We will show that n2 is even n is even by an indirect proof (n is odd
Let n be an odd integer. Then n = 2k + 1 for some integer k.
Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1.
Thus, n2 is odd.
We have now verified that these three propositions are equivalent.
n2 is odd).
Proof by Cases
Ex. Prove that the square of an integer ends with 0, 1, 4, 5, 6, or 9.
Proof:
Let n be an integer which ends in b. We can write n as follows: n = 10a + b.
n2 = (10a + b)2 = 100a2 + 20ab + b2 = 10(10a2 + 2ab) + b2.
The last digit in the decimal expansion of n2 is completely determined by b2. We need to examine b2 for
each possible value of b.
If b = 0 then b2 = 0.
In this case the last digit of the decimal expansion of n2 is 0.
If b = 1 or b = 9 then b2 = 1 or b2 = 81.
In either case the last digit of the decimal expansion of n2 is 1.
If b = 2 or b = 8 then b2 = 4 or b2 = 64.
In either case the last digit of the decimal expansion of n2 is 4.
If b = 3 or b = 7 then b2 = 9 or b2 = 49.
In either case the last digit of the decimal expansion of n2 is 9.
If b = 4 or b = 6 then b2 = 16 or b2 = 36.
In either case the last digit of the decimal expansion of n2 is 6.
If b = 5 then b2 = 25.
In this case the last digit of the decimal expansion of n2 is 25.
Thus, the last digit of n2 must be either 0, 1, 4, 5, 6, or 9.
Existence Proofs
Ex. Use a constructive proof to show that there exists irrational numbers x and y such that x + y is rational.
Proof:
Consider the irrational numbers 2 and  2 .
Note that 2   2  0 , and 0 is a rational number.
Ex. Use a nonconstructive proof to show that there exists irrational numbers x and y such that xy is rational.
Proof:
Consider
If
2
2
2 , which has been shown to be irrational.
happens to be rational then we are done.
Suppose not, suppose
 2


2



2

 2
2
2
2
is irrational. Then consider  2

 2 , so  2

2



2
is rational.
2
2
 .


Ex. Prove that every odd integer is the difference of two perfect squares.
Let n be an odd integer. Then n = 2k + 1 for some integer k.
Note that (k + 1)2 – k2 = k2 + 2k + 1 – k2
= 2k + 1.