19.5 Consecutive integers cannot both be even.
Let x and x+1 be consecutive integers.
Suppose (for the sake of contradiction) that x and x+1 are both even.
Then
 a
 
, x

2 a and
 b
 
, x

1

2 b
Then
2 a
 x
2 b
2 ( a
 b )



2 a
1
1

2 b

1 a
 b

1
. But a
 b must be an integer by closure (contradiction).
2
Hence, x and x+1 cannot both be even.
19.7 If the sum of two primes is prime, then one of them must be 2.
Let x,y be two primes whose sum is prime. Suppose (for the sake of contradiction) that neither x nor y is
2. Since 2 is the only prime that is not odd, and neither x nor y is 2, both x and y must be odd.
Since x is odd,
 a
 
, x

2 a

1
Since y is odd,
Then x
 y


2 a b


1

,

2 b y


1

2 b
2
 a
1

2 b

2

2 ( a
 b

1 ) . Since a+b+1 is an integer, that means that 2 divides x + y, so x + y is even, and clearly x + y is not 2, so x + y is not prime (contradiction).
Hence, either x or y is 2.
19.8 Prove that ( A

B )

( B

A )
 

B )

( B

A )
 
Let A, B be sets. Suppose (for the sake of contradiction) that (
Then
 x

( A

B )

( B

A )
A
Then
So x x

Hence,

( A

B ) and x

( B

A )
A
( A and

B ) x


B
( B and

A ) x

 
B and x

A (contradiction).
19.9 Prove A

B
  iff ( A

B )

( B

A )
 
We can prove each direction using the contrapositive.
Spse ( A

B )

( B

A )
 
Then

( x , y )

So (
So x x ,
 y )
A

( A and
Then x

A
 y and
( A

B )
 x
B

B ) and

( B
( x , and
B , x so y )



B
A )
( B and x

A


B y
A )

.
A
Hence A

B
 
In the other direction:
Spse A

B
 
Then
 x

A

B
Then ( x , x )

A

B and ( x , x )
So ( x ,
Hence x )
( A


( A
B )


B )
( B


( B
A )


A )


B

A .

19.11 a. x

A is the smallest elt of A if
 y

A , x
 y b. E

{ x
 
: 2 | x }
Prove that E has no smallest elt.
Proof by contradiction:
Suppose (for the sake of contradiction) that E has a smallest elt., say a.
Then a

E , a
 y for all y in E.
Hence, 2 divides a, so
 b
 
, s .
t .
2 b
 a
Let c
Then
 a c


2 b
2

2

2 ( b

1 ) , so 2 divides c, so c is in E. But clearly c
 a (contradiction).
Hence, E has no smallest elt. c. Prove: If A
 
has a smallest elt., then it is unique.
Proof by contradiction:
Assume that A
 
has a smallest elt., say a.
Suppose (for the sake of contradiction) that a is not unique. So there is another smallest elt. For A, say b, where b
 a .
Since a is the smallest elt in A and since b is an elt of A,
Since b is the smallest elt in A and since a is an elt of A,
Hence, a
 b (contradiction). a
 b
 b a
So the smallest elt. is unique
.
.