14.1
a. is an equiv. rel.
b. is not an equiv. rel.(fails refl., symm, and trans)
c. is not an equiv. rel. - divides is not symmetric, since 1 divides 2 but 2 does not divide 1.
d. is not an equiv. rel. – not symmetric since 1 is less than or equal to 2 but 2 is not ltoet 1.
e. is an equiv. rel.
f. is not and equiv. rel. since (4,4) is not in R (not reflexive).
g. Depends on what we mean by “rearranging its letters” – we will assume they mean that STOP
is also a rearrangement of the letters of the word STOP; then it is reflexive. Since it is also symm
and trans, it is an equiv. rel.
14.2 Prove that if x and y are both odd, then x  y (mod 2)
Let x and y both be odd. Then  a  , x  2a  1 and  b  , y  2b  1
Then x  y  2a  1  (2b  1)  2a  2b  2(a  b) Hence, 2 divides x-y, so
x  y (mod 2)
A similar argument can be used to show the second part.
14.3 Prove that if a is an integer, then a  a (mod 2)
Let a be an integer. Then a  (a)  a  a  2a so 2 divides a  (a) and hence,
a  a (mod 2)
14.4 Prove that congruence modulo n is transitive.
Let x, y, z   such that x  y (mod n) and y  z (mod n)
WTS: x  z (mod n)
Since x  y (mod n) , n divides x-y, so there is an integer a such that x  y  n a
Since y  z (mod n) , n divides y-z, so there is an integer b such that y  z  n b
But then if we define c=a+b, where c is an integer by closure,
x  z  ( x  y )  ( y  z )  na  nb  n(a  b)  nc
Hence, n divides x-z, so x  z (mod n)
14.5
a. [1]  {1,2}
b. [ 4]  {4}
c. [123]  {120,121,122,123,...,129}
d. [you] = the set of all people who have the same parents as you.
e. [you] = the set of all people who have the same birthday as you.
f. [{1,3}]  {{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}}
14.6 Prove that [1]=[3] for the equiv. rel. that is congruence mod 2.
Proposition 14.12 says that A  B    [a]  [b]
Since 3  [1] (because 3-1=2 is div by 2) and 3  [3] , then [1]  [3]  
So by prop. 14.12, [1]  [3]
14.7 To prove that
[a]  A we want to show that each set is a subset of the other.
aA
Let R be an equiv. rel. on a set A.
Let x 
[a] (WTS: that x is in A)
aA
Then, for at least one a in A, x  [a ] , so by definition, x  A .
In the other direction, let x  A (WTS: x 
[a] )
aA
Since R is an equiv. rel., R is reflexive, so xRx, and therefore x  [x ]
Hence, x 
[a]
aA
14.8 Let R be an equiv. rel. on set A and let a and b be elts. of A.
To prove that a  [b]  b  [a] , we need to prove both directions.
 Let a  [b] (WTS: b  [a ] )
Since a  [b] , by definition, a is in A and aRb.
Since R is an equiv. rel., it is symmetric, so bRa. Hence, b  [a ]
 this direction is identical to the other except we switch the a’s and b’s.
14.10 Let R and S be equiv. rel. on a set A. To prove an iff statement, we need to go both ways.
Let R=S. (WTS: the equiv. classes of R are the same as the equiv. classes of S)
Let a  A Then [a]R  {x  A; xRa}
So x  [a]R , ( x, a)  R. Since R=S, ( x, a)  S , so all such x’s are in [a] S also.
Hence, [a] R  [a] S . Likewise in the other direction, we can show that [a] S  [a] R ,
and hence, [a] R  [a] S , so the equiv. classes of R are the same as the equiv. classes of S.
In the other direction, spse the equiv classes of R are the same as the equiv. classes of S.
(WTS: R=S, by showing that each is a subset of the other)
Let ( x, y )  R . Then xRy so x  [ y] R .
Since R and S have the same equiv. classes, x  [ y ] S also, so ( x, y )  S . Hence R  S
Likewise, we can show that S  R . So R=S.