*Solution:*
We first show that the reflexive law holds.
Given an ordered pair (a,b), we have ab = ba,
and so (a,b) ~ (a,b).

We next check the symmetric law.
Given (a_{1},b_{1}) and (a_{2},b_{2})
with (a_{1},b_{1}) ~ (a_{2},b_{2}),
we have a_{1}b_{2} = a_{2}b_{1},
and so a_{2}b_{1} = a_{1}b_{2},
which shows that
(a_{2},b_{2}) ~ (a_{1},b_{1}).

Finally, we verify the transitive law.
Given (a_{1},b_{1}),
(a_{2},b_{2}), and
(a_{3},b_{3}) with
(a_{1},b_{1}) ~ (a_{2},b_{2})
and (a_{2},b_{2}) ~ (a_{3},b_{3}),
we have the equations a_{1}b_{2} = a_{2}b_{1}
and a_{2}b_{3} = a_{3}b_{2}.
If we multiply the first equation by b_{3}
and the second equation by b_{1}, we get
a_{1}b_{2}b_{3}
= a_{2}b_{1}b_{3}
= a_{3}b_{1}b_{2}.
Since b_{2} 0
we can cancel to obtain
a_{1}b_{3} = a_{3}b_{1},
showing that
(a_{1},b_{1}) ~ (a_{3},b_{3}).
(Click here for a more details.)

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*Note:*
Remember that if z = a+bi is a complex number, then ||z|| =
(a^{2}+b^{2}).

*Solution:*
The reflexive, symmetric, and transitive laws
can be easily verified since ~ is defined in terms of an equality,
and equality is itself an equivalence relation.

*Comment:*
It is also possible to prove that ~ is an equivalence relation
by observing that it is the equivalence relation that comes from
the function f : **C** -> **R** defined by
f(z) = ||z||, for all z in **C**.

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*Solution:*
The reflexive, symmetric, and transitive laws for the relation ~
really depend on an equality, and can easily be verified.
Since **u** has length 1,
**v** · **u** represents the length of the projection
of **v** onto the line determined by **u**.
Thus two vectors are equivalent if and only if
they lie in the same plane perpendicular to **u**.
It follows that the equivalence classes of ~
are the planes in **R**^{3} that are perpendicular to **u**.

*Comment:*
Instead of proving directly that ~ is an equivalence relation,
you could just note that it is the equivalence relation defined
by the function L : **R**^{3} -> **R**,
where L(**v**) = **v** · **u**,
for all **v** in **R**^{3}.

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*Solution:*
The equivalence relation determined by f
is defined by setting a ~ b if f(a) = f(b),
so a ~ b if and only if a^{2} = b^{2},
or, a ~ b if and only if |a| = |b|.

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L(x,y,z) = (x + y + z, x + y + z, x + y + z) ,

for all (x,y,z) in*Solution:*
Since (a_{1},a_{2},a_{3})
~ (b_{1},b_{2},b_{3}) if
L (a_{1},a_{2},a_{3})
= L (b_{1},b_{2},b_{3}),
it follows from the definition of L that
(a_{1},a_{2},a_{3}) ~
(b_{1},b_{2},b_{3}) if and only if
a_{1} + a_{2} + a_{3}
= b_{1} + b_{2} + b_{3}.
For example, { (x,y,z) | L(x,y,z) = (0,0,0) }
is the plane through the origin whose equation is x + y + z = 0,
with normal vector (1,1,1).
The other subsets in the partition of **R**^{3} defined by L
are planes parallel to this one.
Thus the partition consists of the planes perpendicular to the vector (1,1,1).

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*Solution:*
The formula for f is well-defined since if
[x_{1}]_{12} = [x_{2}]_{12},
then x_{1}
x_{2} (mod 12),
and so x_{1}^{2}
x_{2}^{2} (mod 12),
which shows that
f ( [x_{1}]_{12} ) = f ( [x_{2}]_{12} ).

To compute the images of f we have

[0]_{12}^{2} = [0]_{12},

[± 1]_{12}^{2} = [1]_{12},

[± 2]_{12}^{2} = [4]_{12},

[± 3]_{12}^{2} = [9]_{12},

[± 4]_{12}^{2} = [4]_{12},

[± 5]_{12}^{2} = [1]_{12}, and

[6]_{12}^{2} = [0]_{12}.

Thus f (**Z**_{12}) =
{ [0]_{12}, [1]_{12}, [4]_{12}, [9]_{12} }.

The corresponding equivalence classes determined by f are

{ [0]_{12}, [6]_{12} },

{ [± 1]_{12}, [± 5]_{12} },

{ [± 2]_{12}, [± 4]_{12} },

{ [± 3]_{12} }.

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*Solution:*
We have always have A ~ A since IAI^{-1} = A,
where I is the n × n identity matrix.

Next, suppose that A ~ B.
Then PAP^{-1} = B for some invertible matrix P,
and so we get

A = P^{-1}B(P^{-1})^{-1}.

Finally, suppose that A ~ B and B ~ C.
Then PAP^{-1} = B and QBQ^{-1} = C,
for some invertible matrices P and Q.
Substituting gives

Q(PAP^{-1})Q^{-1} = (QP)A(QP)^{-1} = C,
and therefore A ~ C.

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Let S be a set. Then **~** defines an
equivalence relation on S if and only if

**(i)**the**reflexive law**holds :

given any x in S, we always have x**~**x ;**(ii)**the**symmetric law**holds :

given any x and y in S, if x**~**y , then y**~**x ;**(iii)**the**transitive law**holds :

given any x, y and z in S, if x**~**y and y**~**z , then x**~**z .

We should first take a closer look at the way that **~**
is defined. We are given that
(x_{1},y_{1}) ~ (x_{2},y_{2})
if x_{1}y_{2} = x_{2}y_{1}.
Sometimes it helps to put everything into words:
two ordered pairs are equivalent if
the first component of the first pair
times the second component of the second pair
equals the first component of the second pair
times the second component of the first pair.
For example, to find pairs equivalent to (2,3)
we need to solve
2 · y_{2} = x_{2} · 3.
We could have (2,3) ~ (4,6) or (2,3) ~ (6,9), and so on.

Now here is the proof, in three parts.

**(i)**
We first show that the reflexive law holds.
That is, given any x in S,
we must check that x **~** x.

Since the set S consists of ordered pairs of positive integers,
let x = (a,b).
To show that (a,b) ~ (a,b),
we need to use the general condition that
(x_{1},y_{1}) ~ (x_{2},y_{2})
if x_{1}y_{2} = x_{2}y_{1}.
In this case we have
x_{1} = a,
y_{1} = b,
x_{2} = a, and
y_{2} = b,
so after we substitute these values into the general formula,
we end up having to check that ab = ba.
This equation does hold for all choices of a and b,
because multiplication of integers satisfies the commutative law.
This allows us conclude that (a,b) ~ (a,b), so x **~** x,
and we have verified that the reflexive law holds.

**(ii)**
We next check the symmetric law.
We must prove that given any x and y in S,
with x **~** y , we must also have y **~** x .
Let x = (a_{1},b_{1})
and y = (a_{2},b_{2}),
and assume that x **~** y .
In the definition of **~** we take
x_{1} = a_{1},
y_{1} = b_{1},
x_{2} = a_{2}, and
y_{2} = b_{2},
Then by the formula that defines **~** we have the condition
x_{1}y_{2} = x_{2}y_{1},
which translates into
a_{1}b_{2} = a_{2}b_{1}.

Next, we will look at exactly what it is that we must prove.
To show that y **~** x,
in the definition of **~** we need to take
x_{1} = a_{2},
y_{1} = b_{2},
x_{2} = a_{1}, and
y_{2} = b_{1},
and so we must check that
(x_{1},y_{1}) ~ (x_{2},y_{2}), or
a_{2}b_{1} = a_{1}b_{2}.

Now you can see the connection between what we are given
and what we have to prove.
We are given
a_{1}b_{2} = a_{2}b_{1}
and must prove that
a_{2}b_{1} = a_{1}b_{2}.
All we need to do is change the order,
which we can do because we are working with multiplication
of ordinary integers.
We conclude that the reflexive law holds because
(a_{2},b_{2}) ~ (a_{1},b_{1}),
and thus y **~** x.

**(iii)**
Finally, we verify the transitive law.
We can assume that we are given x = (a_{1},b_{1}),
y = (a_{2},b_{2}), and
z =(a_{3},b_{3}) with
x ~ y and y ~ z.
Then
(a_{1},b_{1}) ~ (a_{2},b_{2})
and (a_{2},b_{2}) ~ (a_{3},b_{3}),
and if we interpret the general definition of **~**
for these particular elements, we get the equations
a_{1}b_{2} = a_{2}b_{1}
and a_{2}b_{3} = a_{3}b_{2}.

What do we have to prove?
To check that x **~** z, we need
(a_{1},b_{1}) ~ (a_{3},b_{3}),
and so we must prove that
a_{1}b_{3} = a_{3}b_{1}.

If we multiply the first equation by b_{3}
and the second equation by b_{1}, we get
a_{1}b_{2}b_{3}
= a_{2}b_{1}b_{3}
and
b_{1}a_{2}b_{3}
= b_{1}a_{3}b_{2}.
By rearranging terms in the second equation,
and using the transitive property of equality,
we get
a_{1}b_{2}b_{3}
= b_{1}a_{3}b_{2}.
Since b_{2} 0
(the original assumption was that we are working with positive integers)
we can cancel to get
a_{1}b_{3} = a_{3}b_{1}.
(Note that we had to use the commutativity of multiplication
several times in this chain of reasoning.)
Finally, this shows that
(a_{1},b_{1}) ~ (a_{3},b_{3}),
so x **~** z, and the transitive law holds.

This completes the proof, since we have verified that **~**
satisfies the reflexive, symmetric, and transitive laws.

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