Math 3200: Introduction to Higher Mathematics
Spring 2014, Section 26-761
In-Depth Examples for Week 11
Intersection and Union of Equivalence Relations
Relations, like any other sets, can be combined using the main set theory operations.
Let’s say we have two relations R1 and R2 , both on a set A. Thus, R1 and R2 consist of
pairs from A × A. As a result, R1 ∪ R2 , R1 ∩ R2 , and R1 − R2 also consist of pairs from
A × A, so they are also relations on A!
Let’s look at intersection and union for now. R1 ∩ R2 features pairs that are in both
relations, so x (R1 ∩ R2 ) y means that x R1 y and x R2 y. Similarly, x (R1 ∪ R2 ) y means
x R1 y or x R2 y. Our claim is the following: if R1 and R2 are equivalence relations on A,
then R1 ∩ R2 is also an equivalence relation, but R1 ∪ R2 might not be transitive.
Brainstorming:
First, consider the proof using R1 ∩ R2 . For convenience, we’ll say R = R1 ∩ R2 so
that we don’t have to keep writing the ∩ symbol each time. Remember that x R y means
both x R1 y and also x R2 y. You can treat those two statements separately, and then our
argument should feel just like two very similar proofs done side by side!
Next, what about R1 ∪ R2 ? Reflexivity and symmetry don’t take much work; at worst,
there’s a small proof by cases. However, for transitivity, suppose x (R1 ∪ R2 ) y (R1 ∪ R2 ) z.
This means that either x R1 y or x R2 y (or both, since “or” is inclusive), and similarly we
know y R1 z or y R2 z. The main issue is that the (x, y) pair and the (y, z) pair might not
come from the same relation!
Solution:
Proof that R1 ∩ R2 is an equivalence relation: For convenience, let R = R1 ∩ R2 . Therefore,
for any x, y ∈ A, x R y means (x, y) ∈ R = R1 ∩ R2 , so that x R1 y and also x R2 y. Now,
we assume R1 and R2 are equivalence relations.
First, we prove R is reflexive. Let x ∈ A. Because R1 and R2 are reflexive, we know
x R1 x and x R2 x. Since (x, x) is in both R1 and R2 , x R x.
Next, we prove R is symmetric. Let x, y ∈ A, and assume x R y. This means that x R1 y
and x R2 y. Since R1 and R2 are both symmetric, y R1 x and y R2 x; these together imply
y R x.
Last, we prove R is transitive. Let x, y, z ∈ A, and assume x R y and y R z. Therefore,
x R1 y R1 z and also x R2 y R2 z. By transitivity of R1 and of R2 , this means x R1 z and
x R2 z, so x R z.
Analysis of R1 ∪ R2 : First, R1 ∪ R2 is reflexive. For any x ∈ A, (x, x) is in R1 ∪ R2 since it
is in both R1 and R2 .
Next, to show symmetry, let x, y ∈ A, and suppose (x, y) ∈ R1 ∪ R2 . We want to show
(y, x) ∈ R1 ∪ R2 . We’ve assumed (x, y) is in R1 or R2 . In the first case, if (x, y) ∈ R1 , then
(y, x) ∈ R1 by the symmetry of R1 . Similarly, (x, y) ∈ R2 implies (y, x) ∈ R2 . Either way,
we get (y, x) ∈ R1 ∪ R2 .
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(NOTE: Many people would probably just write “WLOG assume (x, y) ∈ R1 , since the
other case is the same with R2 replacing R1 .”)
However, R1 ∪ R2 might not be transitive. For a counterexample, consider R1 to be
congruence modulo 2 and R2 to be congruence modulo 3, where A = Z. (Thus, R1 is ≡2
and R2 is ≡3 .) We see that 0 ≡2 2 and 2 ≡3 −1, so that 0 (R1 ∪ R2 ) 2 and 2 (R1 ∪ R2 ) − 1.
However, (0, −1) 6∈ R1 ∪ R2 since 0 is not congruent to −1 modulo either 2 or 3.
“Casting out 9’s”: A Nice Gimmick Using Mods
There’s a very well-known parlor trick called “casting out 9’s” which is sometimes used
to check calculations. The essence of the trick is the following result:
Theorem. Let n be a natural number with digit sequence dk dk−1 . . . d0 in base 10 (so d0 is
the ones digit, d1 is the tens digit, and so forth). Then
n≡
k
X
di = d0 + d1 + · · · + dk
(mod 9)
i=0
In other words, to find n’s remainder modulo 9, add up n’s digits and reduce that instead!
As a demonstration, let’s show how this works with a couple examples.
• If n = 4301, then n ≡ 4 + 3 + 0 + 1 = 8 (mod 9), so n has remainder 8 modulo 9.
• If n = 7653987, then n ≡ 7 + 6 + 5 + 3 + 9 + 8 + 7 (mod 9), which equals 45. Thus,
7652987 ≡ 45. You can even repeat the trick again with 45: 45 ≡ 4 + 5 = 9 ≡ 0
(mod 9). Thus, n has remainder 0, so 9 | n.
The actual value of the sum isn’t what matters; its remainder modulo 9 is what matters. Thus, any digits of 9 can be removed from the sum without affecting that remainder,
effectively “casting them out”. Any sum totaling 9 can also be taken out! For instance, in
our second example, you can remove the digit of 9 and also cast out 6 and 3 together (since
their sum is 9)... that will give you a digit sum of 27 instead of 45.
Brainstorming:
First of all, we need to figure out how base-10 digits matter. This requires us to go back
to the very definition of base-10 writing: each digit is multiplied by a power of 10. The 0th
digit (i.e. the ones digit at the rightmost end) is multiplied by 100 . The next digit, the tens,
is multiplied by 101 , and so on. For instance,
321 really means 3 · 102 + 2 · 101 + 1 · 100
Therefore, if our digit sequence is dk dk−1 · · · d0 , then
n=
k
X
di · (10)i ≡ d0 + d1 · 10 + d2 · 102 + · · · + dk · 10k
i=0
We’re trying to simplify modulo 9. It turns out 10 simplifies really well: 10 ≡ 1 (mod 9)!
Let’s use that.
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Solution:
Let n ∈ N have digit sequence dk dk−1 . . . d0 in base 10. By the definition of base 10
(decimal numbers), this means
k
X
n=
di · 10i
i=0
Since 10 ≡ 1 (mod 9), we know that 10i ≡ 1i = 1 (mod 9) for each i from 0 to k. This gives
us
k
k
X
X
i
di · 10 ≡
di · 1 (mod 9)
i=0
i=0
which is what we wanted. Remarks upon the finished solution:
People sometimes use this technique to check calculations by working out the remainders
of their terms or factors modulo 9. For instance, let’s say I do some arithmetic and I compute
256 · 123 = 31478
Let’s cast out 9’s. 256 ≡ 2 + 5 + 6 = 13 ≡ 1 + 3 = 4 (mod 9), so 256 has remainder
4. Similarly, 123 ≡ 1 + 2 + 3 = 6 (mod 9), so 123 has remainder 6. Their product is
4 · 6 = 24 ≡ 6 (mod 9), so if I did my calculation correctly, the right side should have
remainder 6. However, I get 31478 ≡ 3 + 1 + 4 + 7 + 8 ≡ 23 ≡ 5 (mod 9), so there’s a
remainder of 5. Thus, my calculation is incorrect! (The correct answer is 31488, actually.)
As far as the proof technique goes, there’s a lot that can be generalized. Really, there
were only two key facts needed: base 10 was used, and 10 ≡ 1 (mod 9). Any time your base
reduces to 1, this same trick can be used! For example, if you’re writing numbers in base 8
(which happens in computer science occasionally), then you can reduce numbers easily mod
7, since 8 ≡ 1 (mod 7). In this case,
dk dk−1 . . . d0 (in base 8) ≡ d0 + d1 + · · · + dk
(mod 7)
For another variant, there’s a “rule of 11”. Note that 10 ≡ −1 (mod 11). When you look
at the proof for casting out 9’s, you can replace every 10i with (−1)i . The powers of −1
alternate between being 1 and −1, which gives you
dk dk−1 . . . d0 ≡ d0 + d1 (−1)1 + d2 (−1)2 + · · · = d0 − d1 + d2 − · · ·
(mod 11)
In other words, to get the remainder mod 11, alternately add and subtract the digits!
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