Download Marianthi Karavitis - Stony Brook Math Department

Notes: 10/8/07
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Marianthi Karavitis
I)
II)
III)
IV)
I)
Notes on Induction
a. Incorrect methods for proofs by induction and
review of last week’s quiz.
Equivalence relations
a. The first equivalence relations we encounter.
b. Examples of equivalence relations
i. Modular numbers
ii. Rational numbers
Questions
Notes on Induction:
Following last week’s quiz on induction, we reviewed some inconsistencies and
issues with the proofs by looking at a few examples. The first two examples listed below
are illustrations of what not to do in the base case of a proof by induction.
Case 1:
Show true for n=1
1∙2∙3=1(1+1)(1+2)(1+3)
4
6=1(2)(3)(4)=6
4
Or:
Case 2:
1(2)(3)=1(2)(3)(4)
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In both case 1 and case 2, the issue is that we are already assuming the statement we’re
trying to prove is true by setting the left hand side of the statement equal to the right hand
side immediately. It is more appropriate to show first what the left hand side and right
side equal individually and then show that they’re equivalent. Professor Berger cited an
example of a bottom case that accomplishes this.
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Case 3:
Base case: (n=1)
LHS: 1∙2∙3=6
RHS: 1∙2∙3∙4=6
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LHS=RHS
Hence, this proposition is true for n=1
There is another approach that is appropriate for the base case. It is illustrated in case 4:
Case 4:
Base case n=1
When n=1,
1∙2∙3=6=24
4
=1∙2∙3∙4
4
=n(n+1)(n+2)(n+3)
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While both case 3 and case 4 take a correct approach, the two approaches are
different. This difference lies in how equivalence is shown. In case 3, we show what the
left hand side and right hand side equal individually and then show that they are equal to
each other. In case 4, we’re starting with the left hand side and manipulating the terms in
order to show that it is ultimately equivalent to the right hand side. Both methods are
acceptable.
When is a proof not OK?
Whenever you work on a proof, beware of the pitfall of following the procedures
without being aware of what you’re doing. Here is an example of how a proof can go
awry:
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1) Write what you want to prove
a. For example, 1= −1
2) Do some sort of correct math
a. Following the same example, we can square both sides to yield:
(1)²=(−1)² 1=1
3) Obtain the true statement
a. This true statement is what we found in step 2: 1=1
4) Hence, write in that what we initially wanted to prove is true.
a. Therefore, 1=−1
In order to be as clear as possible in a proof, it’s important to write a plan of what
you want to prove and how you want to prove it. Otherwise, the mathematical assertions
stated on paper may be misconstrued and will cause confusion for the reader. Therefore,
it is a good idea to explain, in writing, that you want to prove a statement through the
construction of a certain argument.
We also assessed an example of a proof from the quiz that Professor Berger had a few
objections to. Here is the example:
I want to show that for some k=n
1∙2∙3+ …+k(k+1)(k+2)+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)
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So I start with the original equation of P(k) and
add the (k+1) term to both sides to get P(k+1).
There were two problems with this proof. Firstly, the writer did not assert that
P(k) is true. This is inappropriate for the proof because the purpose of the inductive step
is to first assume that P(k) is true for an arbitrary but fixed k € N and then show that
P(k+1) is true. The other issue with this proof is that the author wrote for k=n whereas we
should be stating k≥n. We ultimately concluded the objections were primarily a matter of
semantics.
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III) Equivalence Relations
While the topic of equivalence relations is broad and advanced, we encounter
equivalence relations early on in our careers as students. We see these relations in
elementary, middle school, and high school math. What is the very first equivalence
relation we encounter? One possible answer to this question is equality. Another possible
answer is parity, since many students learn even and odd numbers early on. Yet another
example could be similar figures, such as squares. In college math courses, we’re
introduced to modular operations. Before we go any further, I need to point out the
defining properties of these relations.
Proposition: Let n be a fixed natural number. Then the binary relation ≡ (mod n)
is defined by:
x ≡ y (mod n) iff (x-y) is divisible by n, is an equivalence relation on Z.
Also, this equivalence relation has exactly n distinct equivalence classes, namely
[0], [1], … , [n-1].
The set of equivalence classes of this function is the set of integers modulo n,
written as Zn or Z/nZ. (Guzman 45)
Remembering the definition of this relation, it’s interesting to think about the
equivalence relations we notice as children that may not even be related to mathematics.
One example of this could deal with names. For example, let’s say having the same last
name is an equivalence relation, R, over the set of all names. This relation R is given by
a R b if and only if a and b have the exact same last name. Then take a name like Roy
Rogers. The people who would be R-related to Roy Rogers could be Jane Rogers, Robert
Rogers, Al Rogers, and so on. By reflexivity, this set would contain Roy Rogers.
Considering symmetry, we can say that if Roy Rogers is R-related to Al Rogers, then Al
Rogers is R-related to Roy Rogers because they both still have the same last name of
Rogers. Finally, we need to assess if this relation is transitive. Well, Roy Rogers is
related to Al Rogers and Al Rogers is R-related to Jane Rogers. Roy Rogers and Jane
Rogers both have the same last name. Therefore, Roy Rogers is R-related to Jane and that
means this condition does satisfy the conditions of an equivalence relation. Ultimately,
the subset of all names that are related to Roy Rogers through the relation R is the set of
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all people whose last name is Rogers. Furthermore, the equivalence class of Rogers≡R
would be the set of all people who last name is Rogers. We could denote this equivalence
class as Rogers/R.
Example:
Returning to the topic of modular operations, here is an example of an equivalence class
using modular operations:
We set up the relation a~b if a and b have the same remainder when divided by
3. This can be expressed in the form of the statement a ≡ b mod 3 or a=b mod 3. It is
sometimes also written as a=b (3). By definition, a ≡ b mod 3 iff 3 | (a-b). While my
proposition asserts that this operation is an equivalence relation, I did not prove my
proposition. So now, I need to find out:
Does a~b satisfy the properties of an equivalence relation?
1) Reflexivity
a. Is a~a?
i. Yes, because a has the same remainder as itself. This can be
expressed mathematically through the statement: a- a=0 and 3 | 0.
2) Symmetry
a. Does a~b imply b~a?
i. Absolutely. Intuitively, this is a clear statement because no
transformations are occurring when we switch b and a. Still, we
need to express this in the form of a mathematical statement.
Therefore, we can write a proof the following way:
We want to show that if 3| (a-b) then 3| (b-a)
From our assumption that 3| (a-b), we can conclude that a-b=3k by
the definition of divisibility. By multiplying negative 1 to our
statement, we get the following:
-1 (a-b) = -1 (3k)
-a+b= b-a= -3k
Now, we have the statement b-a= -3k. Hence, 3| (b-a).
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3) Transitivity
a. If a~b and b~c does that imply a~c?
i. Once again, we can intuitively understand that this condition must
be satisfied, but we need to prove it as well.
(b-a)=3n, n € Z and (c-b)=3m, m € Z
(с-a)=(3m+b)-(b-3n)=3m+3n+0
(c-a)=3(m+n)
3| 3(m+n)
Therefore, yes, this relation does satisfy the condition of
transitivity.
Since a ≡ b mod 3 satisfies the above mentioned conditions, it is an equivalence relation.
In this case, we have three distinct equivalence classes:
_____ if remainder when divided by 3 is zero.  Equivalence class #1: R0
_____if remainder when divided by 3 is 1.
 Equivalence class #2: R1
_____if remainder when divided by 3 is 2.
 Equivalence class #3: R2
We have found that the equivalence classes are: R=0, R=1, and R=2. This set of
equivalence classes is denoted z mod 3z= Z/3Z.
Lets say we want to do math on the set Z/3Z. Does it make sense to multiply or add
elements in Z/3Z?
We can most certainly “do math” on these numbers through the use of modular
multiplication. In the example below, we multiply these numbers:
__ __ _ _
5∙7=2∙1=2=2  Here, we have the remainder of 5 and 7 as 2 and 1 (when divided by 3).
When we multiplied 2 and 1 out, we ended up with the number 2, which has a remainder
of 2 again.
_ _ __
5∙7=35
 This case illustrates that we could have also just multiplied 5 and 7
immediately, without giving thought to the remainder each individual number would
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have when divided by 3. We ended up with the number 35, which reveals something
interesting in the figure below.
__ _
35=2  Now, we have the number 35, which, when divided by 3, gives us the answer of
11 remainder 2. This yields the same answer as the first figure did. If an operation always
works, then we say that the operation on the set as we defined it, Z/3Z is “well defined.”
This is because the result is independent of the representative numbers chosen for the
operations. Also, this clarifies the fact that the order of multiplication doesn’t matter in
modular multiplication. (Guzman 45)
What about rational numbers?
Q: { a/b where a,b € Z b ≠ 0, a/b=c/d iff ad=bc }
Rational numbers are equivalence classes of pairs (a,b) or (a/b) in the set Z x Z*
Let’s prove that the rational numbers form an equivalence relation.
1) Does Q satisfy reflexivity?
a. Since ab=ab, by the reflexive property of the = sign, a/b=a/b. Hence, yes,
Q is reflexive.
2) Is Q symmetric?
a. In this case, we want to know if the following statement is true:
If a/b=c/d then c/d=a/b. We can prove this in the following way:
Since a/b=c/d, ad=bc and bc=ad. Since = is symmetric, cb=da. Therefore,
c/d=a/b.
3) Is Q transitive?
a. Here, we want to show that, if a/b=c/d and c/d=x/y, then a/b=x/y.
ad=bc and cy=xd from our initial assumption.
Therefore, (ad/b)∙y=d∙x
a∙d∙y=d∙x∙b
ay=xb
Hence, a/b=x/y and the relation is transitive.
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As a way of concluding the class, we split into groups to assess the problem of how to
define rational functions in some consistent way as equivalence classes.
We know that, by definition, a rational function is a function that be expressed
in the following form:
P(x)=f(x), where P(x) and Q(x) are rational polynomials and Q≠0 polynomial.
Q(x)
If Q(x) was an equation such as Q(x)=x-16, then f(x) would not be defined for
x=16. However, since Q(x) is not the zero polynomial, we can still utilize this
mathematical statement, provided we have the restriction that x ≠ 16. Thus, by
definition of the rational numbers, we can extend a detail of the definition of
rational functions to say P,Q, Z, G are polynomial functions. P(x)/Q(x)=Z(x)/G(x)
when PG=QZ for Q, G≠0 polynomial. The equivalence classes of the rational
functions would therefore be the individual sets of all polynomial functions that
are equivalent to the original function once they were reduced. For example, 1/(x1) would be in the same equivalence class as (x+1)/( -1).
Questions:
1) Thinking back to my example of last names for equivalence classes, what other types
of equivalence classes can you think of that you encounter every day?
2) We showed modular multiplication is well defined in class. Is modular addition well
defined too? Show proof of this statement.
References:
Anderson, Laura, Ross Geoghan, and Fernando Guzman. Number Systems: Math 330-01.
New York: Binghamton University Department of Mathematical Sciences, 2005.
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Answers:
1) One example could be clocks. Since most of us don’t use military time, we
usually don’t say 13 o’clock. Rather, we say 1 o’clock, which is illustrated by
13 ≡ 1 (mod 12).
2) Yes, modular addition is well defined. The same reasoning can be given for this
as the multiplication. Let a, b, c, d be in Z. If a ≡ c (mod n) and b ≡ d (mod n) then
a + b ≡ c + d (mod n). Pf:
n|(a-c) and n|(b-d) by definition of binary relation ≡
a-c=nx and (b-d)=ny
a+b=nx+c+ny+d
a+b=nx+ny+(c+d)
(a+b)-(c+d)=nx+ny
(a+b)-(c+d)=n(x+y) n|[(a+b)-(c+d)]
Hence, a+b≡c+d (mod n) by definition of our binary relation.