Martha Pujol Beglane
Class Notes MAE 501
Monday, September 12, 2007 3:50-4:50 (Early dismissal – Holiday)
Useful software - LATEX – quicker than using Math Type with Word to produce
documents using math symbols, language, etc.
Discussion about Homework
Homework Scores
9-10 math mostly correct, writing also good
7-8
some math errors and/or problems with writing
4-6
more significant math errors and writing problems
Homework #1 – Problem # 1b – Prove identity element of the group must be unique.
Why is this particular student’s proof not correct?
He states:
Suppose e1 , e2 are 2 identity elements, then pick any g  G
e2 gg1 e1
What is wrong with this approach?
1
1
What the student is saying is that e2  gg and e1  gg . Therefore, by
transitivity e2  e1 . Here the student is implicitly assuming that e2  e1 by
1
1
1
assuming that g2  g1 . For any g  G , g is the unique element such
g1g1ge. The inverse g 1 of g  G is defined in terms of g
that g
1
1
and e. We do not know that g2  g1 , since this is dependent on their
respective ei ’s.
Proof:
Assume there exist two identity elements in G, e1 and e2 .
Then by the definition of an identity element the following are true:
Since e2 is an identity element of G, then
e
e
e
1e
2
1
2e
1
Since e1 is an identity element of G, then
e
e
e
2e
1
2
1e
2
So, e1  e2 .
Therefore, the identity element of a group G must be unique!
Here is an additional example:
1
Let g be an element of group G. Prove that the inverse g is unique.
Assume there exists 2 elements a, b  G that satisfy the property of being
the inverse of g  G .
Then, by definition of the identity of a group a  ae .
By definition and assumption that b is the inverse of g, ae a(gb) .
( b)(agb
).
By associativity, ag
Therefore, ae  eb or a  b .
1
The inverse g is indeed unique!
Proof from Class on Monday, 9/10 clearly stated:

B AB

B
Question: Prove A
Ideas for the proof:
1. To prove two sets are equal you must prove that each one is contained in the
other. A B  Bmeans that both A BB and B A B
2. In order to prove one set is contained in another, take an arbitrary element
from the first set and show that is lies in the second set. If X is contained in Y,
X  Y , take an arbitrary element a  X and show that a  Y .

B AB

B
Now begin: Prove A
1. Assume A  B (assumption may or may not be used in proof), and show
A B B
2. Let x be an arbitrary element of A B , i.e, xA B
3. xA B, therefore, by definition x  A or x  B (you need to consider both
cases)
4. Case 1: If x  B , then we are done.
5. Case 2: If x  A , then since A  B , then by definition, x  B . Then we are
done.
6. Therefore, A BB
7. Now we need to show B A B
8. Let x be an arbitrary element of B , or x  B
9. If x  B , then by definition of union xA B
10. Therefore, B A B
11. Since A BB and B A B, then A B  B
The above question came from our discussion on Monday, 9/10:
 true?
Question: Why is
Explanation: Since all elements of
are in , then 
Therefore,   .
.

Explanation: Since the rationals “live” inside the reals,
to the smaller set which is .

, then
is equivalent
Prove: If A  B , then A B  A.
Is this a bi-conditional (“…if and only if…”) statement? P  Q .
No, it is a conditional (“If…then…”) statement. P  Q
Prove and decide whether the converse is true. Q  P
What is the converse of the above statement? If A B  A, then A  B .
(Prof. gives the students an opportunity to discuss.)
Prof. asks for suggestions on how to go about the proof and a student offers the
following:
Since A  B , then if there exist an element x  A then x  B .
If A B  Athen some element in A is not an element of B, which
contradicts assumption. Because if x  A , then x  B .
What direction did he focus on?
A B A - The student implicitly assumed A B is already contained
in A.
Another student suggests: (thinking is clearer on this one)
Assume A  B
First show that A B A
Let xA B, then x  A and x  B , so x  A
Therefore, A B A
Now show A A B
Let x  A
Since A  B then x  B , so xA B
Therefore, A A B

A

A

B
Is the converse true? AB
One student suggests it is not true and Prof. asks for a counterexample.
Student responds:
x B
Assume A  B, then xA
x  A and x
B x
AB
xA Band x  B
The student stops because she realizes that she started by assuming the
consequent and not the antecedent. She assumes A  B , which is exactly what
she was trying to prove.
She needed to begin by assuming the
antecedent A B  A.
Another student asks a logic question:
Is p  q true if p is false? Yes! A conditional statement is true when the
antecedent is false, regardless of whether the consequent is true or false.
Therefore, in order to prove a statement you need to assume the
antecedent is true and show that the consequent is true.
Another student suggests a counterexample:

2
,3
,4
,B

3
,4




1. A
, A B3,4 This is not a counterexample!
In this example, A B3,4 and A2,3,4, therefore, A B  A.
We are assuming P or the inverse of the antecedent, which will not

A

A

B
help us to prove AB
is a true statement.
2. What if we put a 1 in B? This is not a counterexample for the same
reason the previous example is not a counterexample. The conditional
statement p  q is true when p and q are both false.
Is a picture the only possible proof? You need more than the picture to prove!!
Another
A=B student asks:

A

A

B
A
Is AB
true??
B

A

A

B
Another student suggests the following as a proof that AB
is true:
Let x  A
Since A A Bthen x  B
Therefore, A  B
Prof. comments: Good! It’s a short proof.
Additional Question:
Many times in mathematics there are various ways of proving a particular
statement. The “counterexamples” suggested by some of the students present an idea for

A

A

B
an alternate approach to proving AB
. Look at the “counterexamples”
and see if you can come up with another way of solving this conditional statement.
Functions
Prof. asks students to define functions in their own words.
A student offers the following:
f (x)xy
Prof. states that she is disturbed by the presence of the y.
Another student suggests:
f (x)  y For every x there exists a y.
Another student suggests:
A function is a relationship between something where you have an input
and an output. Like in science you have…
Prof makes an adjustment to the student’s definition:
It is a relationship between two sets.
A function is a mapping between two sets f : AB, such that each input
has exactly one output. There is only one way to get elements of set A to
set B.
Why do functions make you use = signs? Ex: f (x)  y
Equal signs are used to describe a function, or the procedure you need to
follow in order to get a particular element of set A to the set B.
What does it mean to be one to one or injective? The two words are synonymous.
A function f is said to be one to one if and only if whenever
f (x) f (y)then x  y . (i.e., every element of set B appears at most
once.)
2
)f(
1
)1and
For example: f (x)  x is not a one to one function since f(1
1  1