Download James Lynch MAT 501 Class Notes: 10/29/09 Some Exam Problems

James Lynch
MAT 501
Class Notes: 10/29/09
Some Exam Problems Explained
QUESTION 2e:
Below is a problem and solution to a NYS Regents Exam problem:
Solve for x:
=
(i)
=
(ii)
=
(iii)
(iv) -4=11x
(v) X=At which step does the student use injectivity to solve the problem? Explain.
Solution:
The student uses injectivity of the exponential function to solve the equation between steps two and
steps three.
Recall the definition of injectivity:
A function is called injective if: f( )=f( ) →
=
for all
,
in the domain of f.
So in this problem let f(x)=
For this function injectivity means:
f( )=f( ) → =
→ =
In the case of the problem above, let =4(x+2) and =3(5x+4) and it is easy to see that the
student is only allowed to make the jump from step two to step three because the exponential function,
f, is injective.
In general we are not always allowed to make similar jumps.
For example:
1-
Sin(x) = Sin (3x-9) does not allow one to make the conclusion that x = 3x-9, precisely because
the sin function is not injective. (observe the sin function fails the horizontal line test)
2-
f(x)= is an injective function. This is easy to see by checking that the cube root of a number is
equal to the cube root of another number only when those two numbers are equal. ( In
particular we do not have x and –x mapping to the same element in the co-domain)
This function being injective allows us to make the jump:
=
→ =
This example leads us to an interesting idea:
If we have an injective function, this allows us to conclude that there is only one x-value mapping to any
given f(x) value. As in example three, this allowed us to apply the cube root of both sides and have no
fear that we would not lose any information about x. In particular in this example we can “recover” all of
our original x-values from f(x) values. This last statement is the definition of an invertible function.
Indeed a function is invertible if and only if it is injective.
Another important thing to not about injectivity:
Observe the definition of a function:
A map,f, is called a function if and only if:
=
→f( )=f( )
for all x in the domain of f.
The definition of an injective function and the definition of a function are logical converses to one
another. Now we can see that if a function is invertible then the inverse map is also a function. We can
see this on a graph if we see that a function passes both the vertical and horizontal line tests, after a
reflection over the line y=x (graph of the inverse function) we still have a graph that passes the vertical
and horizontal line test.
Question 4c:
The following problem is taken from the NYS Regents Exam:
Express in simplest form:
÷
The solution given is:
Using concepts discussed in class, clearly explain (write) what is correct and incorrect about this
solution.
We went over in class that two rational functions may be equivalent algebraically but not as
rational functions because we have to consider the functions over a certain domain. The
expression as it is given has domain {x|xЄR x≠±3 and x≠0}. To see this simply observe that
these are the only values making the expression undefined. The students solution has domain
{x| xЄR x≠0}. Therefore, although the students solution, and the given expression are
equivalent algebraically they are only equivalent over their shared domain, the rational
functions are only the same when we exclude x=±3 from the domain of the solution. (note that
both rational functions are undefined when x=0 so we can still say they are equivalent at x=0)
Question 3:
In the first part of this question we are asked to show that a given relation is an equivalence
relation. The relation is given: let two polynomials, f(x) and g(x) in Q[x], be equivalent if and
only if some other polynomial, p(x) in Q[x], divides the difference g(x)-f(x).
In this case: p(x) = x-7
Checking that any relation is an Equivalence Relation simply requires the checking of three
axioms:
(i) Reflexivity: f(x) ~ f(x)
This follows from the fact that f(x)-f(x)=0 for any polynomial and
p(x)|0 for all polynomials, and we have that the relation is reflexive.
(ii) Symmetry: f(x)~g(x) implies that g(x)~f(x)
This follows from the definition of being divisible by p(x)
p(x)|g(x)-f(x) ↔ g(x)-f(x)=q(x)p(x) for some polynomial q(x) in Q[x]
multiply the above expression by -1 on both sides to obtain:
f(x)-g(x)=-q(x)p(x) where –q(x) is just some polynomial in Q[x]
Thus p(x)|f(x)-g(x) and g(x)~f(x) and we have that the relation is symmetric.
(iii) Transitivty: f(x)~g(x) and g(x)~h(x) implies that f(x)~h(x)
f(x)~g(x) implies that g(x)-f(x)= (x)p(x)
g(x)~h(x) implies that h(x)-g(x)= (x)p(x)
Add the above two equations and obtain:
h(x)-f(x)=( (x)+ (x)) p(x)
which means that p(x)|h(x)-f(x)
and thus f(x)~h(x) and we have that the relation is transitive.
The second part of the question was:
Let r(x) be equivalent to s(x)
And Let h(x)=r(x)-s(x)
And let p(x)=x-7
Evaluate h(7).
Solution:
Consider:
h(x)=r(x)-s(x)=p(x)q(x) for some q(x) in Q[x]
So to evaluate h(x) we can simply evaluate p(x)q(x)
But p(7)=0 this implies that p(x)q(x)=h(x)=0 and we are done.
Note a common mistake in problem 3: during the proof of the axioms for equivalence relations,
people assumed that if p(x) divided the difference of two polynomials, then it must divide each
polynomial individually. This is not true in general in the polynomial ring, or much simply in the
ring of integers.
Counter example for this type of logic: 2|9-7 does not imply that 2|9 and 2|7.
Question 5a
The definition we came up with in class for an irreducible element in a ring was:
(i) A non-unit element in a ring, P, is called irreducible if whenever P=
of of P, then either
is a unit.
is a factorization
(ii) Based on this definition 3x-9 is irreducible in the ring of polynomials with coefficients in
Q.
Proof:
Remember that a unit in a ring is an invertible element in that ring with respect to
multiplication.
Also remember that we saw in class that the constant polynomial was a unit in Q[x],
correspondingly, every non-zero element in Q[x] of degree 0 is a unit.
If p(x)=3x-9 is irreducible then we must have some factorization:
p(x)=
where
are not units in Q[x]
Therefore the degree of
≥1
And hence the degree of p(x) 2, which is a contradiction because the degree of 3x-9 is 1.
Thus, p(x)=3x-9 is irreducible in Q[x].
(iii) -7 is irreducible over the integers because we have the only factorizations of -7 over the
integers are:
(-1)(7) and (1)(-7)
Both factorizations contain a unit in Z and thus -7 is irreducible in Z.
Note: In the ring of integers Z, in order to check if a given integer p is prime we must check
whether any prime numbers between 1 and
divide p.
Question 5b:
(i) A non-unit element in a ring, d, is called the greatest common divisor of two elements a
and b in a ring if:
d|a and d|b and whenever there exists c in the ring such that c|a and c|b we have that c|d
(ii) d=-2 is a greatest common divisor of 4 and 6 in the integers because d clearly divides
both 4 and 6, and if we have some c that also divides 4 and 6 then c can only be
equal to one of -2,-1,1,2, by virtue that it must divide both 4 and 6. All of these
possible values of c divide -2. Therefore, -2 is a greatest common divisor of 4 and 6.
It is interesting to note that for the same reason 2 is also a greatest common divisor of 4
and 6, so that the greatest common divisor of two integers is not necessarily unique.
Question Three is More Than Meets the Eye
The question defined an equivalence relation on the ring of polynomials with rational
coefficients in the following manner:
f(x)~g(x) if and only if p(x)|g(x)-f(x)
So two polynomials, f and g, are equivalent to one another if p(x) divides their difference.
This is analogous to the equivalence relation defined on the ring of integers defined in the
following manner:
a~b if and only if n|b-a
for a,b,n in Z
One can easily check that this relation does indeed satisfy the equivalence relation axioms.
This equivalence relation is called “equivalence modulo n,” and says that two integers are
equivalence when their difference is divisible by n. After thinking about part b of the
question, we see that the integers can be thought of in the same way as polynomials with
respect to “plugging something in and getting something out”
Recall that we defined h(x)=r(x)-s(x) and said that r(x) was equivalent to s(x). We then said
that h(x)=r(x)-s(x)=p(x)q(x) for some q(x) and evaluated h(7).
In the analogous manner let f(a,b)=b-a and let b be equivalent to a modulo n
Then we have that f(a,b)=b-a=nk for some a,b,n,k in Z.
This map sends two integers, a and b, to the modulus, n, that would make them equivalent.
By comparing these this same equivalence relation defined on two different rings of the
same type, we find a way to evaluate the integers like we would polynomials.
Some More Probability
A little Review:
Our intuition lead us to the fact that if we have (n) ways or event A to occur and (m) ways for
event B to occur then we have (m)(n) ways for A followed by B to occur. Observe that
multiplication is commutative so that it does not matter which event occurs first. We said that
this only hold true when A and B are independent events.
Important Definition:
The pair (Ω,P) is called a Discrete Probability Space if Ω is a non-empty, finite, or countably
finite (discrete) set, and P is a map from the power set of Ω (set of all subsets of Ω) to R that
satisfies the following properties:
(i) P(A) ≥ 0
(ii) P(Ω)=1
(iii) For every sequence of disjoint sets in the power set of Ω, we have that:
P(Σ )= ΣP( )
P is called a set function because it sends subsets of Ω to points on the interval [0,1]
This is an important definition to have because it is the space in which all of the corollaries we
discussed in class hold true:
P(ф)=0
A B → P(A)≤P(B)
P( )=1-P(A)
P(AUB)=P(A)+P(B)-P(A
Some useful facts about probability from today:
P(A followed by B)=P(A) P(B)
for A, B independent events
P(Event)= (# ways for the event to happen) / (total # of possible outcomes)
Some Examples done at the end of class:
1- Flip a coin twice, P(Heads followed by Heads)=P(H) P(H)= ½ ½ = ¼
2- Flip a coin 4 times, what is the probability P(E) of getting exactly two heads?
6 ways to get heads: HHTT, HTHT, HTTH, THHT, THTH, TTHH
Possibilities
P(E)=6/16
3- Roll a die 4 times, what is the probability of getting at least three sixes?
We could get 3 sixes, or 4 sixes in our 4 rolls to fit this requirement, there are four ways to
get three sixes (666x, 66x6, 6x66, x666) , there is one way to get 4 sixes (6666), and there are
possible ways to get anything.
The probability is 3/ + 1/
References:
1-Stony Brook University Math Classes
2-Moeschlin, O, Discrete Stochastics, Springer 2003