Homework due next week

Sections 6.4,8.1, 8.2,8.5, Due Thursday 12/5/2013
Also, I extended the due date for the webwork sections 8.2 and 8.5 to November 30.

webwork question

I am not entirely sure what pairwise disjoint means in the problem
Find the number of elements in A1∪A2∪A3 if there are **** elements in A1, **** elements in A2, and **** elements in A3 in each of the following situations:

(a)   The sets are pairwise disjoint.

(b)  There are ** elements common to each pair of sets and * elements in all three sets
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Pairwise disjoint means that that any pair of distinct sets have an intersection that is empty 

WebWork for next week due 11/26/2013 at midnight

Sections 6.4, 8.2 and 8.5

Office hours cancelled

Office hours are cancelled today Nov 13 and Friday Nov 15.  I have some medical issues to deal with and these were the times I could get.  I'll be available Thursday afternoon if you need to speak to me.

Homework Due Next Week 11/18/2013-11/22/2013

1) Webwork sections 6.2 and 6.3 due Tuesday 11/19/2013 at 11:59pm
2) Book homework sections 6.2 and 6.3 due Thursday 11/21/2013 at the *beginning* of class

Webwork due next week 11/13/13 @ midnight

Sections 5.3 and 6.1

Reminder: review session today. Bring your questions

Question on Euclid's proof of the infinitude of primes

By the fundamental theorem of arithmetic, Q is prime or else it can be written as the product of

two or more primes. However, none of the primes pj divides Q, for if pj | Q, then pj divides

Q −p1p2 ···pn = 1. Hence, there is a prime not in the list p1, p2,...,pn. This prime is

either Q, if it is prime, or a prime factor of Q. This is a contradiction because we assumed that

we have listed all the primes. Consequently, there are infinitely many primes.

This is the proof provided in our textbook (Rosen) that demonstrates the infinitude of primes. I find myself unable to grasp this proof.

Specifically, when it is stated that "if pj | Q, then pj divides Q −p1p2 ···pn = 1." I don't understand why the first proposition implies the second. Can you help me with this?

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....judging from what you didn't put in your quote from the book, you may have missed the part that sets up the proof by contradiction: Suppose that there are a finite number of primes; write them as p_1, p_2, ...  p_n.  Define the integer Q as  Q=p_1p_2...p_n+1.  Here's where what you quoted comes in:

By the fundamental theorem of arithmetic, Q is prime or else it can be written as the product of

two or more primes. However, none of the primes pj divides Q, for if pj | Q, then pj divides

Q −p1p2 ···pn = 1. Hence, there is a prime not in the list p1, p2,...,pn.

Really, you can see from the definition of Q that Q div p_j is the product of all of the p's except p_j, and the remainder Q mod p_j is 1.  This means that Q is not divisible by any of my finite list of primes, so it's not a composite number (i.e. a product of more than one prime), so it must be prime. But Q is also strictly bigger than any of my list of primes p_1, p_2, ..., p_n, so it is not one of p_1, p_2, ..., p_n.  But these were supposed to be all of the primes.   This is a contradiction.  Therefore my supposition that there are only a finite number of primes is false.  Therefore there are an infinite number of primes.

webwork question

Had issues with Section4.1 Rosen7: Problem 5  as practice indicates that

you are rounding up (ceiling) but this is not the case on the actual

homework problem.  The homework problem did not like decimal spots, number

R number, or number plus fraction.

Please let me know how this problem was suppose to be answered.

Last answer:

AnSwEr0001: 2

AnSwEr0002: 0

AnSwEr0003: -6

AnSwEr0004: 2

AnSwEr0005: 7

AnSwEr0006: 2

AnSwEr0007: -4

AnSwEr0008: 0

AnSwEr0009: 3

AnSwEr0010: 8

AnSwEr0011: -5

AnSwEr0012: 0

AnSwEr0013: 9

AnSwEr0014: 28

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OK, first of all, the webwork uses the '÷' sign for what the book calls 'div'.  The mathematical theorem that is the background for all of this is that given any integer a and natural number b, there is an integer d and a natural number r such that 0≤r<b such that

                         a=d*b+r

Note the important fact that r is strictly non-negative, so that d*b has to be smaller than a.  When a is negative, this means that d*b is more negative.  For instance, in problems 3 and 4, which were -26÷4 and -26 mod 4,  this means that d*4 has to be the largest multiple of 4 less than -26, so that d=-7 not -6, while r has to be 2.

Class tomorrow is cancelled. The exam is rescheduled to next Tuesday.

Hi All,
Class tomorrow Nov. 5 is cancelled.  The exam, originally scheduled for Thursday Nov. 7 is rescheduled for Tuesday Nov. 12.  Our in-class review session is rescheduled for Thursday Nov. 7--bring your questions and I will answer them.

Office Hours and Class on Tuesday, Review Session and Exam

1) I'm scheduled to be a workshop all day Tuesday.  Therefore:
       a) I'll have substitute running the in-class review session.  Bring your questions. Better yet, post them here in advance so your fellow students can see what people are wondering about.
       b) My office hours on Tuesday are cancelled.  Instead I'll have office hours 11am-Noon tomorrow.

2) Apparently there will also be a review session at the math center on 11/5/2013 at 7:00pm--this is all the information I have on this.

3) PS: exam 2 will take place on Thursday.