Discrete Mathematics.

LuisRodg · 2008-01-08 10:12:13

So this semester im taking Calculus II, Discrete Mathematics and Physics I with Calc...as well as a Enviromental requirement.

First day of class Discrete Math really freak me out. We started with set theory, something that I never seen before. I understood it all but it was certainly something completely different to Calculus.

How hard is this class? Im looking forward to put a lot of effort into but I just want to know what kind of monster im going to dealing with?

Last edited by LuisRodg (2008-01-08 10:36:33)

luca-deltodesco · 2008-01-08 10:26:37

discrete mathematics is not so much more or less difficult than other areas, it is simply different really.

Ricky · 2008-01-08 11:13:31

Discrete mathematics is very different from all mathematics at the high school level since it is a first glimpse at pure mathematics. Though why you started with set theory has me a bit perplexed. Normal courses start off in logic and proofs, then move on to set theory.

How hard it is depends on who you are, specifically what type of thinker you are. For some, it comes almost naturally. Others it takes a long time to understand. And when I say "comes almost naturally", I mean relatively. Things in math won't come easy, and it takes everyone a lot of work.

In set theory, always immediately expand things you know are true. For example, if x is in A U B, then x is in A or x is in B. Similarly, if x is in A \ B, then x is in A and x is not in B. More often than not, this is enough to get you where you're going.

LuisRodg · 2008-01-09 14:55:31

Thanks a lot for your responses guys.

The Discrete Math book we have starts with logic and proofs like you said but the teacher says she likes to start with Set Theory which is the second chapter. I always like to review stuff before and after class with the book but i find that many theorems and definitions of Set Theory are given using stuff from logic and im basically lost...In class she explains everything without using propositions etc but then the book does it differently... Im going crazy.

To be honest, the class looks challenging and it is certainly very different from anything I have done before BUT I think i can do it but certainly the current teaching method of my professor does not help at all...

Would you recommend I read all of Chapter 1 and then get on with Chapter 2 which is what we are doing in class and that way I can understand stuff better?

Im going to try my best on this class.

Last edited by LuisRodg (2008-01-09 15:00:27)

LuisRodg · 2008-01-10 10:29:17

Please. I would appreciate it if anyone helped me because im kind of confused and I dont want to go to class like this tomorrow.

Theres a definition that says that A is a subset of B iff:

Ax(x e A -> x e B)

is true.

The first A is supposed to be upside down (universal quantifier) and the "e" is the "exists" or "belongs to" symbol.

Now I have a question about this statement. First of all, could anyone please explain me what does the "Ax" mean? I was reading the definition in the book and it says it takes the place of "for every x..." but is that it? Does that symbol just replaces words or does it have a more intricate meaning?

Besides that, that statement could be thought as a conditional statement such as "p -> q" where:
p = x e A
q = x e B

In this sense, if one constructs the truth value of the statement "p -> q" one could see that this statement is true when p is false or when q is true.

Truth table:

p ----- q ----- statement

T ----- F ------ FALSE
T ----- T ----- TRUE
F ----- F ------ TRUE
F ----- T ----- TRUE

Given that (p) -> (q) = (x e A) -> (x e B)

From the truth table one can see that out of the 4 choices, theres 3 choices that yield "true" results. The false results means that in that case A is not a subset of B (we are talking regular subsets, not proper subsets). The other 3 choices yield true which means that given the truth values that p and q takes, A will be a subset of B.

so when p and q are both true, the statement is true which means that for every x in set A theres an x in set B, making A a subset of B.

when p is false and q is false, the statement is true (because p is the hypothesis and when it is false, it makes the statement true) therefore that implies that (x e A) is false and (x e B) is false but since the statement is true, it means that A is a subset of B? How could that be?

This same problem persists with the other 2 truth values....

Would anyone care to explain?

Last edited by LuisRodg (2008-01-10 10:45:08)

luca-deltodesco · 2008-01-10 11:09:26

you can use LaTeX here, (click images to see the code) encase in

tags

This means 'for all' x, i.e. for everypossible 'x', think of it as 'whatever x is'

so the definition can be read as:

whatever x is, if x is in A, it must also be in B

you truth table is a little confusing.

a -> b means only. if a is true, b MUST be true, with the opposite undefined. if you want a truth table you would have something more like:

p->q
p | q
T | T
F | T/F

first case, p is true. that means that x is a member of the set A, because p is true, p->q dictates q is also true. x is also a member of set B. A is a subset of B

second case, p is false. that means that x is not a member of set A, p->q means that q can be either true or false here, x can be in B, it might not be in B.

Last edited by luca-deltodesco (2008-01-10 11:11:50)

LuisRodg · 2008-01-10 11:16:51

What you just said makes PERFECT SENSE.

However, thats not how my book defines "p -> q". In fact, the truth table I gave you is the exact one that goes in my book. In my book it says that given the statement "p -> q", the statement is always true if p is false, and the statement is always true if q is true. It says that literally which didnt make sense to me at all but i just memorized it...

Why is this?

Im SO confused...

Also, thanks for replying.

luca-deltodesco · 2008-01-10 11:38:40

i dont get where you book is coming from to be honest, p -> q tells us only what happens if p is true. if p is false then we cannot infer anything about q.

LuisRodg · 2008-01-10 11:47:02

I dont either.

Im going to copy it down just how it is in the book.

"Let p and q be propositions. The conditional statement p -> q is the proposition "if p, then q". The conditional statement p -> q is false when p is true and q is false, and true otherwise. In the conditional statement p -> q, p is called the hypothesis (or antecedent or premise) and q is called the conclusion (or consequence)"

"Note that the statement p -> q is true when both p and q are true and when p is false (no matter what truth value q has)"

luca-deltodesco · 2008-01-10 12:08:23

ah i see what it means now. although why you need to work it like that is beyond me

its just a long winded (and in my opinion) more complicated way of saying what i said

LuisRodg · 2008-01-10 12:16:40

You just confused me more there... Sorry for bugging you so much but would you care to explain me?

if what the book says is correct then like i said before it leads to conclusions such as this:

p is false, q is true and the conditional statement as a whole is true. If the statement is true it proves that A is a subset of B meaning that (x !e A) and (x e B) and A is a subset of B? How come?

Last edited by LuisRodg (2008-01-10 12:19:35)

luca-deltodesco · 2008-01-10 12:30:30

from the table, using the ones for which the statement was true. you have

none of them give a contradiction to the subset. the defintion was if x is in A, it is also in B. and that is shown by those 3, we only need worry about the first one that tells us what happens if x is in A.

Last edited by luca-deltodesco (2008-01-10 12:31:38)

LuisRodg · 2008-01-10 12:37:10

Ok I kind of get it. It is a confusing class, do you agree?

I mean, i remember when I was introduced to Calc and i thought it was hard but now it all seems natural, does it ever get like that with discrete?

Im doing some serious studying today and trying to let everything sink in, you wouldnt mind if I post more questions if i need to?

Thanks a lot for your help!

Last edited by LuisRodg (2008-01-10 12:48:09)

Ricky · 2008-01-10 13:03:25

What you are confused about is vacuous statements, and it is probably one of the most difficult things to get your head around in logic. It helps if you think of it a certain way.

p -> q or rather, p implies q.

We want to find out the truth value behind this statement. As it is a statement, it must have a value of either true or false, and not both. That is the definition of a statement.

So say p is false. What is the truth value then? Remember you can only pick from true and false. So trying false first, we can reach very bad conclusions. For example:

x is a prime implies x is 2 or odd.

So:

p = x is a prime.
q = x is 2 or odd.

Now let x be 4. We have p -> q with p false and q false. If we were to label p -> q as having a truth value of "false" in this case, that would be saying the above statement doesn't not hold. But it certainly does.

Now let x be 9. We have p -> q with p false and q true. Again, if we were to label such a case as false, we would be saying the above implication is not true.

In general, an "if-then" statement only makes a conclusion when the precondition (the if part) is met. When it isn't met (when p is false), the statement makes no conclusion. You can't possibly regard saying nothing as a false statement, and thus, it must be a true one.

Math Is Fun Forum

#1 2008-01-08 10:12:13

Discrete Mathematics.

#2 2008-01-08 10:26:37

Re: Discrete Mathematics.

#3 2008-01-08 11:13:31

Re: Discrete Mathematics.

#4 2008-01-09 14:55:31

Re: Discrete Mathematics.

#5 2008-01-10 10:29:17

Re: Discrete Mathematics.

#6 2008-01-10 11:09:26

Re: Discrete Mathematics.

#7 2008-01-10 11:16:51

Re: Discrete Mathematics.

#8 2008-01-10 11:38:40

Re: Discrete Mathematics.

#9 2008-01-10 11:47:02

Re: Discrete Mathematics.

#10 2008-01-10 12:08:23

Re: Discrete Mathematics.

#11 2008-01-10 12:16:40

Re: Discrete Mathematics.

#12 2008-01-10 12:30:30

Re: Discrete Mathematics.

#13 2008-01-10 12:37:10

Re: Discrete Mathematics.

#14 2008-01-10 13:03:25

Re: Discrete Mathematics.

Board footer