Fundamental Counting Principle

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Homework Help Overview

The discussion revolves around proving the Fundamental Counting Principle using mathematical induction. The original poster seeks guidance on how to approach this proof, particularly in the context of set cardinalities.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the structure of induction proofs, including base cases and inductive steps. There is a focus on the implications of starting with empty sets and how that affects the proof. Some participants question the validity of using empty sets in the context of induction.

Discussion Status

Several participants are exploring different interpretations of the induction process and its application to the problem. There is an acknowledgment of the need for mathematical justification without relying on the principle being proved. Some guidance on the structure of induction has been offered, but no consensus has been reached.

Contextual Notes

Participants note the challenge of proving the principle without assuming its validity. The discussion includes references to specific set sizes and examples, but there is a clear emphasis on the need for a rigorous mathematical approach.

ElderBirk
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Hey Guys,

So, I am trying to prove the Fundamental Counting Principle using induction. I have no clue where to start or even how to use induction to prove it. I would appreciate some help.

The Question in a mathematical form: Let ^\sharp (A) = m and ^\sharp (B) = n. Proove by induction that ^\sharp (A \times B) = mn
 
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well induction proofs have two parts:
1) select an m and n and proof its true like #A=m=0 and #B=n=0
2) assume there's an n and m that is true
and add an element to the B set which then adds m elements to AxB set right?
 
jedishrfu said:
well induction proofs have two parts:
1) select an m and n and proof its true like #A=m=0 and #B=n=0
2) assume there's an n and m that is true
and add an element to the B set which then adds m elements to AxB set right?

Well, that doesn't really sound right, if ^\sharp (A) = 0, it means A = \emptyset. As far as I understand induction, it has to do with adding our initial assumption to the second assumption and getting to the third assumption. Adding an emptyset doesn't make that much sense.

If I initially start with sets with 1 elements, then assume ^\sharp (A) = m and ^\sharp (B) = n, How can I show if ^\sharp (A) = m +1 then ^\sharp (A \times B) = mn + n without using the fundamental counting principle, which is what I am trying to prove.
 
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if you cross the empty set with the empty you get? drum roll please the empty set right?

hence #A is zero and #B is zero and the empty set is zero right?

suppose the A set is the set of vowels and the B set is the consonants and neither set has Y.

#A=5 and #B=20 so #(AxB) = 100 now adding Y to A (its the hidden vowel of english) changes #A to 6

and since AxB means pairing Y with each element of B how many new elements are added to AxB?

Read up on induction there are two parts to the proof:

http://en.wikipedia.org/wiki/Induction_proof
 
jedishrfu said:
if you cross the empty set with the empty you get? drum roll please the empty set right?

hence #A is zero and #B is zero and the empty set is zero right?

suppose the A set is the set of vowels and the B set is the consonants and neither set has Y.

#A=4 and #B=20 so #(AxB) = 80 now adding Y to A (its the hidden vowel of english) changes #A to 5

and since AxB means pairing Y with each element of B how many new elements are added to AxB?

Haha, thanks for the explanation. I love the drum roll part. Anyhow let's get back to the question:

I obviously know and understand what's going on. The problem is showing whatever is happening mathematically. Remember, Duh or an example is not a proof.

So I know that \emptyset \times \emptyset = \emptyset. Let's look at the structure of an induction outside of this context.

Let A = \left\{n \in A | P(n) \right\}. I want to show if 1 \in A and k \in A and k+1 \in A then A = \mathbb{Z}^+, where p(n) is the condition for A.

The theory of induction stands on the fact that it starts with one. Showing that ^\sharp (\emptyset \times \emptyset) = 0 serves no purpose. Other than that, I understand that if I add 1 to m and make it 1+m the total number will be n+nm. But what is the mathematical justification for that, how can I show this without using the fundamental counting principle, since it is what I am trying to prove?
 
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