Moridin said:P = {x(1), x(2),..., x(p)}
Q = {x(1), x(2),..., x(p), x(p+1)}
HallsofIvy said:Every subset of Q (with x(p+1) added) is of two kinds: those that do not have x(p+1) in them and those that do not. How many subsets of Q do NOT have x(p+1) in them (remember that those would also be subsets of P)? And every subset of Q that does contain x(p+1) is a subset of P union {x(p+1)}.
Kurret said:How many new possible subsets can you create, when you have one more element to pick?
Dick said:Let be concrete here. Take P to be the set of {1,2,3...p}. Take Q to be the set of {1,2,3...p,p+1}. All of the subset of P are also subsets of Q. And there are 2^p of them by the inductive hypothesis. All of the subsets of P are also subsets of Q. The only new subsets consist of a subset S of P that contains also contains p+1. Right? [bb]So there are 2^p new subsets.[/b] What's 2^p+2^p?
Dick said:You have 2^p sets without that element. If you add that new element to each of the 2^p sets, you get 2^p additional sets. If you put them all together you get all sets made from p+1 elements. For a total of 2^(p+1).
Induction is a mathematical proof technique that uses the principle of proving a statement for a base case and then showing that if the statement holds for any given case, it also holds for the next case. In the case of showing 2n subsets in an n element set, we can prove the statement for the base case of n=1, which has 2 subsets (the empty set and the set itself). Then, using the inductive step, we can show that if the statement holds for n=k, it also holds for n=k+1. This way, we can prove that the statement holds for all positive integers, and thus there are 2n subsets in an n element set.
Sure, let's take the base case of n=2. An n=2 element set has 2 subsets: {a,b} and the empty set {}. Now, for the inductive step, we assume that the statement holds for n=k, and we want to show that it also holds for n=k+1. So, for n=3, we can take the k=2 case (n=2, with subsets {a,b} and {}) and add element c to each subset, giving us {a,b,c} and {c}. This gives us 4 subsets, which is indeed 2n for n=3. Thus, the statement holds for n=k+1 and by the principle of induction, it holds for all positive integers.
Yes, there are some limitations. Induction can only be used for statements that can be expressed in terms of integers (like counting or sets). It cannot be used for continuous or uncountable sets. Also, induction can only prove statements for all positive integers, so it cannot be used to prove statements for negative integers or fractions.
Yes, induction can be used to prove statements about sets with any number of elements. The statement may become more complex as the number of elements increases, but the principle of induction remains the same. We can still use a base case and an inductive step to prove the statement for all positive integers.
No, there are other methods for proving this statement. One alternative method is to use the binomial theorem, which states that (a+b)^n has n+1 terms. Using this theorem, we can show that for an n element set, there are 2^n subsets, which includes the empty set and the set itself, giving us a total of 2n subsets. However, induction is a commonly used and efficient method for proving this statement.