The proof to both of these relies on INDUCTION.
The basic idea behind induction is the following:
1) Prove that a statement P(n) is true for n=1.
2) Prove that if P(n) is true, then P(n+1) is true, so long as n is at least 1.
If we can prove the above two statements, then we have...
The probability is actually (3/5)^10, which is about 0.006, which is 0.6%
3/5 is the probability that any given ticket will NOT be given on a Monday or a Friday, and there are 10 independent tickets being given.
The probability of something just as surprising happening is actually 10 times...
If P(A) is small, then the probability of [receiving 10 tickets overall but 0 tickets on Monday or Friday] is small.
If the probability of not receiving any tickets on Monday or Friday is very low under the assumption that tickets are given out uniformly across all days, then this gives us...
Well, normally you define a function to have three parts:
1) The domain.
2) The codomain.
3) A rule assigning to each element of the domain an element of the codomain.
This is why you often write, f: A --> B when f maps the set A into the set B.
f(n) = n^2 between the naturals and the...
Well, n --> n/2 is also a bijection between the two sets, just in the other direction.
In fact we can make this a theorem:
THEOREM: If f is a bijection between A and B, then f-1 exists and is a bijection between B and A.
(Remember, bijection <==> one-to-one and onto).
Now, a mapping like...
Yes. Let f(n) = n/2.
For any even natural number, f(n) is a natural number.
2 -> 1
4 -> 2
6 -> 3
8 -> 4
Notice that it is defined for each and every even natural number. f is clearly one-to-one. Moreover, for each number in N, we can find an even natural number m such that f(m) = n...
Yes, now go check all other 9,999,999,999,999,999,999,999,... one-to-one mappings from (2,4,6,8,...) into N.
Try looking more closely at the logic of that definition you have just given. I'll write it out more formally:
[ card(X) < card(Y) ] \Leftarrow [ \forall \phi mapping X one-to-one into...
Remember, two sets A and B have the same cardinality if and only if you can find a bijection between them.
In order to show that set A has a lesser cardinality than set B, you must show two things:
(i) There DOES NOT exist a bijection between them. (A bijection is a one-to-one and onto map)...
You've made a mistake. You are correct that your mapping is one-to-one, but it is certainly not onto! Remember, part of the definition of a function includes its domain. You can see that no even natural number maps to 3.
More importantly, I think the question was written incorrectly.
What's your reason for wanting this symbol?
While you can certainly have sets of sets of sets of... etc., there is almost always an important distinction between an element of a set and a set (whether or not that element is a set itself). It seems as though you are looking for a way to...