prove that if (S&#92;T) is equinumerous to (T&#92;S), then S is equinumerous to T. \n\nAnd prove that every subset of a finite set is finite..\n\nand prove that every infinite set has a denumerable subset...\n\nwhere the heck do u even start??
199348
Proof's are generally about asking yourself the right question, not making a huge intellectual leap.\n\nProve that every subset of a finite set is finite.  What is the right question?  There are a couple interesting questions to ask here:\n1)  Could there exist an infinite set within a finite set, and if so, what would it look like?  \nBecause the infinite set is within the finite set, every set within the infinite set must be within the finite set.  But the finite set is finite ... (proof follows)\n\n2) If you put a set in another set, can you draw a line (make a function) connecting the things in the inner set with the things in the outer set?  \nYes, you can.  All of the elements of the inner set are in the outer set, so you could draw a line from the elements in the set to themselves.  This probably sounds funny, so let me help you with an example:\nyou have 5 apples and 4 oranges.  Invent 2 sets, the set of apples and the set of fruit.  The set of fruit has 9 elements.  Clearly, the set of apples is within the set of fruit.    You could make a connection between the number of apples in the set and the number of fruit -- eg the set of fruit has nine elements and the set of apples has 5, so you could draw a line from an apple in the set of apples to an apple in the set of fruit.  Can you draw this type of line if the apples were infinite and the fruit were not?  To put this another way, if the apples were infinite, then there must be infinite fruit as fruit is an apple (eg, apples are in the set of fruits.)\n\nMathematically (this proof is way more than needed, but for illustration):\n\nsuppose B is an infinite subset of the finite set A with n elements.  Define F to be a function from an element x in B onto itself.  Clearly F is 1-1 as every element in F maps an element in B to itself.  Because B is infinite, for any nth element in F there exists a (n+1)th element in F mapping the n+1 th element in B onto itself.  But B is a subset of A, so there is a n+1 th element in A that maps to itself, contradiction to the fact that A is finite.\n\nPlain text of what I just said:\nB is within A, A is finite, and B is infinite.  If I count up all the terms in B (using the function), I am also counting all the terms in A (because every element in B is also in A).  If there are 5 terms in A, and I start counting B and get to 6 terms, then there must be 6 terms in A.  But then I can get a 7th term in B, so there must be 7 terms in A.  And so on to infinity (induction).  So A must be infinite, which is a contradiction.  So any subset of a finite set cant be infinite.  So it must be finite.
184040
In your second step, you create a -1 for each and every +1.  Then, in the final step, you miscount, leaving off a -1.  That's the nature of working with infinity.  You could just as easily rearrange the numbers as \n\n= 2 - 2 + 2 - 2...\nor\n= 157 - 157 + 157 - 157...\n\nThus "proving" that every number equals every other number.  Infinity resists such manipulations.
651307
The set {1, 2, 3, ...,} has been closed under multiplication, because any product belongs to that set.\n24 and 25 belong to {1, 2, 3, ...,}. 24 x 25 = 600 as well.\n\nThe identity of the multiplication is 1, because n x 1 = 1 x n = n.\n\nAll numbers n have the inverse 1/n except n=0.\n\nAll the precious answers have no sense.
35254
listen the origine of this is difficult but if want to know contact me\n\nI can tell how you can get them:\nAny year that can satisfy 2 of this 3 conditions (1&2) or\n(1&3) and fail in the third\n1) the year must be divisible by 4\n2) the year must NOT be divisible by 100\n3) the year must be divisible by 400\n\nExample: 1996 this satisfy condtion 1 and 2 >> 1996 is leap\nExample: 2000 this satisfy condtion 1 and 3 >> 2000 is leap\nExample: 1900 this satisfy condtion 1 only >> 1900 is not leap [[[1900 is divisible by 4 and  it is not leap because it is divisible by 100 and not by ]]]\n\nWhy? you can contact me rarebonheur@yahoo.com
292904
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
74494
That is not an example of Russell's Paradox.\n\nRussell's Paradox (on which I should be writing a paper right now) arises from sets or concepts of the form:\nS ={x: x is a set which is not a member of itself.}\n\nThe paradox arises when you try to decide whether S itself is a member of S.  If S is a member of S, then that means that S is a set which is not a member of itself.  This is a contradiction.\nIf S is not a member of itself, then that means S is a member of itself (because it is not in S).  This too is a contradiction.\nSo, S can neither be in nor not be in S.\n\nTo arrive at an example of Russell's Paradox, you need an instance of negated self-reference.
328490
It has dimension 2.\nI will not bother to prove that this subset is indeed a subspace except to note that (0,0,0) is an element.\nThis is most readily seen by constructing a basis.\nI propose that {(1,1,0),(0,1,1)} is a basis for this space.\nLet (a,b,c) be an element of this space. Then (a,b,c)=(a,a+c,c) =  a*(1,1,0) + c*(0,1,1)\nThis our proposed basis Spans this subspace.\nNow I think we can clearly see that our proposed basis is linearly independant since:\nIf:   a*(1,1,0) + c*(0,1,1) = (0,0,0)\nthen: 0 = a, 0 = a+c and 0 = c by equating the components.\nThus we have a linearly independant spanning set.\nThus we have a basis!\nSince our basis is of magnitude 2, our space must be of finite dimension 2.
224643
you arguments are good, however i swear that proving FLT isn't so easy . the best will that you yourself prove FLT in a single page checking out every arguments.\nafter all it took 350 yrs to find a proof !!\nkeep on trying ,good luck.
511344
Is there an infinite number of twin primes? (i.e. N and N+2 are both prime).\n\nFor other problems, see:\nhttp://mathworld.wolfram.com/UnsolvedProblems.html
335037
if the question is f(x)=(2^2n)-1\nthe answer is c,here is the explanation:\nfor all n>1: 2*n = even, (2^even = even) so 2^2n=even, (even -1= odd)\nthat means the term (2^2n)-1 gives an odd value for all n>1\nthen f(x)=odd for all n>1\nNow,\n  a) if n is an odd number, then F(x)  = odd (as we said before that f(x)=odd for all n>1)\nthen n-1=even,\nso f(x) isn't divisible by (n-1) as an odd number can't be divisible by an even one\n  b) if n is an even number, then F(x)  = odd (as we said before that f(x)=odd for all n>1)\nthen f(x) isn't divisible by (n) as an odd number can't be divisible by an even one\n  c) for all n\nF(x) = odd number that's divisible by 3 for all n>1 (you need to try different values for n, whatever n is big,the result Will be divisible by 3)\n\nanswer is c
421526
You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
560985
They are obviously countable.  Remember that the set of perfect numbers is a subset of the set of integers (and more specifically, natural numbers).  Since the set of integers is a countable set, and any subset of a countable set is still countable, the set of perfect numbers must obviously be countable.\n\nAfter rereading your question, I got a little better idea of what you are asking.  It seems that nobody knows if the set of odd perfect numbers is infinite (or even non-null) from the website www.oddperfect.org.  Therefore since I am one of these nobodies, I also don't know.  But I can tell you for certain (as stated above) that the set of odd perfect numbers is countable (I am using the definition of countable in the since that the set can be finite OR infinite).
103730
Hi there,\nWe will always be guaranteed a period of consecutive days when exactly 129 books are sold. The reason the number 129 is important is clear below. The proof uses what is known as the pigeonhole principle. [If n>m pigeons are put into m pigeonholes, then there exists a hole with more than one pigeon.] As usual, the trick is in constructing the "pigeons" and the "pigeonholes".\n\nLet B_i (for i=1,2,...,365) be the number of books sold during the period from the first day to the ith day inclusive (for example, B_3 represents the number of books sold in the first three days). Then we have:\n1 <= B_1 < B_2 < ... < B_365 = 600\nWe have strict inequalities inbetween as at least one book is sold each day. Now for the trick. Add 129 to this sequence of numbers to get:\n130 <= B_1 + 129 < B_2 + 129 < ... < B_365 + 129 = 729.\n\nWe choose B_1, B_2, ..., B_365, B_1 + 129, B_2 + 129, ..., B_365 + 129 as our pigeons. There are 730 of them, each being a positive integer between 1 and 729 inclusive. The numbers from 1 to 729 will be our pigeonholes. Thus by the pigeonhole principle, two of the numbers in our sequence of 730 numbers must be the same. As B_1, B_2, ..., B_365 are distinct, and B_1 + 129, B_2 + 129, ..., B_365 + 129 are distinct, one of the B_i's must be equal to one of the B_i + 129.\n\nSuppose that B_n = B_k + 129, where 1 <= k < n <= 365. Then we have B_n - B_k = 129. So between the (k+1)th day and the nth day inclusive, exactly 129 books were sold.\n\nCheerios,\nMike
132555
First of all, aleph_0 is the cardinality of the natural numbers (also, the cardinality of the integers, the rationals, and the algebraics). The cardinality of the real numbers is usually denoted by C (for the continuum). The continuum hypothesis is the conjecture that C=aleph_1. It turns out that this is independent of the usual axioms of set theory. In other words, whether C=aleph_1 or not is undecidable in the sense of Godel.\n\nEven if C=aleph_1, it is not provable that the collection of subsets of R (which is the same as the collection of geometric curves, suitably defined) is the same as aleph_2. So what you ask for is not only not proven, it is not provable. There is another hypothesis called the Generalized Continuum Hypothesis that states that the collection of subsets of a set with cardinality aleph_n has cardinality aleph_{n+1}. This result turns out to be independent of set theory also.
505962
Hello Rashej\n\nYes we can write that we can prove it by contradiction\nSuppose that our statement is not correct \nthus although 1^(2n+3) = 1^(3n+2)\n2n+3 - 3n+2 is not 0, for some n\nthus for some n , -n +5 , is not 0.\n\nFor n = 1 it is not 0 but clearly 1^(5) = 1^(5)\n\nThis contradicts our assumtion and thus the negation of our assumtion is valid.
152160
Aaaaah... semantics dressed up as logic! Niiice try...\n\nNow I fight back with philosophy...\n\n1. I can't be sure where I am. It (and you) could all be in my  imagination.\n\n2. But given the rather large nature of the Universe, it's as though I never move.\n\n3. Infinity has yet to be proved... and that's just the concept. Whether the Universe is infinite must come much, much later.\n\n4. You assume that infinity is a number when we haven't even proven that it is.\n\n5. Assuming that infinity was real, then the ladder to reach it would start at 1 (the first real number). But even 1 has value no matter how 'large' the value you compare it to. And since infinity is a concept, and math doesn't work with concepts, your last hypothesis is nonsensical in nature.\n\n--\n\nFor the other 'students' here:\nQED = quod erat demonstrandum ("that which was to be demonstrated").
225264
The statement that no such space exists is called the Continuum Hypothesis. It was hypothesized by Cantor, but he gave no proof of it. Godel showed that it is consistent with the usual axioms of set theory to assume CH. Then Paul Cohen showed that it is also consistent with set theory to assume CH is false. In other words, the Continuum Hypothesis is independent of the axioms of set theory. You can either assume it or assume its negation without introducing inconsistencies.\n\nThat said, there are mathematicians that are trying to find other axioms that would either imply CH or its negation. These other axioms would have to have some independent reason for including them in the system, of course. Typically, their attempts have to do with the existence of 'large cardinals'.