let Q= the set of all rational numbers  and N= the set of all natural numbers.....?
360581
One way is to construct the rational numbers so that each is written as a fraction that does not need to be simplified.  \n\nBegin with all fractions whose numerator and denomenator sum to 2.  There is only one such fraction (1/1).  Map this fraction to the integer 1\n\n1/1 --> 1\n\nThen consider all fractions whose numberator and denomenator sum to 3.  There are two such fractions:  2/1 and 1/2.  Map these two fractions to the next available integers:\n\n2/1 --> 2\n1/2 --> 3\n\nThen map the three fractions whose numerators and denominators sum to 4 to the next available integers:\n\n3/1 --> 4\n2/2 --> 5\n1/3 --> 6\n\nAnd so forth.\n\nThis algorithm will map all of the rational numbers onto the set of natural numbers.  However, it is a many-to-1, not a 1-to-1 mapping (it maps 1/1 to 1 and also maps 2/2 to 5, but 1/1 and 2/2 have the same value as rational numbers).  You could get around this by skipping over rational numbers that are not simplified fractions.
295853
It would be countable and uncountable *sets*, not numbers.\n\nWe say that a set is countable if it is in one-to-one correspondence with some subset of the set of natural numbers (1,2,3,...). The collection of rational numbers turns out to be countable.\n\nWe say that a set is uncountable if it is not countable. ;) The set of real numbers turns out to be uncountbale, but there are many, many sets that are even larger.\n\nWe say that a set A is a subset of the set B if every element of A is in B, but not necessarily the other way around. So the set of natural numbers form a subset of the set of real numbers.\n\nA tensor is a multilinear map on a vector space. Way too much theory for this space.\n\nGiven a *function* f, an inverse function g, is one where the compositions fog and gof are both the identity map.\n\nThe gap between +ve and -ve infinity makes no real sense unless you simple mean the set of real numbers.
115808
Since B is a countable set, one can also associate it with the even and odd natural numbers.  Without loss of generality, map the possibilities of 0->b and 1->b, where b is in B, to the even and odd numbers, respectively.  The union of the even and odds is just the natural numbers, which is countable.
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If A is a subset of C and A has the same cardinality as C, then the sets A and C are the same.\n\nI guess as a proof, you could assume that there is an element in C that is not in A...in which case the cardinality of C would be greater than the cardinality of A, which is a contradiction.\n\nThen B gets trapped in the middle.  A = B = C.
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One question in number theory most likely :\n\nfor  all integers a, b, c and n,\n\nprove that :  \n\na^n + b^ n does not equal c^n for n > =3
100604
anything like, n.0... , n.1..., n.2..., n.3..., n.4... will be $n.\n\nanything like, n.5... , n.6..., n.7..., n.8..., n.9... will be $n+1.
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
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Consider the numbers\nA1,\nA1+A2,\nA1+A2+A3\n..\nA1+A2+A3+...+An.\n\nDo the computations modulo n. If none of these is divisible by n, then we have n numbers and n-1 non-zero residue classes (mod n), so there are two of these numbers which are equivalent (mod n) by the pigeon hole principle. But then subtracting gives a sum of the required form that is 0 (mod n), in other words, so that n divides that sum.\n\nLooking at the proof, there is no need for the sequence to be monotonic.
778163
The problem is to show that f is identically 0 using the density of the rational numbers. I would strongly suggest to the asker that they do their own homework at this level. If you don't understand the questions or techniques, you will have a *lot* of trouble later on.
425287
No. A rational number is one that can be written as p/q where p and q are integers. Zero can be written as 0/1. In fact,every integer is rational.
2302
(1) = x\n(2) = / \n(3) = x
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
939187
First, evaluate these functions to understand what they mean. Here is a listing of n and f(n)\n\n0 1 2 3 4 5 6 ...\n0 2 4 6 8 10 12 ...\n\nHere is a listing of n and g(n):\n\n0 1 2 3 4 5 6 ...\n0 0 1 1 2 2 3 ...\n\nEvaluate g • f using g • f (n) = g(f(n)) to get its listing:\n\n0 1 2 3 4 5 6 ...\n0 1 2 3 4 5 6 ...\n\nwhich provides evidence for question (a). g(f(n)) = g(2n) = n, because 2n is even.\n\nThe listing of f • g is\n\n0 1 2 3 4 5 6...\n0 0 2 2 4 4 3 ...\n\nAnalyze f(g(n)) = 2 [n/2]. What happens when n is even, and when n is odd?\n\nA function, F, is injective if different values never map to the same value. Or, the listing does not have duplicates. To prove that F is not injective is easy: just find a counterexample, that is a and b such that F(a) = F(b), but a ≠ b. To prove that F is injective you must show that in general if F(x) = F(y) then x = y. \n\nLook at the listings. If a listing has duplicates then the function is not injective. A proof is just providing a counterexample. Example: f • g has duplicates; the simplest are 0 and 1: f • g(0) = f • g(1), and 0 ≠ 1.The listing of f has no duplicates. To prove that f is injective, prove that if 2x = 2y then x = y. That should be easy!\n\nA function is surjective if the listing contains every number in the domain, that is all the numbers: 0, 1, 2, 3,...\n\nThe listing of f does not contain 1. So f is not surjective. A proof is to show that there is no n such that f(n) = 1. That's easy because f(n) is even!\n\nThe listing of g appears to contain all the numbers. Does it contain 2006? Can you find n such that g(n) = 2006? Such that n/2 = 2006? Yes, n = 4012. Similarly, for any number (instead of 2006).\n\nSo, I have provided some strategies and examples to let you now make your own proofs.
864701
Prove by induction.\n\nWhen k=1, then (n+a)^1 = n+a = Θn, true\n\nAssume true for k.\n\nFor k+1,\n(n+a)^(k+1) = (n+a) * (n+a)^k = Θn * Θ(n^k) = Θ(n^(k+1))\n\nTherefore it holds for all k>0
360700
a = 1, b = 2, n = 3:\n\n1^2 = 1 = 4 = 2^2 (mod 3) but clearly 1 != 2 (mod 3)
64293
No, it is not true that is e^n a transcendental number for all values of n.\n\nConsider the value n=0.\n\ne^0=1, and 1 is a rational number.\n\nSome mathematicians include zero in N, the set of natural numbers, and some do not. If you exclude zero as a possible value for n, then the answer to your question will be different.
1152431
1) Yes\n2) No\n3) Yes\n4) WTH? This is a contradiction\n5) Yes...by definition\n6) Square root of four is rational