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Zhylliolom wrote:

Theorem:

There are infinitely many primes.

Proof:

Define a topology on the set of integers by using the arithmetic progressions (from -∞ to ∞) as a basis. It is easy to verify that this yields a topological space. For each prime p let

A[sub]p[/sub] consist of all multiples of p.A[sub]p[/sub] is closed since its complement is the union of all the other arithmetic progressions with difference p. Now letAbe the union of the progressionsA[sub]p[/sub]. If the number of primes is finite, thenAis a finite union of closed sets, hence closed. But all integers except -1 and 1 are multiples of some prime, so the complement ofAis {-1, 1} which is obviously not open. This showsAis not a finite union and there are infinitely many primes.

Here is a much more elementary proof that there are infinitely many primes:

Let

be prime, and . By the Fundamental Theorem of Arithmetic M has a prime factor which clearly must be greater than .It might help if you post a specific example you're having trouble with.

Very interesting. I believe that you should include the original numbers you're multiplying by in the table (in this example they are crossed out because 76 is even).

Without giving away the details to others that want to prove this, the solution involves powers of 2.

**Challenge problem:**This technique can be generalized by using powers of n for any other fixed positive integer n>2. How would the algorithm change in this case? (Hint: you need one additional column).

If I'm understanding correctly, then what you're saying is that the limit of P_n(R) is a proper class.

This simply isn't true. The limit of a countable sequence of cardinals is a cardinal. In particular, this limit is a set (it's just a countable union of sets which is a set by the union axiom).

You can continue to get much larger sets than this by continuing to take power sets.

cos(x-3pi/2)=cos x cos 3pi/2 + sin x sin 3pi/2 = (cos x)(0) + (sin x)(-1) = 0 - sin x = -sin x

Now assume f is injective. We show f(A) intersect f(B) is a subset of f(A intersect B): If y is in f(A) inter f(B), then y is in f(A) and y is in f(B). So y=f(x) for some x in A, and y = f(z) for some z in B. Since f is injective, x=z. Thus x is in A inter B, and f(x) is in f(A inter B)

Now assume f (A) intersection with f (B) = f (A intersection with B) for all A, B. We show f is injective: Suppose f(x)=f(z)=y. Let A={x}, B={y}. Then f(A inter B)=f(A) inter f(B) = {f(x), f(z)}={y}. So A inter B is not empty. Thus x = y. So f is injective.

Rewrite the expression as

.Now take the square root of each factor.

.You can also simplify the original expression before taking the square root as follows:

.Remark: In general

but we can eliminate the absolute values because x is given to be positive (and similarly for y).Here's something to start you off:

P(6)=3P(1), P(5)=2P(2), P(4)=P(3). Let x = P(1), y = P(2), z = P(3). Since probabilities add to 1, we have

x + y + z + z + 2y + 3x = 1, equivalently 4x + 3y + 2z = 1

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