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#1 Re: Euler Avenue » Interesting proofs » 2011-01-13 02:45:03

Zhylliolom wrote:


There are infinitely many primes.


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 let A be the union of the progressions A[sub]p[/sub].  If the number of primes is finite, then A is a finite union of closed sets, hence closed.  But all integers except -1 and 1 are multiples of some prime, so the complement of A is {-1, 1} which is obviously not open.  This shows A is not a finite union and there are infinitely many primes.

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


be prime, and
. By the Fundamental Theorem of Arithmetic M has a prime factor which clearly must be greater than

#2 Re: This is Cool » Fun math for Eighth Grade » 2011-01-11 07:16:29

I like the Doomsday Algorithm. In case you don't know it, it allows you to compute the day of the week for any date in history. Understanding it requires a little modular arithmetic, but nothing too complicated. Kids love it - they will go around telling their friends the day that they were born afterwards. A quick internet search will provide all the notes you need.

#3 Re: Help Me ! » Intersection between two hyperplanes » 2010-12-31 05:20:55

It might help if you post a specific example you're having trouble with.

#4 Re: Formulas » Russian Multiplication » 2010-12-30 03:31:34

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).

#5 Re: This is Cool » Solve the paradox of set theory V7.2 » 2010-12-27 01:09:32

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.

#6 Re: Help Me ! » Urgent help with cofunction identities » 2010-12-27 00:55:44

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

#7 Re: Help Me ! » Assignment T_T Need help » 2010-12-13 23:51:03

is always a subset of
: If
is in
, then
for some
. So
. So
is in both
, whence

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.

#8 Re: Help Me ! » Help with probability! » 2010-12-10 01:44:03

5) There are 5! = 5*4*3*2*1 = 120 ways to arrange the washers. Two of these ways will be in size order (smallest to largest, and largest to smallest). So the answer is 2/120 = 1/60.

#9 Re: Help Me ! » variable expressions and square roots » 2010-12-09 00:09:41

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).

#10 Re: Help Me ! » Combinatorics - Math :) » 2010-12-03 00:13:36

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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