Math Is Fun Forum

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

A problem

I am trying to prove the set E={4n-1 , n=1,2,3,4....} contains infinite prime numbers.

odd numbers can be written as D={2n-1 , n=1,2,3.4} , E is bounded in D , D is infinite set , and so is E.
Prime numbers besides 2 , can only be divided by themselves and 1 , so Prim numbers besides 2 are odds
And the set A consists of Prime numbers(besides 2) is infinite , but How can I prove the intersection of A and E is infinite?
and How can I prove D has all the elements of E , though it's so obvious.

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Numbers are the essence of the Universe

Hayle

Guest

Re: A problem

i have a problem...:/:/:/:/:/:/:/:/
   
             I have a math problem that i dont understand..ok well it says write each fraction as a percent.  Us equivalent fractions if you can; if not use a over 100.....i have 3 over 500 and 500 wont go into 100..what do i do????

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A problem

How can I prove D has all the elements of E , though it's so obvious.

E={4m-1 , m=1,2,3,4...}
D={2n-1 , n=1,2,3,4...}

Let x be in E.  Then x = 4m - 1.  Let n = 2m.  Then x = 4m - 1 = 2n - 1.  Thus, x is in D.

The other part is pretty tricky.  Let me work on it a bit.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A problem

This is working on the assumption that there for every n in Z, there exists a p in P such that p > n and 4 does not divide p - 1.

Since 4 does not divide p - 1, p is not equivalent to 1 mod 4, and p + 1 is not equivalent to 2 mod 4.

Now it's obvious that p + 1 is not equivalent to 1 mod 4 or 3 mod 4, because then p would be even.  p is a prime, and with 2 aside, no primes are even.

Thus, it must be that p + 1 is equivalent to 0 mod 4.  That is, 4 divides p + 1.  Thus, there exists some integer m such that:

4m = p + 1

So 4m - 1 = p

Now going back to the original assumption, it should be really easy to prove.  Haven't attempted it though.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: A problem

Thanks , I think I have got a clue now ,  I am weak at the part about "mod" , by the way,Ricky,do you have any website that talks about mod , , Thanks

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A problem

Mod is not too hard of an idea.  It is unnatural at first however.  Because of this, you will probably (or at least I did) spend a lot of time playing mental gymnastics with it.  But it gets easier and more familiar the more you use it.

A mod B is simply the remainder when A is divided by B.

We state that two numbers are equivalent in mod n if their remainders are the same after being divided by n.

5 mod 3 = 8 mod 3 because the remainder of 5/3 is 2, and the remainder of 8/3 is 2.

However, to avoid being repetitive, we simply just right the mod on the right side of the equation:

5 = 8 mod 3

Note that all operations: addition, subtraction, multiplication, and division work just like they would if it was an equation.  So for example

5 = 8 mod 3 # Subtract 8 from both sides
5 - 8 = 8 - 8 mod 3
-3 = 0 mod 3

You can do similar things with the other operations listed above.  But I chose this one with a reason.  Simply put, if we have:

a = b mod n

Then we know that:

a - b = 0 mod n

So:

n | (a - b)

Because if a - b is equivalent to 0 in mod n, then it must be that there is no remainder, meaning that n divides a - b.  This simply means that there exists some integer k such that:

nk = a - b

And now we have an equation to work with.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: A problem

Thanks a lot ! Though it will take time to digest , It's very helpful!

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A problem

Now note that if we are using an equation in mod n:

a = b mod n

Then there are exactly n unique values a can have.  That comes directly from the division remainder theorem.  That is, the remainder after division must be somewhere in [0, n] when we are dividing by n.  I used this fact above, and found that p + 1 is not equivalent to 1, 2, or 3 in mod 4.  Thus, there is only one possibility left, p+1 must be equivalent to mod 0.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: A problem

Lol~ it still seems a little bit abstract for me to comprehence ~, I am working on it ~

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Numbers are the essence of the Universe