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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.
Numbers are the essence of the Universe
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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????
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.
"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..."
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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.
"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..."
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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
Numbers are the essence of the Universe
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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.
"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..."
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Thanks a lot ! Though it will take time to digest , It's very helpful!
Numbers are the essence of the Universe
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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.
"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..."
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Lol~ it still seems a little bit abstract for me to comprehence ~, I am working on it ~
Numbers are the essence of the Universe
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