0
Prove that if every even natural number greater than 2 is the sum of two primes then every odd natural number greater than 5 is the sum of three primes?
Wiki User
∙ 14y ago
Given an arbitrary odd natural number greater than five, x, let y = x - 3, then y is an even number greater than 2. By assumption we have that y is the sum of two primes, say y1 and y2, but then x = y1 + y2 + 3 (which is the sum of three primes).
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
Why is 37 square root an irrational number?
The same reasoning you may have seen in high school to prove that the square root of 2 is not rational can be applied to the square root of any natural number that is not a perfect square.
How do you prove that the efficiency C arnot is greater than 1?
You don't. Such an efficiency can be less than 1, but it can't be greater than 1.
Are -14 and -23 negative integers prove your answer?
9
How do you prove that if a real sequence is bounded and monotone it converges?
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
Which number is greater 8.4 or 0.085?
8.4 is greater (larger).0.085 is less than 1. 8.4 is more than 1. 8.4 must therefore be larger.We can prove it is larger by dividing 8.4 by 0.085. If our answer is greater than 1 then it must be larger. This is true for any two positive numbers.8.4 / 0.085 = 98.824 (to 3 decimal places only). This is proof that we are correct.
Use the fact that there are infinitely many primes to prove that there are inifitely many non prime numbers?
For every prime number p greater than 2, p + 1 is composite.
Is there an infinite number of prime number?
Yes. To prove this, we must first assume the answer to be no. If there are a finite number of primes, there must be a largest prime. We'll call this prime number n. n! is n*(n-1)*(n-2)*...*3*2*1. n!, therefore, is divisible by all numbers smaller than or equal to n. It follows, then that n!+1 is divisible by none of them, except for 1. There are two possibilities: n!+1 is divisible by prime numbers between n and n!, or it is itself prime. Either way, we have proved that there are prime numbers greater than n, contradicting our initial assumption that primes are finite, proving that the number of primes is infinite.
What is the proof about multiplying twin primes and adding 1?
The question depends on what it is that you want to prove!
Is there a infinite number of prime numbers?
Yes. To prove this, we must first assume the answer to be no. If there are a finite number of primes, there must be a largest prime. We'll call this prime number n. n! is n*(n-1)*(n-2)*...*3*2*1. n!, therefore, is divisible by all numbers smaller than or equal to n. It follows, then that n!+1 is divisible by none of them, except for 1. There are two possibilities: n!+1 is divisible by prime numbers between n and n!, or it is itself prime. Either way, we have proved that there are prime numbers greater than n, contradicting our initial assumption that primes are finite, proving that the number of primes is infinite.
What is the highest prime number?
There is no highest prime number. Given any set of prime numbers, you can prove that there is at least one prime number that is not in that set. Here's how. First, recall that every natural number is either a prime number, or is a composite number and thus the product of some series of prime numbers. Now, suppose you have a set of n prime numbers, P1 .. Pn. Multiply them all together and add one. Call this number Q. Q is either prime or composite. If Q is prime, then it is obviously not one of P1 .. Pn, and thus is a new prime number not in that set. If Q is composite, then there must be a list of primes that evenly divide into it. Because Q is one greater than the product of P1 .. Pn, dividing by any of those primes will have a remainder of 1. So there must be some new prime, call it R, that divides evenly into Q. In either case, Q or R is a new prime. Now suppose that you have some potential highest prime. Enumerate all of the primes lower than it, and follow the above procedure with that set of primes. You will end up with a new prime not in that list. Since you have listed all primes less than your purported highest prime, any new prime number must be greater than your highest prime. Thus, there is no highest prime.
How do you prove that ten is a composite number?
All even numbers greater than 2 are composite.
Remainder of 4n where n is any natural number when divided by 6 is always 4. Prove this?
It is REM(4n/6)=4. Prove. Correction.
Largest integer that is not the sum of two or more different primes?
Well, the correct answer is 11. At least according to the link I added.Unfortunately, I can't find a proof why. Most theorems about primes run really deep, and it's often hard to find simple intuitive arguemts for questions just like this (the Goldbach conjecture is very similar to this, but has remained unsolved for several centuries.)This question is closely related to the concept of a complete sequence, which is a sequence which 'has enough' terms so that every positive integer can be expressed as the sum of elements in the sequence, without repeating elements.A classic proof about the prime numbers, called Bertrand's Postulate, demonstrates that there is always a prime number between n and 2n, for any positive integer n greater than 1.While I can't give the proof, Bertrand's postulate basically implies that there are 'enough' primes so that, if we include the number 1, the sequence of prime numbers forms a complete sequence.Of course, for this question, 1 is not included. So that means we don't have a complete sequence. However, apparently the sequence is 'complete' if we only consider numbers greater than 11. As the sequence of primes and 1 is complete, for any number there will be a set of unique primes and maybe 1 that add to this number. What the solution to this question says is that, for any number greater than 11, if the sum of primes and possible 1 which adds up to that number does in fact contain 1, then there will exist another sum of primes which does not contain 1. Perhaps we can take away the 1 and turn a 2 into a 3, leaving the same number as the sum. Or maye take away the 1 and add 7, then find a prime number that is in the sum that, if we take away 6, gives another prime number. To be honest I'm not quite sure how to prove this final little bit either...* * * * *Not the correct answer for the question that was asked - which specified two OR MORE primes. 11 = 2+2+7. If the Goldbach conjecture is correct the answer is 3 (nb: 1 is not a prime).
Why is there no direct proof of the infinity of primes naturals sqrt2 after 2000 years?
A direct proof of the infinity of primes would require what is essentially a formula to calculate the Nth prime number; such a formula isn't even guaranteed to exist. It's possible to formulate a proof of the infinity of primes that would be, in a sense, direct. A direct proof that the square root of 2 is irrational is impossible, because the irrational numbers aren't defined in any direct way - just as the real numbers which aren't rational. So to prove that the square root of 2 is irrational, we have to prove that it's not rational, which requires indirect techniques.
Why have there been no direct proofs of the intinity of primes and naturals after 2000 years of study?
Sorry, but both of these have direct and relatively simple proofs. To prove the infinity of primes, let p1, p2, p3, ... pn be the first n primes. Form the product p1 x p2 x p3 ... x pn. Clearly it is divisible by all the prime numbers up to pn. Then add 1 to that sum. The new value cannot be divisible by any of the preceding primes so it must be a new prime number. Because you can do this for any value of n, the number of primes must be infinite. The proof of the irrationality of the square root of 2 is similarly easy and should be available in any high-school math book at the level of Algebra 2 or above.
What did Avogadro's number prove?
no prove....
How will you prove that there will be many more prime numbers from a given set of primes?
As the question is phrased, it is not possible.You need to start with the set of all primes up to some value. Then, you can construct a new prime by multiplying all the numbers in the set together and adding 1.If you divide this new number by any number in the existing list, it will leave a remainder of 1. So none of them will divide it evenly. Since the number has no smaller prime factor, it is a prime.The need to have a complete list of primes up to some value can be seen by considering the set {3, 5}. The above method would then suggest 16, which is not divisible by 3 nor by 5. But it is not a prime, either.
