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How can you prove that a perfect number can never be a prime number?
Wiki User
∙ 13y ago
In a perfect number, the sum of all the factors (including the number itself) is twice the number. E.g., the sum of the factors of 6 is 1 + 2 + 3 + 6 = 12 (equal to 2 x 6).
Every Prime number has two factors: 1, and itself. So, the sum of the factors is only one more than the prime number itself; for any number greater than 1, this can't be twice the number. For example, the prime number 7 has the factors 1 and 7, which add up to 8.
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∙ 13y ago
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How do you prove that 540 plus 219 is a Prime Number?
540 + 219 = 759 759 is evenly divisible by 3, 11, and 23, so it is not a prime number.
How can you prove that the square root of three is an irrational number?
Because 3 is a prime number and as such its square root is irrational
Is 2011 prime?
Yes. To prove it, you can take the square root of 2011 (about 44.8), and divide 2011 by all the prime numbers under this number. You'll see none of them divide, so it's prime.
How can you prove there is no largest prime number?
No. No matter how large of an example you choose, someone always can find a larger number (of any kind), because the upper range of number is infinite. If you take all the known prime numbers and multiply them together, then add 1 to the result, you will have a number that is not divisible by any of the known prime numbers. This number will either be prime or have prime factors that were not previously known. So, in this way, you can always find a new prime number or a number that is a multiple of new prime numbers. If the known prime numbers include all the prime numbers up to the largest known, the new ones must be larger.
How do you find out perfect numbers?
First let us define a perfect number. We call a number perfect if it is a positive integer and is the sum of all it proper divisors. The proper divisors part means we do not include the number itself even though any number divides itself. So for example 6 is perfect because it is a positive integer and 1+2+3=6. Not we excluded the number 6 even though 6 does divide 6.Now we could look at any number and find all the divisors of the number except the number itself. Next add them all and if the sum is equal to the number itself. If so then that number is a perfect number. While this method will always work, it is quite tedious.Euclid found another way thousands of years ago. He discovered that you could find the first four perfect numbers by using the formula 2p-1(2p − 1) where p is a prime such as 2,3,5 or 7. So let us consider the first prime 2 and plug it in Euclid's formula. 22-1 (22 -1)is equal to 21 (4-1)=2x3=6 which is the first perfect number. It is important to notice that the (2p − 1) portion is always giving us a prime number. The perfect number we would obtain with the primes 2,3,5, and 7 are also even. For example 23 -1=7 which is prime and we multiply that by 22 so 4x7=28 which is perfect and even.Everything was great until we tried the 5th prime which is the number 11. When we plug that into the (2p − 1) part just as we did 2 and 3, we get the number 211 − 1 = 2047 = 23 × 89 so it is not a prime number like the values we get with 2,3,5, and 7. Why do we care about the fact the this is not prime? We were looking for perfect numbers not primes?Because when we multiply 2047 by 211-1 we find out the the number we get is not perfect.So Euler was correct only for those first few values.When (2p − 1) is prime it has a special name. It is called a Mersenne prime named after Marin Mersenne who was a monk in the 17th ( a prime number by the way) century who studied math and primes and a field known as number theory. In order for (2p − 1) to be prime is has been found that it is necessary but not sufficient for p to be prime. In math a necessary condition is one that must hold true for the statement to be true. So p must be a prime. However a sufficient condition is one that says if this is satisfied the statement is always true. So the prime 11 showed us the (2p − 1) is not always prime and this means the condition is not sufficient. However, if p is not a prime (2p − 1) will not be prime either.A century later, Euler proved that all perfect numbers are generated by (2p-1 )(2p − 1).Euclid has tried but could not prove it. We can find the first 39 perfect numbers by using the primes: 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917Note that all these number are even. We do not know if there are any odd perfect numbers. It has been shown that there are no odd perfect numbers in the interval from 1 to1050 but we can't say in general.We don't even know how many perfect numbers or Mersenne primes there are. While we suspect the number is infinite, that has not been proven yet.So here are the first few perfect numbers and you will see they become quite larger very fast!6,28,496,8128,33550336,8589869056,137438691328,2305843008139952128,2658455991569831744654692615953842176,191561942608236107294793378084303638130997321548169216,13164036458569648337239753460458722910223472318386943117783728128,14474011154664524427946373126085988481573677491474835889066354349131199152128,23562723457267347065789548996709904988477547858392600710143027597506337283178622239730365539602600561360255566462503270175052892578043215543382498428777152427010394496918664028644534128033831439790236838624033171435922356643219703101720713163527487298747400647801939587165936401087419375649057918549492160555646976,141053783706712069063207958086063189881486743514715667838838675999954867742652380114104193329037690251561950568709829327164087724366370087116731268159313652487450652439805877296207297446723295166658228846926807786652870188920867879451478364569313922060370695064736073572378695176473055266826253284886383715072974324463835300053138429460296575143368065570759537328128,54162526284365847412654465374391316140856490539031695784603920818387206994158534859198999921056719921919057390080263646159280013827605439746262788903057303445505827028395139475207769044924431494861729435113126280837904930462740681717960465867348720992572190569465545299629919823431031092624244463547789635441481391719816441605586788092147886677321398756661624714551726964302217554281784254817319611951659855553573937788923405146222324506715979193757372820860878214322052227584537552897476256179395176624426314480313446935085203657584798247536021172880403783048602873621259313789994900336673941503747224966984028240806042108690077670395259231894666273615212775603535764707952250173858305171028603021234896647851363949928904973292145107505979911456221519899345764984291328You can find every know Mersenne prime and hence every know perfect number on a site called GIMPS. I have included a link. But even more than seeing them, you can help discover the next one and become quite famous!Now this nice answer on perfect numbers will conclude with an important theorem.If 2k-1 is a prime number, then 2k-1(2k-1) is a perfect number and every even perfect number has this form. I will include a proof by Chris Caldwell. It uses the sigma function which is a function that finds the sum of the divisors. I will give link for the sigma function for those who are interested.Proof: Suppose first that p = 2k-1 is a prime number, and set n = 2k-1(2k -1). To show n is perfect we need only show sigma(n) = 2n. Since sigma is multiplicative and sigma(p) = p+1 = 2k, we knowsigma(n) = sigma(2k-1)(sigma(p)) = (2k-1)2k = 2n.This shows that n is a perfect number.On the other hand, suppose n is any even perfect number and write n as 2k-1m where m is an odd integer and k>2. Again sigma is multiplicative sosigma(2k-1m) = sigma(2k-1)(sigma(m)) = (2k-1)(sigma(m)).Since n is perfect we also know thatsigma(n) = 2n = 2km.Together these two criteria give2km = (2k-1)(sigma(m)),so 2k-1 divides 2km hence 2k-1 divides m, say m = (2k-1)M. Now substitute this back into the equation above and divide by 2k-1to get 2kM = sigma(m). Since m and M are both divisors of m we know that2kM = sigma(m) > m + M = 2kM,so sigma(m) = m + M. This means that m is prime and its only two divisors are itself (m) and one (M). Thus m = 2k-1 is a prime and we have prove that the number n has the prescribed form.It is even easier to prove that if for some positive integer n, 2n-1 is prime, then so is n.So this question about perfect numbers has generated some interesting discussion of Mersenne primes and included some exciting number theory. This is a beautiful topic in math that is often not covered.
Why would you divide 61 using prime numbers only?
To prove that 61 is a prime number.
How do you prove that 540 plus 219 is a Prime Number?
540 + 219 = 759 759 is evenly divisible by 3, 11, and 23, so it is not a prime number.
Is the square root of 11 a whole number?
It is true that the square root of a prime number like 11 is never a whole number. But to say that that has never been proven is incorrect. The square root of any positive integer that is not a square number (the square of an integer) is always irrational, and that is relatively easy to prove. To prove that prime numbers are not square numbers is even easier. That is basically true by definition. If a number greater than 1 were a square number, its square root would be a factor other than 1 and itself; therefore, it would not be a prime number.Answer 1No - the square root of 11 is not a whole number. 11 is prime so it has no factors except itself and 1, anyways.Any prime number has no square root that is a whole number or integer. (That postulate has not been proven, but it has not been disproven so it is accepted as true.)
How can you prove that the square root of three is an irrational number?
Because 3 is a prime number and as such its square root is irrational
How can you prove that 57 and 111 are not a prime number?
57 = 3 x 19 111 = 3 x 37 Thus they are not prime numbers
How prove that 3208 is a prime number?
Show that only itself and one divide into it. Seriously how old are you? this is easy ...
How do you prove keaghan prime number theory wrong?
Unfortunately, there is no evidence of someone called "Keaghan" (or something similar) who proposed a prime number theory. I suggest you check your spelling and resubmit the question.
Is 2011 prime?
Yes. To prove it, you can take the square root of 2011 (about 44.8), and divide 2011 by all the prime numbers under this number. You'll see none of them divide, so it's prime.
Is 1637143983 prime or composite?
1637143983 is not prime. Its prime factorisation is: 3 × 3 × 7321 × 24847.In the case of this number and one ninth of all integers, the fact that it is composite can be quickly determined by calculating the core digit value, the sum of all digits reduced to one digit. The core digit value of this number is 9, and every single integer with this core digit value is a multiple of 9. This can only prove that a number is composite. A core digit value other than 9 does not prove that a number is prime.
Is 216 prime or composite?
It is composite. It ends in a 6, which is an even number, and the only even prime is 2. You can prove it to be composite by simply dividing it by two.216/2=108
What is the past perfect tense of prove?
Had proven.
How can you prove there is no largest prime number?
No. No matter how large of an example you choose, someone always can find a larger number (of any kind), because the upper range of number is infinite. If you take all the known prime numbers and multiply them together, then add 1 to the result, you will have a number that is not divisible by any of the known prime numbers. This number will either be prime or have prime factors that were not previously known. So, in this way, you can always find a new prime number or a number that is a multiple of new prime numbers. If the known prime numbers include all the prime numbers up to the largest known, the new ones must be larger.
