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Yes. I would say it’s culturally biased, just being a normal/good citizen of that society. I don’t think there is an ultimate meaning of life, just personal meanings of life. It’s a question I’ve been most concerned about.

Common themes might include not being on drugs, getting a fairly good job, looking attractive to the opposite gender, having a tidy house, not breaking the law of that country etc. I also think soldiers would be particularly proud of themselves for serving their country, but that’s down to the individual. Contrarily, being an unemployed drug addict, not having good personal hygiene, living in a very messy house and having a criminal record might leave you feeling none too proud of yourself. I don’t think your family would be too proud of you either, but anyone can change!

The revelation to me was that they don’t actually want to live like that, they want to be proud of themselves. I always thought drug addicts etc. don’t want to change sometimes, but if they’re truly honest with themselves they actually do. (According to the dictionary).

Why you get pleasure and satisfaction from doing what makes you proud is another question entirely. Who cares? If you know what you want to do in life the only problem is how you’re going to do it.

I thought the tool of logic was universal?

What are you saying then? You use the existence of things that don’t exist to prove things that do exist?

How does that work?

If A then B

No A

Therefore B

Wouldn’t your reasoning just be false then?

.........It’s very interesting.

**Primenumbers**- Replies: 3

According to the Cambridge English Dictionary:

**Proud **means feeling pleasure and satisfaction because you or people connected with you have done or got something good

**Satisfaction **means a pleasant feeling that you get when you receive something you wanted, or when you have done something you wanted to do

**Happy **means feeling, showing, or causing pleasure or satisfaction

**Enjoy **means to get pleasure from something

Let:

P=Proud

S=Satisfaction

L=Pleasure

W=Getting/Doing what you want

H=Happy

E=Enjoy

If P then L and S

If S then W

If L or S then H

Therefore If P then W

Therefore if P then H

Therefore If P then W and H

If E then L

If L then H

Therefore If E then H

So, if you’re proud of yourself you will have got what you wanted and you will be happy. As opposed to if you’re just enjoying yourself you will be happy but it’s not what you want to do.

So, the meaning of life according to the dictionary is to do good things, not necessarily good things like helping people or being a good person, but good things that make you feel proud of yourself, like achieving things or giving up drugs or getting a good job etc. Whatever makes you feel proud of yourself! It will make you happy and it’s what you truly want! It will also make people around you happy too because they’ll be proud of you too. And it’s what they truly want for you.

**Primenumbers**- Replies: 0

**Whatever done to a Mersenne number to get to zero must be able to be done to it’s factors. **

Mersenne Number= 2^11 -1

Factors = 23 and 89

2^11 -1 takes eleven goes to get to zero if I continually minus one and divide by 2. Watch what happens to 23 and 89 acting as remainders as I do the same to them.

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

8(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

They both take eleven goes to get to zero too.

Working backwards it can be seen that this must be the case. Starting with zero I multiply by 2 and add one continually until I get to 2^11 -1. The same must be done to the** remainders **which will then become 23 and 89 or zero, the factors of 2^11 -1.

Let the number of goes an odd number, y, takes to get to zero by continually minusing one and dividing by 2, (adding y when even) =z.

The z for 2^11 -1, 23 and 89 =11.

**Whatever z equals for y, 2^z -1 must be factorable by y. **When we use the method for 2^z -1, we never add 2^z -1 as it never equals an even number. This could potentially alter remainders for potential factors, but for Mersenne Numbers, this is not the case.

If z is the smallest Mersenne number that an odd number, y, will be a factor of.

Does anyone know what “z” a composite can’t be?

If z= a factor of (y-1), we know that 2^(y-1) -1 is factorable by y, as the process starts again for a multiple as seen above. So y is either a prime or a Fermat pseudoprime, which according to Wikipedia are rather rare.

Another way is to take any odd number, y, you want to know if it's prime. So count the number of times it takes you to get to zero by minusing 1 and dividing by 2, adding y if even, this number of times =z. If z= a factor of (y-1) then y is either a prime or a Fermat pseudoprime. Otherwise it is composite.

Thanks Bob

Yes you are right. 11 has z=10 which is not prime although is above (square root 11), so we could try using the method again, but we can't because 10 is even. So we can't determine if 11 is prime, and in turn 23 using this method. But if we know 11 is prime and above square root 23, we know 23 is prime, using this method.

The method does not prove that all primes are prime. You seem to have the method down, so I'll expand a little on the proof.

I forgot to mention that for mersenne Numbers, 2^p -1, z=p as you can see below;

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

So, as 2^10 -1 is factorable by 11 according to Fermats Little Theorem, I know that the z for 11 must be 10 or smaller. So as z's can never be bigger than their corresponding y's, if z is bigger than the square root I know that that exact number did not occur in any potential Factors below square root y. And if y has no factors below it's square root, I know it's prime. HOWEVER, if z is a multiple, y could be a factor of something with a smaller z. If you can imagine z being prime and when you get to z goes you turn it into zero and then it takes 2z goes to get to zero again, then 3z, 4z, 5z etc.

Example:

7 is a factor of 49

49 takes 21 goes to get to zero

7 takes 3 goes to get to zero

49's z= a multiple of 3

Therefore 49 has the potential to be factorable by other y's with z=3

7 has z=3, 7 has the potential to be a factor of 49

(0x2) +1=1 First go

(1x2) +1=3 Second go

(3x2) +1=7 Third go

7-7=0 Process starts again

(0x2) +1=1 Fourth go

(1x2) +1=3 Fifth go

(3x2)+1=7 Sixth go

7-7=0 Process starts again

(0x2) +1=1 Seventh go

.................

(3x2) +1=7 Twenty first go

They are only ever potential factors, which is why 11 can have z=10, which has the potential to be factorable by z=2 or z=5, the factors of 10.

Hope this helps......?

(I must admit I'm not very good at explaining things.)

**Primenumbers**- Replies: 4

THEORY

Take any odd number, y, count the number of times it takes you to get to zero by minusing 1 and dividing by 2, this equals, z. If this action results in an even number, simply add y. If z is a prime number >(square root of y), then y is prime. Example:

y=23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

Therefore z=11

z>(square root 23) AND z is prime THEREFORE 23 is prime.

Also if you don't know if z is prime , you can repeat the process!

PROOF

Take any odd number, y, minus 1 and divide by 2 until you get to zero, adding y as necessary when it results in an even number. The factor on the other hand will operate the same way. You can see that this must be the case if you work backwards. What is the remainder for n for 0/n? That's correct, zero. What have I done to y after that? I've multiplied it by 2 and added 1, so I must do the same to the remainder to get the new remainder. What happens when I add y? Well, you will effectively be adding a multiple of n, so you would end up just misusing them again to get the remainder. Eventually on the last number of goes I will end up with y and the remainder n, for y/n where n is a factor.

If y is a prime number the number of goes, z, will never be greater than (y-1). Proof of this is in the fact that, according to Fermats Little Theorem, if y is prime, 2^(y-1) -1 will be factorable by y. So y will always be a factor of 2^(y-1) -1 making z=y-1 or smaller, having occurred beforehand in a smaller z.

Therefore if z= a prime >(square root of y) no potential factor, ( primes below the square root of y) can possibly have the same z because they can't be that big.

**Primenumbers**- Replies: 0

Mersenne numbers are of the form 2^p -1.

If I minus 1 and divide by 2 it will take p goes to get to zero.

Example:

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

Similarly if 2^p -1 has a factor of the form, n, it should take p goes to get to zero. Only, if minus 1 divided by 2 results in an even number, I should be allowed to add n to the number to continue. As this is what happens to n, as a remainder for 2^p -1, as each go takes place.

Examples:

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

Both 23 and 89 take eleven goes to get to zero, as they must do, as they are factors of 2^11 -1, and 2^11 -1 takes eleven goes to get to zero also.

**Primenumbers**- Replies: 0

Let x= ANY ODD "+" INTEGER >3

Let y=(x/3) Rounded down

IF any of the below are TRUE then x is composite OTHERWISE x is prime.

x+1=EVEN triangular number

x+1+2= ODD triangular number

x+1+2+3=EVEN triangular number

x+1+2+3+4= ODD triangular number

x+1+2+3+4+5=EVEN triangular number

x+1+2+3+4+5+6= ODD triangular number

........etc.

Until x+1+2+3......+y

EXAMPLE

x=35

x/3=11+2/3 rounded down =11

Therefore y=11

x+1=35+1=36 36 is an EVEN triangular number

Therefore 35 is composite

EXAMPLE

x=17

x/3=5+2/3 rounded down =5

Therefore y=5

x+1=17+1=18

x+1+2=18+2=20

x+1+2+3=20+3=23

x+1+2+3+4=23+4=27

x+1+2+3+4+5=27+5=32

Therefore 17 is prime

**Primenumbers**- Replies: 2

The primes there are in a certain range can be estimated because there are;

1 No. not factorable by 2 in (2)

There are 2 No.'s not factorable by 2 or 3 in (6)

There are 8 No.'s not factorable by 2 or 3 or 5 in (30)

There are 48 No.'s not factorable by 2 or 3 or 5 or 7 or in (210)

times 48 by (prime -1) to get the next number of no.'s. i.e. =480 no.'s in (2310) not factorable by 2,3,5,7, or 11.....and so on.

So just work out the percentages and apply the correct one to your number x. For example you will not be concerned with No.'s not factorable by 13 if x<13 squared=169. This should be a far more accurate method than the Prime Number Theorem that approximates the number of primes upto x as x/log(x). You will need a computer to do this.

**Primenumbers**- Replies: 1

Triangular numbers are the sum of 1+2+3+4+5+6+7+8..........

a and b= triangular numbers where a>b

Any odd composite= a or (a-b)

This is because the sum of any group of numbers all separated by +1, I.e. 3,4,5,6,7 will be an odd composite with factors of the middle number and the length.

Examples:

2+3+4+5+6+7+8

Middle number=5

Length=7

Therefore=5*7=35

14+15+16+17+18+19+20

Middle number=17

Length=7

Therefore=7*17=119

116+117+118

Middle number=117

Length=3

Therefore=3*117=351

Therefore I would have thought we would be able to compute prime numbers faster by minusing the potential prime, p, off triangular numbers <p to see if they equal another triangular number. If they don't p is prime. This surely must be faster than seeing if p is factorable by all possible factors......................?

Does anyone know how people are using computers to test if very, very, VERY large numbers are prime?

**Primenumbers**- Replies: 0

p=Any number

A=2x3x5x7x11x13x17........ up to

m=Any multiple

A-pm= A number factorable by a factor of p

if p has one.To get A-pm, go through the primes, subtracting p as many times as you like.

Example:

p=129

A=2x3x5x7x11

2x3x5x7=210 210-129=81

81x11=891 891-(129x6)=117

A-129m=117

117 will have a factor the same as p if it has any.

117/129=39/43 129 is NOT prime (Common denominator =3)

In this way we can find out if p is composite without ever having to use a number

. Might be useful for computers.**Primenumbers**- Replies: 0

Basically I take my number p, square root it, Rd. Up to the nearest prime, then the next prime greater than that -1= largest prime gap below p.

Example:

p=130

Rd. Up to nearest prime= 13

Next prime after that = 17

17-1=16

Largest prime gap <130 = 16 (Correct)

This works because the greatest number of composites between two primes occurs when factors are not combined. So what could have been two composites is actually just one, like 15=3x5. To create the greatest possible number of composites I start at 2 not 0. 0 has an infinite number of prime factors, and so the greatest gap between the next repeat will occur after 0. Starting with the smallest composite which is NOT combined factors, I move up. Deleting all numbers factorable by primes less than the square root. The first time I attain TWO primes is when I reach the second prime after the square root. So this -1 is the gap required to create two non-composites with greatest possible occurrence of composites.

Largest prime gap <p =

Rd. Up to second nearest prime -1.**Primenumbers**- Replies: 7

A new discovery about prime numbers is that they are not truly random. Prime numbers that are not factorable by 2 or 5 only end in 1,3,7or 9. If prime numbers were truly random we would expect a prime number ending in 1 to be followed by another prime ending in 1 about 25% of the time but apparently using the first billion prime numbers the likelihood is more like 18%. This is a bias and they discovered biases for prime numbers ending in 3,7 and 9 as well.

Am I missing something here? Don't we know that there is a bias?

We know that any prime not factorable by 2, 3 or 5 = 30m + 1 or 7 or 11 or 13 or 17 or 19 or 23 or 29. Surely we can work out biases from this, for example a number ending in a number is never followed by another number ending in the same number. I.e. A number ending in 1 never follows another number ending in 1. This means the other numbers are more likely. And that's what Soundararajan and Oliver discovered; an 'anti-sameness' bias.

zetafunc wrote:

It isn't particularly difficult to understand -- I can explain it to you if you just tell me what it is that you need help with.

I get the beginning but not the end of it where you've put (-1/p)=-1.

zetafunc wrote:

Just to make sure, this is what you want to know, correct?

-You are considering only numbers of the form any non-zero positive integer.

-You would like to know something about the prime factorisations of these numbers.

First point is correct but I just wanted to show that by using

we can reduce the number of possible factors which should help compute primes faster. And now you have shown that will never be factorable by the primes of the form so thank you. I thought about 50% of primes could not possibly be factors but now you're saying that not only would there be 50%, but that there would be a bias to more than 50% due to Chebyshev's bias. Have I got this right?I am also still trying to work out if there's anything else we can use this scenario for to do with primes. If there was a wheel of prime factors that could factor

, it would be 20m + 1 or 9 or 13 or 17, not including 2 and 5.zetafunc wrote:

You can immediately rule out any primes such that whenever is not itself prime.

But after you have ruled out primes where

all that is left are p's For instance n=12 =145 rule out/divide by 5 and you get 29, a primezetafunc wrote:

You can also rule out primes

by Euler's criterion (or more specifically, by the corollary I mentioned in my previous post).

I don't get Euler's criterion, and probably never will it's just too advanced for me, don't worry. So what you're basically saying is we can rule out possible prime factors where p=4m+3 for

?zetafunc wrote:

may only be factorable by factors of where x<nPrimenumbers wrote:Potential factors for

may only be factored by previous values ofForgive my ignorance, but I have no idea what this means. Could you explain this?

zetafunc wrote:

Because.............

Let p= potential factor

Let p=n-y

(n-y)(n+y) =

Therefore remainder when is divided by p =And remainder when

is divided by p =That's correct. But I still don't know what a potential factor is (or what you need this part for).

y=n-x

zetafunc wrote:

y will be less than p because y=n-p and p>0

You mean y will be less than n, not p.

Yes

zetafunc wrote:

p will not be greater than n because we are only concerned with primes

So I am guessing by "potential factor" you mean "prime factor"? Also, assuming you want n^2 + 1 to be composite (which isn't true in general, as your next post shows), you need a ≤ sign.

Basically x<n because we are only concerned with factors below the square root, and because y=n-x, y cannot=n because x>0.

zetafunc wrote:

This should reduce potential factors for

You mean reduce the possibilities for the prime factors of n^2 + 1?

Yes

zetafunc wrote:

I have tried up to n=43. Prime factors <44 that were missing were 3, 7, 11, 19, 23, 31 and 43 which exactly halved the no. of potential prime factors. Any value for n to infinity will not be factorable by these no.'s.

then they will never occur because ones above p are just repeats of ones <p.

The factors repeat themselves every +p value for n. So if they don't occur beforeI still don't really understand what you are trying to do, from your post or from your examples in the post above. If you're wondering why 3, 7, 11, 19, 23, 31 and 43 don't appear in the prime factorisations of n^2 + 1, then this is because of quadratic reciprocity. (Let me know if you haven't come across this term before, and I'll be happy to explain it.) In other words, consider the congruence for p = 3 mod 4:

but this doesn't have any solutions, because if , then , where is the quadratic reciprocity symbol. (Which is just another way of saying that -1 isn't a square modulo p, where p is 3 modulo 4 -- in other words, the congruence has no solutions for p = 3 mod 4.)

Basically I am trying to make it so that instead of a computer having to try to factor all the primes below the square root for an integer, it only has to try and factor a lot less. Also because the factors repeat themselves this could make things easier. Maybe could do a wheel or something? No, I have never come across Euler' Criterion and I don't get it.

**Primenumbers**- Replies: 4

I was going to go to a Taekwondo class but then I found out online that martial arts are completely useless in a real fight apparently. And I wanted self-defence training.

As I thought about this I realised it was completely true..................

I am trained in Judo, (no punching and kicking), and my identical twin is one belt below black in karate. We had lots of fights when I was mentally ill and it would usually end in a standoff because we were both pretty evenly matched. I didn't use any of my throws or holds and that means my twin's brown belt in karate was useless against me who would actually be a white belt in karate. How amazing is that!

Apparently things like boxing, wrestling and MMA are better because you actually get hit and experience the real ruff-and-tumble of a real fight.

7 is prime so only factorable by 1 or 7.

the only squares less than 6 are 1 and 4 therefore a=1, r=2

a=7, r=0

73 is prime so only factorable by 1 or 73. 73 too big so must be 1. Only smaller than 73 is r= 1 or 2. 1 is too small so a=1, r=2.

Examples: [Formulas followed by prime factorization, as you can see factors 3,7,11,19,23,31 and 43 are missing]

n=1

n=2 5

n=3 10=2,5

n=4 17

n=5 26=2,13

n=6 37

n=7 50=2,5

n=8 65=5,13

n=9 82=2,41

n=10 101

n=11 122=2,61

n=12 145=5,29

n=13 170=2,5,17

n=14 197

n=15 226=2,113

n=16 257

n=17 290=2,5,29

n=18 325=5,13

n=19 362=2,181

n=20 401

n=21 442=2,13,17

n=22 485=5,97

n=23 530=2,5,53

n=24 577

n=25 626=2,313

n=26 677

n=27 730=2,5,73

n=28 785=5,157

n=29 842=2,421

n=30 901=17,53

n=31 962=2,13,37

n=32 1025=5,41

n=33 1090=2,5,109

n=34 1157=13,89

n=35 1226=2,613

n=36 1297

n=37 1370=2,5,137

n=38 1445=5,17

n=39 1522=2,761

n=40 1601

n=41 1682=2,29

n=42 1765=5,353

n=43 1850=2,5,37

**Primenumbers**- Replies: 9

New Theorem:

Potential factors for

Because.............

Let p= potential factor

Let p=n-y

(n-y)(n+y) =

Therefore remainder when is divided by p =

And remainder when is divided by p =

y will be less than p because y=n-p and p>0

p will not be greater than n because we are only concerned with primes

This should reduce potential factors for

and reduce computing space for finding primes.I have tried up to n=43. Prime factors <44 that were missing were 3, 7, 11, 19, 23, 31 and 43 which exactly halved the no. of potential prime factors. Any value for n to infinity will not be factorable by these no.'s.

The factors repeat themselves every +p value for n. So if they don't occur before

**Primenumbers**- Replies: 3

Fermat's Last Theorem states that no three positive integers a, b, and c satisfy the equation

for any integer value of n strictly greater than two.Let

where n >2Then

must equal:must have opposite remainders for

Adding y to one and minusing y from another will keep them opposite if they are already opposite. So let's see if the two squares have opposite remainders for .

If we add them together they should=0 if they are opposite.

No!

Fermat's Last theorem is True.