Help What is wrong with this proof?

[mathew3]

An Arithmetic Solution to the Goldbach Conjecture
Prove that any and every even integer >4 may be expressed as the sum of at least some 2 prime integers.

Proof D.
1. We may regard prime integer multiplication as being equivalent to prime integer summation, i.e. 2x3=2+2+2
2. Therefore, given the Fundamental Theorem of Arithmetic, any and every even integer >4 may be expressed as the summation of a series of prime integers, i.e.,
I = Pa+…+Pb+…+Pc+…
where I is any integer >1 and P is some prime integer
3. Any and every even integer must equal the summation of some two odd integers, therefore
E = Oa +Ob
where E is any and every even number and O is some odd integer.
4. Given the FTOA it must also be the case that for any even integer >4

E = Pa+…+Pb+…+Pc+…

5. Therefore for any and every even integer >4

Pa+…+Pb+…+Pc+…= E = Oa +Ob [4]
6. Therefore the sum Oa + Ob must equal a summation of a series of primes.
7. Since there are at least two addends comprising the Oa +Ob summation then each addend is allowed to be a prime number.
8. E, in this case, must meet two conditions:
a. E must be composed of 2 and only 2 odd integers.
b. E must be a summation of primes.
9. In order to satisfy both conditions a and b then it must be the case that the two odd integers, Oa and Ob must sum as primes where

Pa+Pb = E= Oa +Ob [5]
10. Therefore any and every even integer >4 may be expressed as the sum of at least some 2 prime integers.

 

[pwsnafu]

9 does not follow from 8.

Eg. Let's do 12. 12 = 2² times 3 = 2² + 2² + 2² = 2 + 2 + 2 + 2 + 2 + 2.

Now write 12 as a sum of two odd integers. I can choose anything. 12 = 9+3.
So
2 + 2 + 2 + 2 + 2 + 2 = 9 + 3

If 9 is to be true we get 2 = 9 = 3, which is nonsense.

 

[mathew3]

Your reply is of course true on its face. However, it seems that this implies that 8 is false or at least the two conditions cannot exits simultaneously. Otherwise 9 has to follow, or so it seems to me. So why is 8 wrong?

 

[pwsnafu]

Here is 8a.

E must be composed of 2 and only 2 odd integers.

12 = 3 + 3 + 3 + 3

Edit: Further, take a closer look at what I previously wrote:
12 = 2 + 2 + 2 + 2 + 2 + 2. This satisfies 8b, but not 8a.
12 = 9+3. This satisfies 8a, but not 8b.
Why should 9 follow 8?

 

[mathew3]

Again you are correct but your cited case doesn't apply in this case. Only the 2 integer set is under consideration. eg.
1. Fact: I have 1 dollars worth of change in my pocket. That could be 100 pennies or 10 dimes... up to or 1 silver dollar. (E)
2. Fact: I only have two coins in my pocket. (Oa +Ob)
3. Conclusion: Each coin is a half dollar (Pa+Pb=Oa +Ob)

 

[pwsnafu]

Again you are correct but your cited case doesn't apply in this case.

Please quote where precisely you require the two integer set as a requirement.

Edit: look, Steps 1, 2 and 4 talk about one possible way to write out the sum, in this case every term is prime. Step 3 describes a different sum, in this case only two odd numbers. At no point in Step 3 do you require primality, and at no point in 1 do you require the number of terms to be two. You can't just say at Step 9 "oh, they were the same sum all along".

 

[mathew3]

step3

 

[pwsnafu]

Bleh, I was editing my post while you posted your reply.

Here is Step 3 again

3. Any and every even integer must equal the summation of some two odd integers, therefore
E = Oa +Ob
where E is any and every even number and O is some odd integer.

 

[mathew3]

Please quote where precisely you require the two integer set as a requirement.

Edit: look, Steps 1, 2 and 4 talk about one possible way to write out the sum, in this case every term is prime. Step 3 describes a different sum, in this case only two odd numbers. At no point in Step 3 do you require primality, and at no point in 1 do you require the number of terms to be two. You can't just say at Step 9 "oh, they were the same sum all along".

I find myself agreeing with most of your statements. I wholeheartedly agree with you up to your last sentence. I'm pretty sure equation [4] is pretty much irrefutable. The question is is there anything that requires that the the number of prime addends must be greater than 2. It seems to me equation [4] indicates it is true, covering a range of 2 to 300,000 (Schnirelmann?) prime addends. The conjecture doesn't say 5 primes or ten primes or a thousand primes but 2 primes. Equation [4] certainly allows this. Equation [5] states this.
Again, [5] doesn't demand that the two odds must always and exclusively be primes.(Neither does the conjecture) But it does demand that there must always be a summation of primes that equals the summation of the two odds. Here I bring up the loose change analogy once again.

 

[pwsnafu]

Again, [5] doesn't demand that the two odds must always and exclusively be primes.(Neither does the conjecture). But it does demand that there must always be a summation of primes that equals the summation of the two odds.

*facepalm*

From http://en.wikipedia.org/wiki/Goldbach%27s_conjecture" :

Every even integer greater than 2 can be expressed as the sum of two primes.

Now that I look back at your first post

Prove that any and every even integer >4 may be expressed as the sum of at least some 2 prime integers.

I should have picked that up on the first read.

 

[mathew3]

Oh, by the way. The greater than 2 holds if we consider 1 as a prime, which was the case when the conjecture was first presented. The greater than 4 holds if 1 is considered as non prime which is the case in modern number theory.

 

[ramsey2879]

I find myself agreeing with most of your statements. I wholeheartedly agree with you up to your last sentence. I'm pretty sure equation [4] is pretty much irrefutable. The question is is there anything that requires that the the number of prime addends must be greater than 2. It seems to me equation [4] indicates it is true, covering a range of 2 to 300,000 (Schnirelmann?) prime addends. The conjecture doesn't say 5 primes or ten primes or a thousand primes but 2 primes. Equation [4] certainly allows this. Equation [5] states this.
Again, [5] doesn't demand that the two odds must always and exclusively be primes.(Neither does the conjecture) But it does demand that there must always be a summation of primes that equals the summation of the two odds. Here I bring up the loose change analogy once again.

I don't get what you are saying. The Goldbach conjecture required that every even number greater than 4 be the sum of two odd primes. You haven't proven that by statements 8 and 9 since a string of more than 2 primes does not equate to a string of 2 primes.

 

[mathew3]

Fact: every even number>4 must be the sum of some 2 odd integers
Fact: every even number >4 must be the sum of at least 2 primes

Before I go further are we agreed on these 2 facts?

 

[ramsey2879]

Fact: every even number>4 must be the sum of some 2 odd integers
Fact: every even number >4 must be the sum of at least 2 primes

Before I go further are we agreed on these 2 facts?

yes but 1) what do you mean by "some 2 odd integers?" and 2) how do you get to prove the Goldbach conjecture from that?

 

[mathew3]

Great. We are agreed on those two, I think. It is critical to the proof that we establish than any and every even number greater >4 is the sum of some 2 odd numbers. I don't know the theorem offhand but this has to be true.

The proof proceeds in 4 critical steps
a. establish that any even integer>4 can be composed of at least 2 prime addends; thus step 2.
b. establish that any even integer>4 can be composed of 2 odd integers; thus step 3.
c. equate a and b. Thus step 5, equation [4]
d. By considering the special case where our number of odd integer addends is restricted to 2 and only 2 addends then this imposes steps 9, equation [5]. Again I refer you to the previous loose change analogy.

Thus any even integer>4 can be expressed as the sum of two primes.

The proof is remarkably simple. The problem seems to be is it logically legitimate to advance from equation [4] to equation [5]. I think it is but I would of course like to hear reasons to the contrary.

 

[ramsey2879]

Again you are correct but your cited case doesn't apply in this case. Only the 2 integer set is under consideration. eg.
1. Fact: I have 1 dollars worth of change in my pocket. That could be 100 pennies or 10 dimes... up to or 1 silver dollar. (E)
2. Fact: I only have two coins in my pocket. (Oa +Ob)
3. Conclusion: Each coin is a half dollar (Pa+Pb=Oa +Ob)

You have Pa + Pb + ... Pe = Oa + Ob , How do you get from that to Pa + Pb = Oa + Ob? Step 5 does not do that.

 

[mathew3]

You have Pa + Pb + ... Pe = Oa + Ob , How do you get from that to Pa + Pb = Oa + Ob? Step 5 does not do that.

By restricting our numbers of addends to be 2 and only 2 addends then the number of primes are necessarily restricted to 2 primes thus Pa + Pb = E= Oa + Ob.

 

[ramsey2879]

By restricting our numbers of addends to be 2 and only 2 addends then the number of primes are necessarily restricted to 2 primes thus Pa + Pb = E= Oa + Ob.

You can't do that without first establishing there corresponds at least one set of 2 prime addends that add up to each possible value of Oa + Ob. Thus you have no proof of the conjecture.

 

[pwsnafu]

Again I refer you to the previous loose change analogy.

Except that it's not an analogy. Or at least an analogy relevant to the problem. And anytime someone resorts to analogies rather than straight logic to prove mathematics, it's time to walk away.

The "loose change analogy" relevant to this thread is:
1. Fact: I have 1 dollars worth of change in a pocket. That could be 100 pennies or 10 dimes... up to or 1 silver dollar. (E)
2. Fact: I only have two coins in a pocket. (Oa +Ob)

At no point do you say the change is in the same pocket. See post #6.

 

[mathew3]

You can't do that without first establishing there corresponds at least one set of 2 prime addends that add up to each possible value of Oa + Ob. Thus you have no proof of the conjecture.

Step 3 and the two equal signs allows me to place this restriction.

 

[pwsnafu]

Step 3 and the two equal signs allows me to place this restriction.

Proof required.
Even if it was true, that's extremely non-trivial.

 

[mathew3]

Except that it's not an analogy. Or at least an analogy relevant to the problem. And anytime someone resorts to analogies rather than straight logic to prove mathematics, it's time to walk away.

Touche.

All primes greater than 2 are odd.
The proof necessarily excludes 2 as being a prime under consideration
Thus any prime under consideration must be odd
Thus the two equal signs place the odds and primes in the same universe of integers under consideration.
To consider more than 2 primes necessarily includes the consideration of more than 2 odds which violates the conditions of the stated proof.

Thus the two odds can be expressed as 2 primes. There is nothing mathematically that prohibits this.

If the two odds cannot be expressed as 2 primes then equation [4] is false.

 

[ramsey2879]

Step 3 and the two equal signs allows me to place this restriction.

Step 3 merely states that an even number can be expressed as the sum of two odd numbers. Nowhere does step 3 establish that for each even number greater than 2 there is a set of two "primes" that sum to the even number. Odd numbers are not necessarily prime. Again you have not proven the conjecture until you prove that for "each" and "every" even number greater than 2, there is a set of two odd "primes" that sum to the even number. This is the only requirement of a proof that you have failed to even address in your statements. This is not a simple requirement. That is why the Goldbach conjecture is so hard to prove or disprove in the first place.

 

[mathew3]

Thank you gentlemen. I will certainly take your comments and critiques under consideration.

 

[pwsnafu]

All primes greater than 2 are odd.
The proof necessarily excludes 2 as being a prime under consideration
Thus any prime under consideration must be odd

No problem here.

Thus the two equal signs place the odds and primes in the same universe of integers under consideration.

Aha, no. The equal sign does no such thing. The equal sign says the expression
$P_a + P_b + \ldots + P_z$ (it's a finite sum so I'll call the last term P_z) is equal numerically to $O_a + O_b$ in the ring of integers.

If the two odds cannot be expressed as 2 primes then equation [4] is false.

3+3+3+3 = 12 = 9 + 3.

That satisfies [4]. Left hand side is a sum of primes. Right hand side is the sum of two odds.
Sure 12 = 5 + 7, but even if we didn't know that 3+3+3+3 = 9 + 3 is true.

 

[mathew3]

3+3+3+3 = 12 = 9 + 3.

That satisfies [4]. Left hand side is a sum of primes. Right hand side is the sum of two odds.
Sure 12 = 5 + 7, but even if we didn't know that 3+3+3+3 = 9 + 3 is true.

Again we can only consider two odd integers thus you can't consider 3+3+3+3 because the number of addends exceeds 2. You can consider 5+7 because it meets the condition of being composed of 2 odd addends.

Suppose we were to say
a.3+3+3+3=12=the sum of two odd numbers
b.therefore 3+3+3+3=the sum of two odd numbers.

Is statement b. true or false? Depends on the conditions imposed.

 

[pwsnafu]

Again we can only consider two odd integers thus you can't consider 3+3+3+3 because the number of addends exceeds 2.

The right hand side is odd integers sure. And yes there are two.
The left hand side? No. You require only primality. P_a can be 2 (that's a prime!). The number of terms can be 108.

Here's step 2:

2. Therefore, given the Fundamental Theorem of Arithmetic, any and every even integer >4 may be expressed as the summation of a series of prime integers, i.e.,
I = Pa+…+Pb+…+Pc+…
where I is any integer >1 and P is some prime integer

Look, you claim, and I use this very loosely, that we can find two primes that sum to E in Step 9. But [4] is before this. The "number of addends" in the sum $P_a + P_b + \ldots$ is not mentioned in Steps 1~5.

You can consider 5+7 because it meets the condition of being composed of 2 odd addends.

I'm not interested in them being odd I'm interested in them being prime!

Suppose we were to say
a.3+3+3+3=12=the sum of two odd numbers
b.therefore 3+3+3+3=the sum of two odd numbers.

Is statement b. true or false? Depends on the conditions imposed.

What does that have to do with anything?

 

[mathew3]

The right hand side is odd integers sure. And yes there are two.The left hand side? No. You require only primality. P_a can be 2 (that's a prime!). The number of terms can be 108.

Agreed

Look, you claim, and I use this very loosely, that we can find two primes that sum to E in Step 9. But [4] is before this. The "number of addends" in the sum $P_a + P_b + \ldots$ is not mentioned in Steps 1~5.

Agreed. But then again I don't need to. Steps 1-7 are merely statements of fact. They form the predicate for steps 8 and 9. It is the imposition of of the conditions posed in step 8 that really forms the basis of the proof.

I'm not interested in them being odd I'm interested in them being prime!

But that's just it. You can't divorce the two, at least in this proof. It is the imposition of oddness and pairing that forces the prime addends to assume a stance of pairing. The imposition of two odds ensures both pairing and primality since 2 is obligatorily excluded. If you can't see the critical importance of the addends being odd then you miss one of the critical aspects of the proof.

What does that have to do with anything?

True: 3+3+3+3=the sum of two odd integers eg.3+3+3+3 = 5+7
False:3+3+3+3=the sum of two odd integers eg. 3+3+3+3 is the sum 4 odd integers not two odd integers.

Conditions must be defined as to what meaning is to be applied.

At step 8 we impose the dual conditions upon E that it must, simultaneously, form a sum of primes and that E must be composed of 2 and only 2 odd addends. There is nothing mathematically prohibiting me from posing these two conditions. The imposition of these two conditions are not mutually exclusive. These two conditions can only be met, simultaneously, when the primes are both odd and there is only two of them.

The only objection that is relevant is that can the two conditions exist simultaneously. There is nothing mathematically that says they can't. Indeed, they must, otherwise steps 3 and 4 are false which is certainly not the case.

 

[micromass]

OK, maybe an example is in order. I take the even number 14532.

We can write this as

$14532=\sum_{i=1}^{4844}{3}$

so it is a sum of primes.

But we can also write

$14532=5+14527$

Please indicate how you use these two decompositions to write 14532 as a sum of two primes.

 

[mathew3]

OK, maybe an example is in order. I take the even number 14532.

We can write this as

$14532=\sum_{i=1}^{4844}{3}$

so it is a sum of primes.

But we can also write

$14532=5+14527$

Please indicate how you use these two decompositions to write 14532 as a sum of two primes.

It must be the latter ;
$14532=5+14527$
Note however that just stating this alone is not sufficient. To do so declares the summation of pairs of primes by fiat. It must also be linked to the sum of a pair of odd integers.

The former is excluded from consideration because it utilizes more than two odd addends.

 

[micromass]

It must be the latter ;
$14532=5+14527$
Note however that just stating this alone is not sufficient. To do so declares the summation of pairs of primes by fiat. It must also be linked to the sum of a pair of odd integers.

The former is excluded from consideration because it utilizes more than two odd addends.

14527 is not prime...

OK, please indicate exactly how you would find the decomposition of 14532??

 

[mathew3]

14527 is not prime...

I stand corrected. I was lazy and assumed you had included two primes since that is the point of the conjecture.

OK, please indicate exactly how you would find the decomposition of 14532??

Why? As you have indicated it has nothing to do with the problem at hand. And as you well know I could decompose a thousand such numbers\examples and it wouldn't get us one iota closer to proving or disproving the conjecture.

 

[micromass]

I stand corrected. I was lazy and assumed you had included two primes since that is the point of the conjecture.


Why? As you have indicated it has nothing to do with the problem at hand. And as you well know I could decompose a thousand such numbers\examples and it wouldn't get us one iota closer to proving or disproving the conjecture.

You want to know what is wrong with the proof, right? So I present you with a method how you can get to know that something is wrong: a concrete example.

I want you to apply your method on 14532 to show you that your method does not work. So please, explain how you would apply your method to 14532.

 

[mathew3]

You want to know what is wrong with the proof, right? So I present you with a method how you can get to know that something is wrong: a concrete example.

I want you to apply your method on 14532 to show you that your method does not work. So please, explain how you would apply your method to 14532.

I thought you were aware of this but apparently not. Even if I supply the two primes it means nothing! I does not and cannot prove the conjecture. All it does is encourage you to keep asking me to solve yet another example.

 

[ramsey2879]

the two odds can be expressed as 2 primes. There is nothing mathematically that prohibits this.

Yes, but a rigid proof is alleged here and to make a proof you must logically show how your steps lead to the one and only conclusion that for each even number greater than 4, that there necessarily exists at least one pair of primes that sum to each such even number. Myself,
pwsnafu, and Micromass have shown that your proof is lacking in this regard. That is exactly what is wrong with your proof.