Help What is wrong with this proof?

[mathew3]

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.

I've yet to see how I haven't shown this however I realize this may be tendentious on my part. Again that's why I presented it here on the forum for criticism. I will certainly take your criticisms into consideration and I appreciate your input. Thanks.

 

[micromass]

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.

Just solve the example with your method... It will tell you why your method is bunk.

 

[ramsey2879]

Just solve the example with your method... It will tell you why your method is bunk.

Really, Isn't it possible to establish the numericaL existence of something without actually being able to determine its value? For instance there are tests that show a number to be composite that do not provide any factor of the number.

 

[mathew3]

just solve the example with your method... It will tell you why your method is bunk.

14519 +13=14532=14529 +3

 

[micromass]

Really, Isn't it possible to establish the numericaL existence of something without actually being able to determine its value? For instance there are tests thast show a number to be composite that do not provide any factor of the number.

Indeed, this is possible. But the method of the OP seems to be constructive. Or at least he claims that.

I normally resist these juvenile challenges however you need to be taught a lesson...

14519 +13=14532=14529 +3

How did you arrive to this solution?? How did your method help you in getting the solution?? THAT's what I want to know.

Of course, you can find the decomposition easily. I found it very easily using a computer program. But I want you to solve it with your method. You can't because the method is bunk.

Please give me ALL the steps you did in solving this problem. If you can do this, then I'll agree that your method works. You can't however.

(oh, yes, 14529 isn't a prime)

 

[pwsnafu]

Really, Isn't it possible to establish the numericaL existence of something without actually being able to determine its value?

Yup. But that still requires a proof. And it's very hard to do this using only elementary techniques, which is what OP asserts.

OP's argument is basically:

1. For all even numbers greater than 4, there exists a finite set of primes which sum to it. But not necessarily two primes.
2. For all even numbers greater than 4, there exist two odd positive integers which sum to it. But not necessarily primes.
3. The two sums are equal to each other numerically, i.e. the "equal sign" is from PA.
4. Therefore, for all even numbers greater than 4, there exists two odd primes which sum to it.

It's simply a case of not understanding the "there exists" quantifier properly. His coin analogy is evidence for this.

 

[mathew3]

Yup. But that still requires a proof. And it's very hard to do this using only elementary techniques, which is what OP asserts.

OP's argument is basically:

1. For all even numbers greater than 4, there exists a finite set of primes which sum to it. But not necessarily two primes.
2. For all even numbers greater than 4, there exist two odd positive integers which sum to it. But not necessarily primes.
3. The two sums are equal to each other numerically, i.e. the "equal sign" is from PA.
4. Therefore, for all even numbers greater than 4, there exists two odd primes which sum to it.

It's simply a case of not understanding the "there exists" quantifier properly. His coin analogy is evidence for this.

Close. And if that is the logical progression I have portrayed (which I don't think I have) then I have been remiss. There is a huge gaping chasm logically between your 3 and 4. Let's try this :
Your 1,2, and 3
3a. We consider the case where E can only be composed of two and only two odd integers
3b. To this case we then apply your 1.
Therefore your 4.

I'm not familiar with the there exists quantifier and will look it up.

 

[micromass]

Close. And if that is the logical progression I have portrayed (which I don't think I have) then I have been remiss. There is a huge gaping chasm logically between your 3 and 4. Let's try this :
Your 1,2, and 3
3a. We consider the case where E can only be composed of two and only two odd integers

This case is void. Every nontrivial integer can be decomposed in multiple way as such a sum.

3b. To this case we then apply your 1.

This is false, I'm sorry. You can't apply 1 to anything. I have asked you to write down the method in an example so you could see where it went wrong. You didn't do this.

I'm asking myself this:
- Do you want us to give you a reason why your "proof" is false?
- Do you want to convince us that your "proof" is true?

Your attitude seems to indicate the latter.

If it is the former (which I doubt), then please do as I told. Please write it out in an example. You'll see immediately what goes wrong!

 

[pwsnafu]

Close. And if that is the logical progression I have portrayed (which I don't think I have) then I have been remiss.

I took your original post, cut all the waffle, and got that.

3a. We consider the case where E can only be composed of two and only two odd integers
3b. To this case we then apply your 1.

More waffle.

3a is meaningless because it says nothing about primality, so its just 2.
3b is meaningless because it is no different to 4.

Note that if you had written

3a. Take 2 and consider the case where E can only be composed of two and only two odd integers

that is a lot worse.

Sorry, everything else you write, whether its analogies, solving examples etc is nothing more than window dressing. You write so much of it to create an illusion of more logic. And you have convinced yourself that there is logic there.

 

[Diffeomorphic]

I think this guy must be a troll...