# Is this a valid proof? (proof that if 2|n and 3|n then 6|n)

• TalkOrigin
In summary, the proof for 6 | n is as follows:"Suppose 2 | n and 3 | n. Then there is an integer k such that (2)(3)k=n. Thus, (6)k=n and therefore 6 | n."
TalkOrigin
Hi, I wasn't sure whether to post this here or in the pre-calc forum, so apologies if this is in the wrong section. I'm working through Vellemans 'How to prove it' and he gives a proof that if 2|n and 3|n then 6|n (note that a|n means a divides n, just in case this is not standard notation). I think I proved it a different way, but as he did it differently, I'm half assuming my proof is incorrect. His proof is as follows:

"Suppose 2 | n and 3 | n. Then we can choose integers j and k such that n=2j and n=3k.
Therefore 6(j−k)=6j−6k=3(2j)−2(3k)= 3n − 2n = n, so 6 | n."

My "proof" is as follows:

"Suppose 2 | n and 3 | n. Then there is an integer k such that (2)(3)k=n. Thus, (6)k=n and therefore 6 | n."

If my proof is wrong, then could you tell me why? I was wondering if it's because my proof has something to do with the fact that every number can be written as a product of prime numbers and maybe I need to state this or something. Thanks in advance.

TalkOrigin said:
Suppose 2 | n and 3 | n. Then there is an integer k such that (2)(3)k=n.

The definition of "2 | n" implies there is an integer k_a so that 2 k_a = n. The definition of "3 | n" implies there is an integer k_b such that 3 k_b = n. The compound statement "2 | n and 3 | n", doesn't assert anything about an integer k such that (2)(3)k = n.

Hi, thanks for the reply. I'm not sure why "2 | n and 3 | n" does not assert there is an integer k such that (2)(3)k = n. I forgot to say at the beginning that n is an arbitrary integer. I would think that if 2 and 3 both divide n, then obviously n is not prime, but can be expressed as a product of primes (2)(3) and an integer k.

The fact that 2 and 3 divide n, and 2 and 3 are prime, that is a valid proof.

HallsofIvy said:
The fact that 2 and 3 divide n, and 2 and 3 are prime, that is a valid proof.

Ah ok, so I have to say ""Suppose 2 | n and 3 | n. As 2 and 3 are both prime, then there is an integer k such that (2)(3)k=n. Thus, (6)k=n and therefore 6 | n."

EDIT: Also, could this be generalised to "Suppose a | n and b | n. As a and b are both prime, then there is an integer k such that abk=n. Thus, (ab)k=n and therefore ab | n."?

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Stephen Tashi said:
The definition of "2 | n" implies there is an integer k_a so that 2 k_a = n. The definition of "3 | n" implies there is an integer k_b such that 3 k_b = n. The compound statement "2 | n and 3 | n", doesn't assert anything about an integer k such that (2)(3)k = n.
Actually, if 2 and 3 divide a number ##n##, then since ##gcd(2,3)=1##, ##n## also divides ##2\times 3=6##.

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TalkOrigin said:
Ah ok, so I have to say ""Suppose 2 | n and 3 | n. As 2 and 3 are both prime, then there is an integer k such that (2)(3)k=n. Thus, (6)k=n and therefore 6 | n."

EDIT: Also, could this be generalised to "Suppose a | n and b | n. As a and b are both prime, then there is an integer k such that abk=n. Thus, (ab)k=n and therefore ab | n."?
Yes, this is correct. But ##a## and ##b## don't need to be prime. As long as, ##gcd(a,b)=1## and ##a|n## and ##b|n## then, ##ab|n##.

certainly said:
Actually it does. If a number ##n## divides 2 and also divides 3, then since ##gcd(2,3)=1##, ##n## also divides ##2\times 3=6##.
It's actually that 2 divides n, and 3 divides n, not that n divides 2 and n divides 3.

certainly said:
Yes, this is correct. But ##a## and ##b## don't need to be prime. As long as, ##gcd(a,b)=1## and ##a|n## and ##b|n## then, ##ab|n##.
whoops, sorry i replied with a brainfart

Ah, yes! silly error on my part. Edited post accordingly.

TalkOrigin said:
I would think that if 2 and 3 both divide n, then obviously n is not prime, but can be expressed as a product of primes (2)(3) and an integer k.

You can try an argument along those lines, but you have to state it! You can't just assert that there exists a k so that (2)(3)k = n without saying why.

Stephen Tashi said:
You can try an argument along those lines, but you have to state it! You can't just assert that there exists a k so that (2)(3)k = n without saying why.
Ok thanks, this is my first dip into proof based mathematics, so I am still learning how to formulate proofs and everything. This book is great though! Thanks for all the help

## 1. How can I determine if a proof is valid or not?

To determine if a proof is valid, you need to carefully examine the logical steps and assumptions made in the proof. You should also check if the proof follows the accepted rules of logic and mathematics. If there are any errors or inconsistencies in the proof, then it may not be valid.

## 2. What are the key elements of a valid proof?

The key elements of a valid proof include clear and logical reasoning, accurate and appropriate use of mathematical notation, and the use of accepted axioms, definitions, and theorems. A valid proof should also be able to be replicated and understood by others.

## 3. Can a proof be considered valid if it contains errors or incomplete steps?

No, a proof cannot be considered valid if it contains errors or incomplete steps. Even one small error or missing step can invalidate the entire proof. It is important to carefully check and verify each step in a proof to ensure its validity.

## 4. Are there specific criteria for a valid proof in mathematics?

Yes, there are specific criteria for a valid proof in mathematics. These include logical consistency, completeness, and correctness. A valid proof should also be clear, concise, and able to be understood by others.

## 5. How can I improve the validity of my proof?

To improve the validity of your proof, you should carefully check and double-check each step to ensure accuracy and logical consistency. You can also seek feedback and assistance from other mathematicians or experts in the field. Additionally, studying and understanding common proof techniques and strategies can help improve the validity of your proof.

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