Verify that the integers 1949 and 1951 are twin primes

[Math100]

Homework Statement

Verify that the integers 1949 and 1951 are twin primes.

Relevant Equations

None.

Proof:

Consider all primes $p\leq \sqrt{1949} \leq 43$ and $q\leq \sqrt{1951} \leq 43$.
Then we have $p\nmid 1949$ and $q\nmid 1951$ for all $p\leq 43$.
Thus, $1949$ and $1951$ are both primes.
By definition, twin primes are two prime numbers whose difference is $2$.
Note that $1951-1949=2$.
Therefore, the integers $1949$ and $1951$ are twin primes.

 

[Mark44]

Homework Statement:: Verify that the integers 1949 and 1951 are twin primes.
Relevant Equations:: None.

Proof:

No. The problem is not asking for a proof, just a verification that both 1949 and 1951 are prime. If so, given that they are 2 apart, they would then be twin primes.

Consider all primes $p\leq \sqrt{1949} \leq 43$ and $q\leq \sqrt{1951} \leq 43$.
Then we have $p\nmid 1949$ and $q\nmid 1951$ for all $p\leq 43$.

Do we? In the verification you should show which primes you considered. It's not sufficient to go from "Consider all primes..." to "Then we have ... " with no actual justification shown.

Obviously, you don't need to consider any even number, including 2, the only even prime. Nor do you need to consider 3 or any of its multiples, since the sets of digits of 1949 and 1951 do not add to a multiple of 3. You also don't need to consider 5.

Please show us the primes that you actually considered.

Thus, $1949$ and $1951$ are both primes.
By definition, twin primes are two prime numbers whose difference is $2$.
Note that $1951-1949=2$.
Therefore, the integers $1949$ and $1951$ are twin primes.

 

[PeroK]

You could check that the remainer of $1951$ upon division by all the primes up to $43$ is not $0$ or $2$. If it were $2$ then the prime in question would divide $1949$.

That's if you want to do it without a computer.

 

[epenguin]

You have given a bad answer here, possibly not having understood the question.

Mark 44 has explained to you what the question is, and reminded you of the only known way, roughly speaking, of finding out whether a given number is prime, in the absence of other information.

After which, after knowing how you verify the first prime the question you should ask yourself is do I have to go through the same thing again for the second? - or can the fact that the second number is 2 greater than the first then be used to economise on calculation? PeroK has given you the answer to that.

Only remains to do the said calculation. No one here will blame you for not doing it!

You are getting rather a lot of your answers to these many problems you are bringing us wrong. And most of the problems seem to me trivial, although they might be hard work sometimes.

it is at least debatable whether these exercises have any value at all. And I suspect you are getting them wrong because fundamentally they are boring you out of your mind. I do not know what your position is but if I were in it I might if possible (a) ask your Prof when are we going to start some real math? or (b) quit this particular course in favour of almost any other. I guarantee, you are never going to need this stuff.

(True I could never get very interested in numbers, even so I believe number theory is something better than this rubbish.).

 

[fishturtle1]

You have given a very bad answer here, possibly not having understood the question.

What's wrong with it? In the OP, they use the theorem that to show $n$ is prime, it is sufficient to show all primes $p \le \sqrt{n}$ do not divide $n$.

In fact, I really don't see anything wrong with the OP. Maybe other than writing something like $\lfloor \sqrt{1949} \rfloor = 44$.

 

[epenguin]

What's wrong with it? In the OP, they use the theorem that to show $n$ is prime, it is sufficient to show all primes $p \le \sqrt{n}$ do not divide $n$.

In fact, I really don't see anything wrong with the OP. Maybe other than writing something like $\lfloor \sqrt{1949} \rfloor = 44$.

I cannot see that the OP shows what your 2nd sentence says. It at most asserts that the two numbers are primes.

I cannot believe that the question is as trivial as 'verify that the two primes, 1949 and 1951 are twin'!

 

[Mark44]

What's wrong with it? In the OP, they use the theorem that to show $n$ is prime, it is sufficient to show all primes $p \le \sqrt{n}$ do not divide $n$.

I cannot see that the OP shows what your 2nd sentence says. It at most asserts that the two numbers are primes.

I agree. The problem entails a bit more work than merely saying "Thus, 1949 and 1951 are primes."

I cannot believe that the question is as trivial as 'verify that the two primes, 1949 and 1951 are twin'!

I don't have access to the textbook in use here, so I'll assume that the problem was given exactly as written in the OP.

 

[vela]

What's wrong with it? In the OP, they use the theorem that to show $n$ is prime, it is sufficient to show all primes $p \le \sqrt{n}$ do not divide $n$.

In fact, I really don't see anything wrong with the OP. Maybe other than writing something like $\lfloor \sqrt{1949} \rfloor = 44$.

The OP's reasoning is correct; it's just a matter of showing the appropriate level of work. If @Mark44 were to assert that he tested divisibility by all of the primes less than 44, I'd be inclined to trust that he did it correctly. However, for a student learning about primes, I'd want to see evidence that they actually did the divisions, did them correctly, included all of the prime numbers, etc. There are a lot of places students can make mistakes.

I cannot believe that the question is as trivial as 'verify that the two primes, 1949 and 1951 are twin'!

This problem does seem to involve a bit of busy work.

 

[fishturtle1]

The OP's reasoning is correct; it's just a matter of showing the appropriate level of work. If @Mark44 were to assert that he tested divisibility by all of the primes less than 44, I'd be inclined to trust that he did it correctly. However, for a student learning about primes, I'd want to see evidence that they actually did the divisions, did them correctly, included all of the prime numbers, etc. There are a lot of places students can make mistakes.

I had not thought of that. Rereading this thread, I completely agree with the points made by Mark and Perok and hope the OP understands them. I think i was mainly just reacting to some of the things in post #4.

Apologies and hopefully the thread gets back on track.