5
$\begingroup$

I'm a noob. Is the following valid?

$$ N = (P_1 \cdot P_2...P_n)+1 $$

where $Pi$ is the $i^{th}$ consecutive prime number and $n$ is any natural number. If $N$ is even, $\dfrac{N}{2} \in \Bbb{N}$, which can be put into the first equation like so:

$$ \frac{N}{2} = \frac{(P_1 \cdot P_2...P_n)}{2}+\frac{1}{2} $$

where it is expected that $ \dfrac{(P_1 \cdot P_2...P_n)}{2}+\dfrac{1}{2} \in \Bbb{N} $ also. $P_1 = 2$, being the first prime number and thus

$$ \frac{(P_1 \cdot P_2...P_n)}{2} = \frac{2(P_2 \cdot P_3...P_n)}{2} = P_2 \cdot P_3...P_n $$

where $P_2 \cdot P_3...P_n \in \Bbb{N}$. Because $\dfrac{1}{2} \notin \Bbb{N}$, $\dfrac{(P_1 \cdot P_2...P_n)}{2}+\dfrac{1}{2} \notin \Bbb{N}$ because any natural number plus a non-natural number is not a natural number.

Because the RHS of the first equation does not belong to $\Bbb{N}$, $\dfrac{N}{2} \notin \mathbb N$ also. Thus $N$ is odd.

$\endgroup$
  • 1
    $\begingroup$ Note that $N$ as you define it will always be odd since the product $(P_1 \cdot ... \cdot P_n)$ is necessarily even since it contains $P_1 = 2$. $\endgroup$ – Zubzub Apr 13 '17 at 9:35
5
$\begingroup$

That is really complicated, but yes, you are right. An easier way to see it would be the following:

As $P_1 = 2$, we have that $$P_1\cdot P_2 \cdot \ldots \cdot P_n = 2\cdot P_2 \cdot \ldots \cdot P_n$$ is even. Now $1$ is odd, and you might know that even plus odd always gives odd, case closed. :)

$\endgroup$
1
$\begingroup$

If $P_1 = 2$, obviously $N = 2k + 1$ is odd, so your argument after "If $N$ is even.." is invalid.

$\endgroup$
  • $\begingroup$ No, it is not. He is assuming that $N$ is even and the proving by contradiction that this case can not be true. Thus the argument is valid, even though a proof by contradiction is maybe not the easiest choice here. $\endgroup$ – Dirk Apr 13 '17 at 9:50
  • $\begingroup$ He edited the question to make it a proof by contradiction. The original post didn't conclude that $N$ was odd. $\endgroup$ – Aryaman Jal Apr 13 '17 at 9:51
1
$\begingroup$

This reminds me of Conway's PRIMEGAME, in which a list of fractions is used to obtain powers of 2 with prime indices. It makes the sieve of Eratosthenes look very sophisticated by comparison. But your question assumes we already know the prime numbers.

You state that $P_1 = 2$. That's obviously even. So is $P_1 P_2$. And $P_1 P_2 P_3$. And any $P_1 P_2 \ldots P_n$ for $n > 2$. These numbers are all divisible by 2 but not by 4: 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870, 6469693230, etc. See Sloane's A002110.

So halving $N$ is an entertaining but unnecessary sidetrack. Since $P_1 P_2 \ldots P_n$ is even, it immediately follows that $N = P_1 P_2 \ldots P_n + 1$ is odd.

This is not to say that there don't exist situations where halving a number is useful. There are such situations. It's just that this one is not such a situation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.