# Prime Number Problem

Standard

Following is a problem about prime factorization of the sum of consecutive odd primes. (source: problem 80 from The Green Book of Mathematical Problems)

Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeated) prime factors.

The first thing to observe is that sum of odd numbers is even, hence the sum of two consecutive odd primes will be divisible by 2. Let’s see factorization of some of the examples:

$3 + 5 = 2\times 2 \times 2$
$5 + 7 = 2 \times 2\times 3$
$7+11 = 2 \times 3\times 3$
$11+13 = 2 \times 2 \times 2 \times 3$
$13+17 = 2 \times 3 \times 5$
$17+19 = 2\times 2\times 3 \times 3$
$19+23 = 42 = 2\times 3\times 7$
$23+29 = 52 = 2\times 2 \times 13$

Now let $p_n$ and $p_{n+1}$ be the consecutive odd primes, then from above observations we can conjecture that either $p_n+p_{n+1}$ is product of at least three distinct primes or $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ such that $k+\ell \geq 3$.

To prove our conjecture, let’s assume that $p_n+p_{n+1}$ is NOT a product of three (or more) distinct primes (otherwise we are done). Now we will have to show that if $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ then $k+\ell \geq 3$.

If $\ell = 0$ then we should have $k\geq 3$. This is true since $3+5=8$.

Now let $\ell > 0$. Since $k\geq 1$ (sum of odd numbers is even), we just need to show that $k=1, \ell=1$ is not possible. On the contrary, let’s assume that $k=1,\ell = 1$. Then $p_n+p_{n+1} = 2p$. By arithmetic mean property, we have

$\displaystyle{p_n < \frac{p_n+p_{n+1}}{2}} = p

But, this contradicts the fact that $p_n,p_{n+1}$ are consecutive primes. Hence completing the proof of our conjecture.

This is a nice problem where we are equating the sum of prime numbers to product of prime numbers. Please let me know the flaws in my solution (if any) in the comments.

# Real Numbers

Standard

Few days ago I found something very interesting on 9gag:

There are lots of interesting comments, but here is a proof from the comments:

…. Infinite x zero (as a limit) is indefinite. But infinite x zero (as a number) is zero. So lim( 0 x exp (x²) ) = 0 while lim ( f(X) x exp(X) ) with f(X)->0 is indefinite …

Though the statement made in the post is very vague and can lead to different opinions, like what about doing the product with surreal numbers, but we can safely avoid this by considering the product of real numbers only.

Now an immediate question should be (since every positive real number has a negative counterpart):

Is the sum of all real numbers zero?

In my opinion the answer should be “no”. As of now I don’t have a concrete proof but the intuition is:

Sum of a convergent series is the limit of partial sums, and for real numbers due to lack of starting point we can’t define a partial  sum. Hence we can’t compute the limit of this sum and the sum of series of real numbers doesn’t exist.

Moreover, since the sum of all “positive” real numbers is not a finite value (i.e. the series of positive real numbers is divergent) we conclude that we can’t rearrange the terms in series of “all” real numbers (Riemann Rearrangement Theorem). Thus the sum of real numbers can only be conditionally convergent. So, my above argument should work. Please let me know if you find a flaw in these reasonings.

Also I found following interesting answer on Quora:

The real numbers are uncountably infinite, and the standard notions of summation are only defined for countably many terms.

Note: Since we are dealing with infinite product and sum, we can’t argue using algebra of real numbers (like commutativity etc.).