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Let $p_i$ be the $i$th prime number. Each number $x$ in $\Bbb{Z}$ can be expressed as a finite sum $\sum\limits_i (k_i p_i), \ k_i \in \Bbb{Z}$, in many ways. But define $\|x\| = \min\{ \sum_i |k_i| : \sum\limits_i (k_i p_i) = x\}$. Then $\|x\|$ satisfies properties that almost make it a norm:

  1. $\|x\| \geq 0$
  2. $\|x\| \iff x = 0 = \text{the empty sum}$
  3. $\|xy\| \leq |x|\cdot \|y\|$
  4. $\|x + y\| \leq \|x\| + \|y\|$
  5. $\|x\| = \|-x\|$

The last one is true since if $x = k_1 p_1 + \dots k_r p_r$ then $-x = (-k_1) p_1 + \dots (-k_r)p_r$. So $\|-x\| \leq |x|$. By symmetry the reverse holds.

A weakened version of Goldbach's conjecture is then for all even numbers $x$, $\|x\| = 2$. So it includes $p - q = x$ or negatives of primes, which makes it weaker.

You can turn it into Goldbach's conjecture if you restrict the $k_i$'s to $\Bbb{N}$ in all the above.

Questions

  1. Has this almost norm popped up before?
  2. Isn't $\Bbb{Z}$ automatically a topological group under the pseudometric induced by this norm?

Proof. We need to show that each of the functions $g(z) = -z$, and $f_a(z) = a + z, \ a \in \Bbb{Z}$ is continuous.

We need to show in $\epsilon$-$\delta$ style that each of these are continuous or that $\forall z \in \Bbb{Z}$, $\forall \epsilon \gt 0$ there exists $\delta \gt 0$ such that $\forall y \in \Bbb{Z}$, $\|y - z\| \lt \delta \implies \|g(z) - g(y)\| \lt \epsilon$ for instance.

But in the first case of $g$, the above is true for a simple choice of $\epsilon = \delta$ since then $\|g(z) - g(y)\| = \|-z - (-y)\| = \|y - z\|$. But for that we need $\|z - y\| = \|y - z\|$.

For the case of $f_a$ similarly we have $\|f_a(z) - f_a(y)\| = \|a + z - (a + y)\| = \|z - y\|$.

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  • $\begingroup$ I'd say the main hole is that It works for any sequence of integers with a not too low density, not only the primes $\endgroup$ – reuns Sep 6 '17 at 4:44
  • $\begingroup$ @reuns which only widens its range of applicability $\endgroup$ – I Said Roll Up n Smoke Adjoint Sep 6 '17 at 4:59
  • $\begingroup$ I think by 3. it also forms a topological ring. Interesting $\endgroup$ – I Said Roll Up n Smoke Adjoint Sep 6 '17 at 5:01

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