0
$\begingroup$

What is this set describing?

{$n∈\mathbb{N}|n\ne 1$ and for all $a∈\mathbb{N}$ and $b∈\mathbb{N},ab=n$ implies $a= 1$ or $b= 1$}

Is it describing a subset of natural numbers, excluding 1, that is the product of two other natural numbers, of which one must be 1? Isn't that just every natural number except 1?

$\endgroup$
  • 1
    $\begingroup$ Best way is to test which of the numbers $2$, $3$ or $4$ are in the set. $\endgroup$ – Orest Bucicovschi Sep 11 '17 at 23:46
3
$\begingroup$

Prime numbers only have two positive divisors, of which one of them is $1$. Hence the set is describing the set of prime numbers.

It doesn't include composite numbers such as $6$. If $6 = ab$, we can't conclude that $a=1$ or $b=1$, since it is possible that $a=2$ and $b=3$.

$\endgroup$
2
$\begingroup$

!

"Is it describing a subset of natural numbers, excluding 1, that is the product of two other natural numbers, of which one must be 1? Isn't that just every natural number except 1?"

Almost. It is saying that if $n = ab$, no matter which $a$ or $b$ you choose, it must be that either $a$ or $b$ is $1$.

Example: $6$. Is $6$ in the set? Well $6 \ne 1$ so that's a start. And "if $6 = a*b$..." Okay, that could be $(a,b) = (1,6),(2,3),(3,2)$ or $(61)$ "it must be that $a = 1$ or $b = 1$". Well... we could have $a = 2$ and $b=3 $ and neither $a$ nor $b$ must be $1$. So $6$ is not in the set.

SPOILER ALERT: There is a word for numbers that must be in the set--- that is numbers for whom every pair of divisors must include $1$... numbers that can not be a product of any pair of numbers of which neither are $1$....

$\endgroup$
  • $\begingroup$ The word is also given in the other answer that's already on this page..... $\endgroup$ – G Tony Jacobs Sep 11 '17 at 23:35
0
$\begingroup$

Might it help if I translate to ordinary English?

  1. $n$ must be a natural number other than $1$
  2. one must be able to write $n$ as the product of two natural numbers $a$ and $b$
  3. one of the aforementioned $a$ or $b$ must equal $1$
  4. the prior statements must hold for all natural $a$ and $b$ that equal $n$ when multiplied; if there exists a pairing of $ab=n$ where one of $a$ or $b$ is not $1$, then $n$ goes out the window
$\endgroup$

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.