0
$\begingroup$

The set given to me was this

$B = \{x \in \mathbb{N} | \ \ x \; mod \ y = 0 \Rightarrow y = 1 \ or \ y = x, x > 1, y \in \mathbb{N}\}$

I just couldn't wrap my head around as to why this set describes a set of prime numbers.

Anybody could explain this in further detail?

Thanks for help.

$\endgroup$
1
  • $\begingroup$ The conditions tell us that any (positive) divisor of $x>1$ must be either $1$ or $x$, which is the definition of a prime. $\endgroup$ – lulu Feb 18 '19 at 20:38
0
$\begingroup$

"If natural number $x > 1$ is divisible by a natural number $y$, then $y$ is either equal to $1$ or equal to $x$."

$\endgroup$
0
$\begingroup$

$x \mod y = 0$ means that $y$ divides $x$. So this is the set of all numbers $x > 1$ such that the only numbers $y$ that divide $x$ are $y=1$ and $y=x$.

$\endgroup$
0
$\begingroup$

It says that a natural number $x$ (presumably natural numbers start at 1 in this text) is prime if (1) it's greater than $1$, and (2) whenever it's divisible by a natural number $y$, then either $y$ is $1$ OR $y$ is $x$.

Since the definition of "$x$ is prime" is that it's a natural number greater than $1$ that divisible only by $1$ and itself, that's the same thing.

The only tricky thing here is remembering that $x \bmod y = 0$ means that $y$ divides evenly into $x$ (i.e., $y$ is a factor of $x$, or $x$ is divisible by $y$).

$\endgroup$

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.