0
$\begingroup$

Prove that for every natural number $n\ge 2$, $n$ does not divide $n+1$.

Proof: Suppose for every natural number $n\ge 2$, $n$ does divide $n+1.$ However, for natural numbers $a$ and $b,$ $a$ divides $b$ or goes into $b$ if $b=ka$ for some natural number $k \ge 2$. Thus, there is only some $n$ that divides $n+1.$ Therefore, this is a contradiction.

$\endgroup$
  • $\begingroup$ What is the contradiction, exactly? I can imagine where it's going, but it's best to spell it out explicitly. $\endgroup$ – Eevee Trainer Mar 7 at 7:42
  • $\begingroup$ @EeveeTrainer that there is some n that divides n+1 but it is assumed every natural number 𝑛≥2 divides n+1. $\endgroup$ – k.rudin Mar 7 at 7:45
  • 1
    $\begingroup$ Your assumption should be, "Suppose there exists a natural number $n_0\geq2$ such that $n_0\mid n_0+1$." $\endgroup$ – Thomas Shelby Mar 7 at 7:46
3
$\begingroup$

A direct proof seems to be easier. If $n\geq 1$ divides $n+1$ then $n$ divides also the difference $n+1-n=1$ (note that $n$ divides $n$) which implies that $n=1$.

You may also say. If $n\geq 2$ divides $n+1$ then $n$ divides also the difference $n+1-n=1$ (note that $n$ divides $n$) which implies that $n=1$. Contradiction.

$\endgroup$
  • $\begingroup$ but aren't we assuming 𝑛≥2? $\endgroup$ – k.rudin Mar 7 at 7:51
  • $\begingroup$ @k.rudin See my edit. $\endgroup$ – Robert Z Mar 7 at 7:55

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.