# Checking Proof for the following Divisibility

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.

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

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.