Dashawn Clark

2022-04-24

Proof by contradiction: ${n}^{3}-n-6=0$ means ${m}^{3}-m-6\ne 0$ when $m\ne n$

drenkttj9

Beginner2022-04-25Added 20 answers

It looks like your proof would be assuming ${n}^{3}-n-6=0,m\ne n$ (where m, $n\mathrm{\setminus}\in \mathbb{Z}$ ) showing a contradiction if we had ${m}^{3}-m-6=0$ as well. You have shown by factoring and the fact that m, $n\in \mathbb{Z}$ , that m, $n=2$ . Hence $m=n$ , a contradiction.

Felicity Carter

Beginner2022-04-26Added 16 answers

We have

${n}^{3}-n-6=0\iff$

$(n-1)n(n+1)=1.2.3\iff n=2$

thus$(m-1)m(m+1)\ne 6\iff m\ne 2$ .

thus

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