Twitter proof: irrational high-order roots of 2


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For this twitter proof we will be using a piece of mathematics straight from the 17th century.

Claim: the number $\sqrt[n]{2} $ is irrational for $n \geq 3$.

Twitter proof: suppose that $n \geq 3$ and $\sqrt[n]{2}$ is rational, i.e. $\sqrt[n]{2} = \frac{a}{b}$ for some integers $a, b $. Then taking the $n $-th power of both sides we get $2 = \frac{a^n}{b^n} \iff b^n + b^n = a^n $, contradicting the well-known Fermat's Last Theorem.
Para esta prova num tweet vamos usar um pedaço de matemática do século 17.

Proposição: o número $\sqrt[n]{2} $ é irracional para $n \geq 3$.

Prova num tweet: suponhamos que $n\geq 3$ e que $\sqrt[n]{2} $ é racional, i.e. $\sqrt[n]{2} = \frac{a}{b} $ para alguns inteiros $a, b $. Se for esse o caso, elevando os dois lados da igualdade a $n $, obtemos $2 = \frac{a^n}{b^n} \iff b^n + b^n = a^n $, contrariando o famoso Último Teorema de Fermat.

  - RGS

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