> [!NOTE] Proposition > There is no [[Rational Number|rational number]] whose square is $2$. >*Proof*. BWOC suppose $p,q \in \mathbb{Z}/\{ 0 \}$ with $\gcd(p,q)=1$ such that $\left( \frac{p}{q} \right)^{2}=2.$We have $\frac{p^{2}}{q^{2}} \implies p^{2}= 2q^{2}$ so $2 \mid p^{2} \implies p= 2m$ for some $m \in \mathbb{Z}$. But then $(2m)^{2}=4m^{2} = 2q^{2} \implies q^{2} = 2m^{2},$so $2 \mid q^{2} \implies q =2n$ for some $n\in \mathbb{Z}$. This contradicts our initial assumption that $\gcd(p,q)=1$ so it must be wrong. >*Proof*. Consider $a_{n} = (\sqrt{ 2 }-1)$. Note that $1<2<4 \implies 1< \sqrt{ 2 }-1 <1$. So $a_{n}\to 0$ as $n\to \infty$. >Now $a_{1}= \sqrt{ 2 }-1$. Now suppose inductively that $a_{k}=A_{k} \sqrt{ 2 } +B_{k}$ for some integers $k\in \mathbb{N}, A_{k},A_{k}\in \mathbb{Z}$. >Then $a_{k+1}=(\sqrt{ 2 }-1)(A_{k} \sqrt{ 2 } +B_{k})=(B_{k}-A_{k})\sqrt{ 2 }+ 2A_{k}-B_{k}$. >Hence by mathematical induction $a_{n}=A_{n}\sqrt{ 2 }+B_{n}$ where $A_{n},B_{n}\in\mathbb{Z}$ for all $n\in \mathbb{N}$. >Suppose $\sqrt{ 2 } = \frac{p}{q}$ for some $p,q\in \mathbb{N}$ then $\left\lvert A_{n} \frac{p}{q} + B_{n} \right\rvert = \left\lvert \frac{A_{n}p+B_{n}}{q} \right\rvert \geq \frac{1}{q} $which contradicts the fact that $a_{n} \to 0$.