FALSE: the rationals are complete

For each $n$ let $k_n$ be the largest natural with $k_n^2\le 2\cdot {10}^{2n}$, and set $s_n=k_n\mathrm{/}{10}^n$; then $s_n^2\le 2<(s_n+{10}^{-n}{)}^2$, and $s_n\le 2$ (since $k_n^2\le 2\cdot {10}^{2n}<(2\cdot {10}^n{)}^2$).

$(s_n)$ is Cauchy: for $m\ge n$, $({10}^{m-n}{k}_n{)}^2\le 2\cdot {10}^{2m}$ gives ${10}^{m-n}{k}_n\le k_m$, so $s_n\le s_m$; and $s_m^2\le 2<(s_n+{10}^{-n}{)}^2$ forces $s_m<s_n+{10}^{-n}$; hence $0\le s_m-s_n<{10}^{-n}$, and ${10}^{-n}$ eventually falls below any $\epsilon >0$.

$(s_n^2-2)$ is null: $0\le 2-s_n^2<(s_n+{10}^{-n}{)}^2-s_n^2={10}^{-n}(2s_n+{10}^{-n})\le 5\cdot {10}^{-n}$.

If $(s_n)$ converged to a rational $q$, then $(s_n-q)$ would be null; since $(s_n+q)$ is Cauchy, $(s_n^2-q^2)=(s_n-q)(s_n+q)$ would be null; adding the null $(2-s_n^2)$, the constant $2-q^2$ would be null, forcing $q^2=2$.

No rational squares to $2$, so $(s_n)$ is a Cauchy sequence of rationals with no rational limit: the claim is refuted.