a)It follows from $\sqrt{n-1}+0.011<\sqrt{n}$ for every $n \in M$ b) $(44+0.0115)^2>1937>44^2$, so for $n=1937$ is true

RagvaloD wrote: a)It follows from $\sqrt{n-1}+0.011<\sqrt{n}$ for every $n \in M$ b) $(44+0.0115)^2>1937>44^2$, so for $n=1937$ is true Can someone please elaborate on part a Also for part b is there a way to actually solve for $n$ Thanks

Let`s prove , that $\sqrt{n-1}+0.011<\sqrt{n}$ for $n \in M$ $0.011<\frac{1}{\sqrt{n}+\sqrt{n-1}}$ $\frac{1}{\sqrt{n}+\sqrt{n-1}}>\frac{1}{2\sqrt{n}}$ So we will prove $0.011<\frac{1}{2\sqrt{n}}$ or $\sqrt{n}<\frac{1000}{22}$ $n\leq 2015$ so we will prove that $\sqrt{2015}<\frac{1000}{22}$ $2015<\frac{1000^2}{22^2}$ $121*2015<250000$ $121*15<250000-2000*121=8000$ $121*15<3000<8000$ so $\sqrt{n-1}+0.011<\sqrt{n}$ for $n \in M$ As $n$ is not square, than it means, that $ n=a^2+b,0<b<2a$ $\{\sqrt{n}\}=\sqrt{n}-[\sqrt{n}]=\sqrt{a^2+b}-a>\sqrt{a^2+b}-\sqrt{a^2+b-1}=\sqrt{n}-\sqrt{n-1}>0.011$ For part b) $44^2<2015<45^2$ Let $n=44^2+1$ $\{\sqrt{n}\}=\sqrt{n}-[\sqrt{n}]=\sqrt{44^2+1}-44=\frac{1}{\sqrt{44^2+1}+44}<\frac{1}{2*44}$ But $88*0.0115=1.012>1 \to \frac{1}{88}<0.0115$ So $\{\sqrt{n}\}<0.0115$