We seek an irrational $x\in\mathbb{R}$ such that $\frac{3}{4}<x<\frac{7}{8}$. You changed the denominators so that they would be the same - this was overly complicated. Instead, change the numerators: $$\frac{3}{4} = \frac{21}{28}<x<\frac{21}{24}<\frac{7}{8}.$$ Notice that $24^2<24^2+1<28^2$, so that $24<\sqrt{577}<28$. This gives $$\frac{1}{24}>\frac{1}{\sqrt{577}}>\frac{1}{28},$$ and hence, $$\frac{7}{8}>\frac{21}{\sqrt{577}}>\frac{3}{4}.$$ Since $577$ is one more than a square, it follows that $577$ is not square itself, and so $\sqrt{577}$ is irrational.

We seek an irrational $x\in\mathbb{R}$ such that $\frac{3}{4}<x<\frac{7}{8}$. You changed the denominators so that they would be the same - this was overly complicated. Instead, change the numerators: $$\frac{3}{4} = \frac{21}{28}<x<\frac{21}{24}=\frac{7}{8}.$$ Notice that $24^2<24^2+1<28^2$, so that $24<\sqrt{577}<28$. This gives $$\frac{1}{24}>\frac{1}{\sqrt{577}}>\frac{1}{28},$$ and hence, $$\frac{7}{8}>\frac{21}{\sqrt{577}}>\frac{3}{4}.$$ Since $577$ is one more than a square, it follows that $577$ is not square itself, and so $\sqrt{577}$ is irrational.