TITLE: How to find irrational numbers between rationals. (And is my method correct?)
QUESTION [23 upvotes]: I have a question from an A-level revision book:

Find an irrational number which lies between $\frac34$ and $\frac78$.

What is the correct method for doing this? Here is my method:

Square numerators and denominators of the (in this case: both) fractions.
Find LCD (Lowest Common Denominator) for denominators and convert all fractions to this LCD.
The numerators of the fractions are now perfect squares. Write a new fraction with a numerator that is not a perfect square and between the original fractions' numerators.
Put the new fraction inside a square root. The new fraction (inside a square root sign) is now a surd, and an irrational number (also an irrational fraction).

Using this method, $\sqrt{\frac{37}{64}}$ would be one such irrational number between $\frac34$ and $\frac78$.
Please let me know if I've made any glaring lapses in logic. If I'm correct, a fraction with an irrational (surd) numerator is itself an irrational number. And non-perfect square numbers are all irrational. Therefore a fraction with an irrational numerator is irrational. (Pardon me if I'm not using all the correct terminology yet.)
Many thanks in advance
PS. I want to check I understand these relatively basic concepts correctly. (I apologize in advance if I have made a schoolboy error or stupid mistake. I'm trying to take the A-level mathematics exams in nearly 4 months, the reason is a rather long story, and I'm refreshing all my mathematics which I did at school. I did well at Math GCSE, although it was the intermediate not higher tier, and I'm considering blitzing all the A-level maths in a little over 3 months to take exams starting mid-May.)

REPLY [2 votes]: 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.