How to find irrational numbers between rationals. (And is my method correct?) 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.)
 A: Come on, it's easier than that. Just interpolate (a generalization of what @TonyK has done). If $a$ and $b$ are distinct rationals, then $a-b$ is also rational, and rational $\times$ irrational = irrational, so any irrational number $r\in (0,1)$ will make the following irrational:
$$c=a(1-r) + br$$
Sure, you can choose $r=1/\sqrt2$. But any irrational between 0 and 1 works. And this procedure is not at risk of accidentally hitting a rational number - you can't. There are no conditions on $a$ and $b$ other than they have to be rational.
A: You are making this more complicated than it needs to be. If $p,q$ are any two distinct rational numbers, then $$p + \frac {q-p}{\sqrt 2}$$ is an irrational number between $p$ and $q$.
A: Although this isn't well suited for time-intensive tasks, there is a nice tangentially related mathematical curiosity:

For any two real numbers $x < y$ there exist two integers $a$ and $b$ such that $$x < a + b\sqrt{2} < y.$$

To see why this works consider $$x < m\cdot(-1+\sqrt{2})^n < y$$ for some natural number $n$ and integer $m$. Observe that by making the $n$ large, the parenthesized expression can be as small as we want (i.e. strictly smaller than $y-x$), and then it is enough to adjust $m$ to fit the number between $x$ and $y$.
In fancy terms $\mathbb{Z}[\sqrt{2}]$ is dense in $\mathbb{R}$. 
In your case $\frac{3}{4} < -2+2\sqrt{2} < \frac{7}{8}$. 
I hope this helps $\ddot\smile$
A: I would note that $3 < \pi < 3.5$ which would mean that
$${3\over 4} < {\pi \over 4} < {3.5\over 4} = {7\over 8}$$
Not a general technique, but it popped into my mind instantly.
A: 
Given two real numbers $a$ and $b$, with $a<b$, there exists a rational number $q$ such that $a<q<b$.

The proposition is obvious if $a<0$ and $b>0$. If we prove it for $0<a<b$, then also the case $a<b<0$ follows by considering $0<-b<-a$.
By the Archimedean property, there exists an integer $n$ such that $n(b-a)>1$. Let $m$ be the least positive integer such that $na<m$. If $m\ge nb$, then
$$
m-na\ge n(b-a)>1
$$
so also
$$
m-1>na
$$
which is a contradiction. Therefore $na<m<nb$ and so
$$
a<\frac{m}{n}<b
$$
Now, if $a$ and $b$ are rational, consider your favorite irrational number $r$, say $r=\sqrt{2}$, and a rational number $q$ such that
$$
a+r<q<b+r
$$
Then
$$
a<q-r<b
$$
and $s=q-r$ is irrational.

If you're in the different situation where you don't know whether $a$ and $b$ are rational and still want to find an irrational number between them, you can proceed for steps.
First find a rational number $q$ such that $a<q<b$; then find a rational number $q'$ such that $q<q'<b$. Finally, the result above provides an irrational number $s$ such that $q<s<q'$ and you're done.
A: A minor modification of @Stella's answer gives a general method for explicitly defining an irrational number $x$ satisfying $p<x<r$ where $p\in\Bbb Q$ and $r\in\Bbb R$ with $p<r$.
The numbers of the form $p+\frac{\sqrt 2}n$ for $n\in\Bbb N$ are all irrational and approach $p$, so for some $n\in\Bbb N$, $p+\frac{\sqrt 2}n<r$; then we can define $x=p+\frac{\sqrt 2}n$ for the smallest such $n$.
You can still find explicit definitions of irrational numbers between any $r<r'$ reals, but the expression is not as clean. Choose some enumeration $(q_i)_{i\in\Bbb N}$ of the rational numbers. Then $\{q_i+\sqrt2:i\in\Bbb N\}$ is a dense set, so there is some $i$ such that $r<q_i+\sqrt2<r'$. (Equivalently, there is a rational number $q_i$ between $r-\sqrt2$ and $r'-\sqrt2$.) Then we can define $x=q_i+\sqrt2$ where $i$ is the smallest natural number satisfying $r<q_i+\sqrt2<r'$.
A: 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.
A: This is a correct way to solve be problem, and this technique works in general.
Another technique is to take some irrational number (maybe $\pi$) and divide it by a ration number that's so big that the quotient is tiny (compared to the difference between the numbers you want to bound it by). $\pi$ is between $3$ and $4$, so $$\frac{1}{100}<\frac{\pi}{300}<\frac{1}{75}$$ which is smaller than $\frac{7}{8}-\frac{3}{4}=\frac{1}{8}$. Thus $\frac{3}{4}+\frac{\pi}{300}$ falls within the range you want.
Since the product of a rational number and an irrational number is irrational, and the sum of a rational number and an irrational number is irrational, this too is irrational. This method is very quick to find answers with, because it's easy to pick monstrously large denominators. For example, dividing $\pi$ by $2^x$ where $x$ is the larger denominator will always work.
EDIT: As a comment notes, there are reasons to prefer $\sqrt{2}$ to $\pi$ here, in case you are asked to prove that the number is irrational. Any irrational number will work with this method, it's just a question of making sure you can easily demonstrate that the number produces is irrational. The comment about $2^x$ still holds for $\sqrt{2}$, but if you were using a larger irrational number you might have to pick a bigger base than $2$. $\lceil \alpha\rceil$ should work as a base for any irrational $\alpha>1$
A: Let $0<a<b$. There are an non countable infinity of irrationals between $a$ and $b$; in particular, for the given numbers $a$ and $b$ the number $\sqrt{ab}$ is irrational as it is proven below.
Is it $\sqrt{ab}<b$? Yes because $\sqrt{ab}<b\iff ab<b^2\Rightarrow a<b$, the same reasoning giving $a<\sqrt{ab}$.
Consequently $a<\sqrt{ab}<b$.
When $a=\frac34$ and $b=\frac 78$, the number  $\sqrt{ab}=\sqrt{\frac{21}{32}}$ is an irrational and $\frac 34<\sqrt{\frac{21}{32}}<\frac78$ as it is easily seen.
