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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:

  1. Square numerators and denominators of the (in this case: both) fractions.
  2. Find LCD (Lowest Common Denominator) for denominators and convert all fractions to this LCD.
  3. 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.
  4. 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.)

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Yes, this is right. However, I would recommend simplifying the fraction to $\frac{\sqrt{37}} 8$. – Noble Mushtak Jan 24 at 15:43
Hello, thanks very much for the reply. Yes, I think in a normal exam they would ask for it to be simplified. With this practice question though, they just ask for an "irrational number", but don't really specify the format. This part of the book is a bit simplified, although I want to actually understand what it is I'm doing. Hence the finding an exact method to find an irrational number/fraction between fractions. I haven't actually found this anywhere in the book (maybe it's beyond GCSE and A-level level), and had to work it out on my own. – Peter S Jan 24 at 15:51
Can you just take any known irrational number and add or subtract a rational number to it to get a result in the range? Given that sqrt(2) is irrational and roughly equal to 1.4, subtract 0.6 from it. The number sqrt(2) - 0.6 is irrational and in the range. – user2023861 Jan 25 at 15:50
Yes, your method is correct. Basically, given $a<b$, we know that if $a^2<r<b^2$, then $a<\sqrt r < b$, so the problem is to show that there is always at least one $r$ which is not a rational square, which you've done in probably the simplest way possible. I prefer your construction to the ones in the answers, which are only "simpler" if you don't think about the amount of machinery that has to go into defining what it means to multiply irrationals by rationals, what it means for one real number to be greater than another, etc. Your method is simpler and more direct. – Jack M Jan 25 at 16:22
Thanks for explaining like that. You've actually helped me understand better what I did myself! Your equation does it perfectly I think. – Peter S Jan 25 at 21:41

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 $q$.

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Thanks. I suspected I was oversimplifying, but I need to start from the relative basics again to make sure I actually understand what I'm doing each step of the way. I suspect I may be overcomplicating my studies because the book itself doesn't seem to ask for something so worked out. But I myself want to know what I'm doing. Your formula is interesting, I'm going to think about that now. – Peter S Jan 24 at 16:05
Or, more generally, p + (q-p)*x, where x is any irrational in (0,1). Here, x is (1/sqrt(2)), but it can be (pi/4), (log_2(3)/2), etc – dberm22 Jan 25 at 13:19

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 $\alpha>1$

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Thanks for the reply. I more or less understand your reply (because I'm refreshing all my maths including the relative basics), and I suspect it will be much clearer in a few days. – Peter S Jan 24 at 16:02
Better to pick $\sqrt 2$ instead of $\pi$ in case you might get asked to prove that $\pi$ is irrational,which is not as easy as proving that $\sqrt 2 $ is.:) – user254665 Jan 25 at 7:40

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.

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Thanks for the reply. This is an interesting method too. I more or less understand, but it'll become clearer in the next few days. I'm going to think about this too. It's all a bit clearer now. – Peter S Jan 24 at 16:18
This doesn't always work. $\sqrt{3/2*2/3} = 1$. – Ramchandra Apte Jan 24 at 17:30
$\sqrt{2/1\times 8/1} = 4$, and you have neither $2$ nor $8$ perfect squares. – TokenToucan Jan 24 at 18:17
@CuddlyCuttlefish So then, this method does not work? – Peter S Jan 24 at 18:31
@Vlad: this does not work if the product $ab$ is a square. – Piquito Jan 24 at 22:47

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.

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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 $\ddot\smile$

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Very nice, and useful for application to my answer as well. I see one problem, though: your statement does not actually imply the OP's (for general rational bounds), since $b$ might be zero! – Mario Carneiro Jan 25 at 16:44
The proof provided proves that $b\neq 0$. – Stella Biderman Jan 25 at 20:39

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.

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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.

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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 $r<q_i+\sqrt2<r'$.

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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.

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