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This is a question which seems intuitively true but I have yet to come across its mathematical proof. Suppose you have two consecutive positive integers x and y. x and y can be as small or as big as you want them to be. The gap between x and y is choke full of rational numbers. However, what is the proof that the gap between x and y has at least one number $t^2$, where t is a rational number? Most likely t is going to be a decimal. So what proof can you present that any gap between two consecutive integers must have this number which is a square of a rational number? It is obvious that t cannot be an integer.

User4576283
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Let $x\geqslant 0$ be some integer, then $r=x+\tfrac12 > 0$ and $\sqrt r\in\Bbb R$.

Because $\Bbb Q$ is dense in $\Bbb R$ there is a sequence of rational numbers $(r_i)_i$ such that $$\lim_{n\to\infty} r_n = \sqrt r$$ Now because $x\mapsto x^2$ is continuous, it follows that $$\lim_{n\to\infty} r_n^2 = r = x+1/2$$ In particular, there exists some index $N$ such that $$|r_n^2 - (x+1/2)| < 1/2\quad\text{ for all }n \geqslant N$$ This means all such $r_n^2$ have distance less than 1/2 from the midpoint of interval $[x,x+1]$, and thus $r_n^2\in(x,x+1)$.

Whilst the proof is easy enough, I don't like that it uses topology. Maybe there is some number-theoretic approach that's also more explicit.

The proof above is more generic, though: It can be easily adjusted such that:

Between any two positive real numbers $0\leqslant x < y$ there are infinitely many rational $n$-th powers, i.e. numbers of the form $r^n$ with $r\in \Bbb Q$ and $n\in\Bbb N$.

emacs drives me nuts
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It is possible to give a constructive proof and exhibit such a rational.

Firstly for $n=0$ then $0<\left(\frac 12\right)^2<1$, I consider this case separately to avoid division by $0$.

You can now use this approximation for $a,b$ integers.

$$\sqrt{a^2+b}\approx\left(a+\frac b{2a}\right)-\frac{\left(\frac b{2a}\right)^2}{2\left(a+\frac b{2a}\right)}$$

See here for details: https://math.stackexchange.com/a/2866233/399263.

This approximation is a Taylor expansion of order $1$ followed by one step of Newton's method.


Here we want $n<q^2<n+1$, so we will approximate $\sqrt{n+\frac 12}$ so as to be sure our number is strictly between the bounds.

We can take $\begin{cases}a=\lfloor\sqrt{n}\rfloor\\b=n+\frac 12-a^2\end{cases}$

Note that in our case $b$ is not integer as in the original post, but that constraint can be lifted with no issue, since the approximation will stay rational anyway when $a$ integer and $b$ rational.

1 .. 2 : (49/40)^2 = 1.500625
2 .. 3 : (89/56)^2 = 2.525829
3 .. 4 : (137/72)^2 = 3.620563
4 .. 5 : (577/272)^2 = 4.500014
5 .. 6 : (713/304)^2 = 5.500876
6 .. 7 : (857/336)^2 = 6.505536
7 .. 8 : (1009/368)^2 = 7.51773 
8 .. 9 : (1169/400)^2 = 8.541006
9 .. 10 : (2737/888)^2 = 9.500001
10 .. 11 : (337/104)^2 = 10.500092
11 .. 12 : (3337/984)^2 = 11.500645
12 .. 13 : (3649/1032)^2 = 12.502254
13 .. 14 : (147/40)^2 = 13.505625
14 .. 15 : (4297/1128)^2 = 14.511507
15 .. 16 : (4633/1176)^2 = 15.520652
16 .. 17 : (8449/2080)^2 = 16.5     
17 .. 18 : (8969/2144)^2 = 17.500018
18 .. 19 : (9497/2208)^2 = 18.500128
19 .. 20 : (10033/2272)^2 = 19.500465
20 .. 21 : (10577/2336)^2 = 20.501202
zwim
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