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So I attempted to prove that between every rational numbers is an irrational as an exercise, and wanted to see if there are problems in my solution.

Proof:
Suppose $n \in \mathbb{N}$ and $x$ is a positive irrational real number and $y \in \mathbb{Q}$
By the archimedian property there is a $n>0$ such that nx>y
Let $x$ and $y$ be such that $y<nx<x+y$
As $nx>y$, $(n+1)x>y$ , and similarly,$nx<x+y$, so $(n-1)x<y$
so $$(n-1)x<y<(n+1)x$$
Diving throughout by $2m$ and $x$, where $m \in \mathbb{N}$
we get $$\frac{n-1}{2m} < \frac{y}{2mx} < \frac{n+1}{2m} $$
in the case that $n$ is odd, then $n = 2k+1$ where $k \in \mathbb{N}$
which simplifies the inequality to $$\frac{k}{m} < \frac{y}{2mx} < \frac{k+1}{m}$$
AS $\frac{y}{2mx}$ is irrational (can be proven by contradiction), this completes the proof

A similar argument can be constructed if $n$ is even, and if $x$ is negative.

Is this proof valid? can i just say this "Let $x$ and $y$ be such that $y<nx<x+y$"?

Proof that between a pair of rationals $\frac{a}{b}$ and $\frac{c}{d}$ there exists a pair of rationals $\frac{k}{m}$ and $\frac{k+1}{m}$ such that $\frac{a}{b} \leq \frac{k}{m} \lt \frac{k+1}{m} \leq\frac{c}{d}$ where $a,b,c,d,k,m \in \mathbb{Z}$

First consider the case where $b=d$. Then $\frac{a}{b} <\frac{c}{b}$.

So $(c-a)\geq 1$ so for any $n>1 \in \mathbb{N}$, $n(c-a) \gt 1$, and $nc>na+1$
Thus we can say $\frac{na}{nb} \lt \frac{na+1}{nb} \lt\frac{nc}{nb} $ and the pair of rationals are $\frac{k}{m} =\frac{a}{b}$ and $\frac{k+1}{m} = \frac{na+1}{nb}$

For the case where $b \neq d$ , we simply multiply the RHS by d/d and the LHS by b/b and the proof follows from the one above.

Burger crazy
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    This approach requires to seperate the case where the smaller number is negative. An easier way is to define $$r:=a+\frac{b-a}{\sqrt{2}}$$ if the rational numbers $a,b$ with $a<b$ are given. It is easy to show $a<r<b$ and that $r$ is irrational. – Peter Jun 05 '20 at 09:46
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    How does showing there's an irrational between $k/m$ and $(k+1)/m$ show that there's an irrational between each pair of rationals? Not every pair of rationals is of that form. – Gerry Myerson Jun 05 '20 at 10:32
  • @GerryMyerson my understanding was that all rationals are in the form k/m and k+1/m will be the next closest rational, so if there is an irrational between the closest possible pair, which is when m is as big as needed, then it will be the same for any other pair. Am i right to say this? where did i go wrong? – Burger crazy Jun 05 '20 at 11:00
  • @Peter what is the line of thinking that led you to define r in such a manner? – Burger crazy Jun 05 '20 at 11:07
  • First to Gerry's comment : Between two distinct rationals $a$ and $b$ , there are always infinite many rationals (the easiest way is to take $\frac{a+b}{2}$ and this can be continued forever). Concerning $r$, we need a number larger than $a$ , but smaller than $b$ (which we get if we add less than $b-a$), and we have to ensure that it is also irrational. – Peter Jun 05 '20 at 11:11
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    There is no such thing as "the next closest rational", Burger. Between $2/7$ and $3/7$ (for example) there's $3/8$ and $1/3$ and $2/5$ and infinitely many others. – Gerry Myerson Jun 05 '20 at 13:06
  • @Peter I'm confused, so is showing there exists a rational between k/m and k+1/m sufficient? – Burger crazy Jun 05 '20 at 13:18
  • @GerryMyerson sorry i still don't quite get it. like for the example, suppose Ive shown that between 2/7 and 3/7 theres an irrational, then we say what about between 2/7 and 3/8 but then i showed that between 16/56 and 17/56 there is an irrational, which takes care of the above case, and this can continue as the rationals become closer and closer. so thats like what i meant by 'next closest' – Burger crazy Jun 05 '20 at 13:26
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    Have you shown that between any two rationals $a/b<c/d$ there's $k$ and $m$ such that $a/b\le k/m<k/(m+1)\le c/d$, Burger? – Gerry Myerson Jun 05 '20 at 13:29
  • @GerryMyerson i edited the post with such a proof at the bottom . Does this complete the proof? Or did i go wrong somewhere? – Burger crazy Jun 06 '20 at 02:08
  • @Cheesecake it helps thanks ! but i was more of looking for errors in my line of reasoning. – Burger crazy Jun 06 '20 at 02:55
  • If $y$ is negative, then there may not be an $n>0$ such that $nx<x+y$. – Gerry Myerson Jun 06 '20 at 05:54
  • @GerryMyerson ahh okay thanks! So specifying that y > 0 completes it then? As i don't think that limiting y>0 impacts the end result – Burger crazy Jun 06 '20 at 08:28
  • $k$ in your argument is determined by $n$, and $n$ isn't arbitrary, it's the one positive integer such that $(n-1)x<y<nx$, so you've proved that between the very special rational $k/m$ and the rational $(k+1)/m$ there's an irrational. So it's not good enough to prove that between $a/b$ and $c/d$ there are rationals of the form $k/m$ and $(k+1)/m$; you have to prove that you can take $k=(n-1)/2$ where $(n-1)x<y<nx$. – Gerry Myerson Jun 06 '20 at 12:22
  • Let us continue this discussion in chat. – Burger crazy Jun 06 '20 at 12:36
  • Sorry, I don't do chat. Also, it's going on 11 o'clock at night here. – Gerry Myerson Jun 06 '20 at 12:39
  • @GerryMyerson no problem thanks for the help anyways! – Burger crazy Jun 06 '20 at 12:41

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