# Alternative way of thinking about irrationality of $\sqrt{2}$

A user on stack exchange suggested to think of the following problem as a good way to distinguish between algebraic mindset and an analysis mindset:

a) Prove $\sqrt{2}$ is irrational by expressing it is equal to $\frac{m}{n}$ and then come up with some irreducibility argument (lowest terms). (We did this in my introduction to maths course)

b) Prove that whenever some rational $a^2 <2$, there is some other rational $b>a$, such that $b^2 <2$

I found this second way of thinking about it extremely interesting and have been messing around with inequalities like $0< a^2 -1<1$, but I think I'm completely missing the point on the second one, to be quite frank, I'm kind of lost. I do get that the point is that you will never reach $\sqrt2$ since it's a limit point of the set of integers leading up to it, but it's not contained within $\mathbb{Q}$. This is the feature that makes $\mathbb{Q}$ not closed (does not contain all of its limit points).

• Do you mean $a^2≤2$ maybe? I mean, it is also true that whenever we are given some rational $a$ with $a^2<4$ we can find a rational $b>a$ with $a^2<b^2<4$. – lulu May 15 '18 at 19:53
• It's not clear how (b) indicates that $\sqrt{2}$ is irrational. If $a^2<1$ then there is a $b>a$ so that $b^2<1.$ – Thomas Andrews May 15 '18 at 19:54
• Yeah, you need $a^2\le2\implies\exists b>a,b^2\le2$ – Akiva Weinberger May 15 '18 at 19:55
• This answer shows how to prove by descent, without using unique factorization, that if $m^2<n<(m+1)^2$ for $n,m$ non-negative integers, then $\sqrt{n}$ is not rational. math.stackexchange.com/questions/5/… – Thomas Andrews May 15 '18 at 19:59
• – Przemysław Scherwentke May 15 '18 at 20:00

If $a>0$ and $a^2<2$ then $b^2>2$ when $b=2/a$. If $a$ is close to $\sqrt2$ we'd expect $b$ to be roughly as close to $\sqrt2$ so $c=\frac12(a+b)$ should be closer. But $c^2>2$. So try $d=2/c$. Your task: prove $d^2<2$ and $d>a$.

• Now notice that $d=\frac{4}{a+b}>\frac{4}{b+b}=\frac{2}{b}=a$, So indeed $d>a$, also $d^2=\frac{4}{c^2}<\frac{4}{2}=2$ $\blacksquare$ – Wesley Strik May 16 '18 at 21:04
• We can also express our final number explicitly in terms of a: for every $a>0$, with $a^2 <2$, there exists a unique number $d=\frac{4a}{a^2+2}$, such that $d>a$ and $d^2<2$ – Wesley Strik May 16 '18 at 21:10

You can take $$b=a+\frac {1}{n}$$ with $n$ such that

$$a <a+\frac {1}{n}<\sqrt {2}$$

or

$$n=\lfloor \frac {1}{\sqrt {2}-a} \rfloor +1.$$

• Downvoting. What does this answer mean? How can $b = a + \frac1n$ but also $a + \frac1n < b$? – BallBoy May 15 '18 at 20:11
• @Y.Forman Thanks. i fixed it. – hamam_Abdallah May 15 '18 at 20:44
• Thanks for editing! – BallBoy May 15 '18 at 20:45
• If $a^2 < 2$ then we want $n$ such that $2 > (a+1/n)^2=a^2+2a/n+1/n^2$ which is true if $2 > a^2+3a/n$ since $1/n^2 < a/n$ is the same as $na > 1$ which is true for large enough $n$. Therefore $n$ works if $3a/n < 2-a^2$ or $n > 3a/(2-a^2)$. So this answer is fine. Just needs more elaboration. – marty cohen May 15 '18 at 20:52