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I have trouble in proving this, that is if $t\in Q\land t^2<2$, then there exists another rational number $q$ such that $q>t\land q^2<2$. May I ask how to find this rational number?

Andes Lam
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  • Are you allowed to use $\sqrt{2}$? – zkutch Jan 11 '21 at 06:05
  • No, because $\sqrt{2}$ is irrational. – Andes Lam Jan 11 '21 at 06:06
  • There are a number of posts on this site that give an explicit example. – Ross Millikan Jan 11 '21 at 06:07
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    $q=t+\frac1 n$ with $n >\frac 5 {2-t^{2}}$ will do. – Kavi Rama Murthy Jan 11 '21 at 06:07
  • Try $q=\frac{4t}{t^2+2}$ (assuming $t>0$, otherwise use $q=1$), constructed using some AM-GM intuitions. A different intuition (say, Newton's method) will also make a possibly different $q$. –  Jan 11 '21 at 06:08
  • @KaviRamaMurthy Do you mean expanding the term? The term is complicating though. – Andes Lam Jan 11 '21 at 06:13
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    After expanding the square you can use the following facts: $t <2$ (becasue $t^{2}<2<4$) and $\frac 1 {n^{2}} <\frac1 n$. – Kavi Rama Murthy Jan 11 '21 at 06:15
  • @KaviRamaMurthy Sorry for being a downer. Was a bit frustrated after many failed attempts. May I ask how did you come up with this trick? The reason for me to study real analysis is because I am bad at algebraic tricks. Any texts recommended for improvements? – Andes Lam Jan 11 '21 at 08:31
  • @AndesLam: there is not much of a trick. Just take a number $t+h$ where $h$ is positive and try to analyze the equation $(t+h) ^2<2$ ie $0<h(2t+h)<2-t^2$. The thing to note is that you are not supposed to solve this inequality but rather find any $h$ which satisfies this. Attempts to solve inequalities is to miss the point of analysis. – Paramanand Singh Jan 11 '21 at 13:34
  • @ParamanandSingh Interesting critique of my answer, which I have just deleted. I must admit that you have a point. Although the idea that the decimal expansion of $\sqrt{2}$ is infinite is fairly elementary, the OP did comment that he is not allowed to use $\sqrt{2}$. I simply didn't notice that comment. Also, the OP did emphasize the premise that $t \in \mathbb{Q}$, which is irrelevant to my approach. – user2661923 Jan 11 '21 at 13:41
  • Another approach which shows the existence of $q$ without finding it explicitly is given in this answer. The argument in that answer is typical of many arguments used in analysis proofs. – Paramanand Singh Jan 11 '21 at 13:41
  • @user2661923: although you have deleted your answer, I hope you have taken my comment in positive spirit. You will observe that the problem becomes more interesting if one remains within $\mathbb {Q} $. Also the point of the problem (my guess) is to highlight issues with $\mathbb{Q} $ and the need to extend it to $\mathbb {R} $. – Paramanand Singh Jan 11 '21 at 13:46
  • @WillJagy: yes Pell equation and the related continued fractions help a lot here to find nice approximations. – Paramanand Singh Jan 12 '21 at 16:17
  • Check this answer for an exact procedure how to construct such a $q$ from $t$. – rtybase Jan 12 '21 at 18:19
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    @rtybase note $$ \left( \begin{array}{cc} 1&2 \ 1&1 \ \end{array} \right)^2 = \left( \begin{array}{cc} 3&4 \ 2& 3 \ \end{array} \right) $$ as in my answer below – Will Jagy Jan 12 '21 at 18:36
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    @rtybase: So this question is a duplicate of https://math.stackexchange.com/q/2069310/42969, isn't it? – Martin R Jan 12 '21 at 20:28
  • @MartinR the questions themselves sound different, although yes, probably a duplicate on a very careful reading. – rtybase Jan 12 '21 at 20:33

2 Answers2

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If $t$ is negative or zero, we can use $1$ as bigger.

If $t$ is already positive, we can write it as $ t = \frac{p}{q}$ with positive integers $p,q.$ The requirement that $ \left( \frac{p}{q} \right)^2 < 2$ tells us that $$ p^2 - 2 q^2 < 0 $$ We choose $$ w = \frac{3p+4q}{2p+3q}= \frac{3t+4}{2t+3} . $$ As $$ \left( 3p+4q \right)^2 - 2 \left( 2p+3q \right)^2 = p^2 - 2 q^2 < 0$$ we see that $w^2 < 2$ At the same time $$ w - t = \frac{3p+4q}{2p+3q} - \frac{p}{q} = \frac{-2(p^2 - 2 q^2)}{q(2p+3q)} > 0 $$

The reasons for doing this: every quadratic form, in this case $x^2 - 2 y^2,$ has an automorphism group; we use a matrix that generates the group, which is a recipe once we find the first nontrivial Pell solution, $3^2 - 2 \cdot 2^2 = 1$

$$ \left( \begin{array}{cc} 3&2 \\ 4&3 \\ \end{array} \right) \left( \begin{array}{cc} 1&0 \\ 0&-2 \\ \end{array} \right) \left( \begin{array}{cc} 3&4 \\ 2&3 \\ \end{array} \right) = \left( \begin{array}{cc} 1&0 \\ 0& -2 \\ \end{array} \right) $$

note: many binary quadratic forms also have an automorphism of negative determinant, in this case we could negate one variable while leaving the other fixed. The entire group is defined once we add a single instance of that onto the infinite cyclic group of automorphisms with positive determinant

Will Jagy
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Let $q=t+h$ where WLOG $0<h<1$ and $q>0$. Then we want $q^2+2qh+h^2 <2$ but $h^2\leq h$ so it is enough to have $q^2 +2qh+h<2$ or $$h<\frac{2-q^2}{1+2q}$$

MATHS MOD
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