Find a rational number that is greater than $t$ where $t^2<2$

Question

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?

Answer 1

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

Answer 2

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 $$h<\frac{2-q^2}{1+2q}$$