We have
\[\begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} p + 2q \\ p+q \end{bmatrix}\]
By squaring the rational number \(\frac{p}{q}\) with p and q being positive integers, thus \(\frac{p^2}{q^2} < 2\)
After multiplication we have a new rational number lets call it \(\frac{r}{s}\) and \(\frac{r^2}{s^2} > 2\) \[\begin{bmatrix}
p^2 \\ q^2
\end{bmatrix}<2 ,
\begin{bmatrix}
p^2 + 4pq +4q^2 \\ p^2+2pq+2q^2
\end{bmatrix} > 2\]
We can use 1 to be our rational number. Cube root of 1, 1^2, is 1 and is less than 2.
\[\begin{bmatrix}
1 & 2 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ 1
\end{bmatrix} =
\begin{bmatrix}
3 \\ 2
\end{bmatrix}\] \[=\frac{3}{2}\] \[(\frac{3}{2}) ^2 = \frac{9}{4}\]
Here \(\frac{9}{4} > 2\) therefore our proof is true.
Lets say a = 0.5 and b = 0.6
\( a = 0.5 > 0 \\ (0.5)^2 < 2 \\ 0.25 < 2 \)
That works, Now for b = 0.6 does it apply?
1. \(a < b, 0.5 < 0.6\)
2. \(b^2 < 2, 0.6^2 < 2, 0.36 < 2 \)
It satisfies our proof therefore being true.