In the exercise below I need to check whether the relation below is a preference relation ( need to be transitive (if $x>y$ and $y>z$ then $x>z$) and connected ). But I cannot find an example of why it would not be transitive.
$$ f\succeq g \hspace{0.5cm}\Leftrightarrow\hspace{0.5cm} f(q) \geq g(q) \hspace{1cm} \textit{ and }\hspace{1cm} q\in Q\cap[0,1] $$
If this $\succeq$ is transitive and connected then how I can find an utility function that represents it?
Addition: Sorry i mean $ \mathbb{Q} \cap [0,1]$ Ok: so in other example i proved that the relation is not transitive, so it cannot be preference. I had $$ f\succeq g \hspace{0.5cm}\Leftrightarrow\hspace{0.5cm} \int_{0}^{1} f \geq \int_{1}^{2}g $$ I took as a $f(x)= x$, $g(x)=-x+2$, $h(x)=1$.
$$ \int_{0}^{1} f(x) = \int_{0}^{1} x dx= [\frac{1}{2}x]_{0}^{1} =\frac{1}{2} $$ $$ \int_{1}^{2}g(x) = \int_{1}^{2}( -x+2 )dx= -[\frac{1}{2}x^{2}]_{1}^{2} + 2[x]_{1}^{2} = -\frac{3}{2} + 2 = \frac{1}{2} $$
so it is true for $$ \int_{0}^{1} f(x) \geq \int_{1}^{2}g(x) $$ now
$$ \int_{0}^{1} g(x)= \int_{0}^{1} (-x+2)dx = -[\frac{1}{2}x^{2}]_{0}^{1} + 2[x]_{0}^{1}= -\frac{1}{2} +2= \frac{3}{2} $$ $$ \int_{1}^{2}h(x) = \int_{1}^{2} 1dx = 1 $$
it is true for $$ \int_{0}^{1} g(x) \geq \int_{1}^{2}h(x) $$ but it is not true for $$\int_{0}^{1} f(x) \geq \int_{1}^{2}h(x), (\frac{1}{2} \not \geq 1) $$