Showing that a preference relation admits a utility function representation Setting: We have two choices of goods $(x_1,y_1)$ and $(x_2,y_2)$ from the set of choices $[-1,1]^2$. Moreover, we have the following preference relation $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff |x_1|\geq|x_2|\&gt;\&gt;\text{or}\&gt;\&gt; |y_1|\geq|y_2|$$

Question: We have to check if there exists a utility function reprensation of this preference relation.

My attempt: So from what I have learned, we know that a preference relation admits a utility function representation if it is rational (reflexive, complete, transitive) and continuous. I have found that this preference relation is not transitive, but this does not mean that there does not exist a utility function representation, because the aforementioned statement is not an if and only if statement.

Moreover, I thought we could try to derive a contradiction from the fact that if there exists a utility function $u$ representation of the preference relation, then we have $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff u(x_1,y_1)\geq u(x_2,y_2)$$
I tried to use the fact that the relation is not transitive to derive a contradiction by using the statement above, but was unsuccessful.

Sadly, these are the two main theorems/propositions that I've learned to solve these problems.

Any help is appreciated!
You just need to use the violation of transitivity and proceed by contradiction. 

Suppose you have  that $(x_1,y_1)R(x_2,y_2)$ and $(x_2,y_2)R(x_3,y_3)$ and a utility function, $u:[-1,1]^2\rightarrow \mathbb{R}$, exists, then $u(x_1,y_1)\geq u(x_2,y_2)$ (these are two reals) and $u(x_2,y_2)\geq u(x_3,y_3)$. (another two reals). Since the reals are transitive, we conclude  $u(x_1,y_1)\geq u(x_3,y_3)$ which in turn implies that $(x_1,y_1)R(x_3,y_3)$. However this is a contradiction (if you choose carefully your three bunddles).