Proof with Cartesian coordinates.

Question

Let $S_b := \{(x,y) \in\mathbb R^2 | y = 3x + b\}$ where $b\in\mathbb R$. Give a direct proof that if $(r,s)\in\mathbb R^2$, then there exists a $b\in\mathbb R$ such that $(r,s) \in S_b$.

I have not worked a proof with Cartesian Coordinates before, so this problem is just a little confusing, but I was able to figure out that $b = s-3r$, but I don't know where to show that or prove it.

Take a look at the notation help page to see how to use MathJax on this site. Also, the $5$ is probably meant to be an $s$: $b=s-3r$ — AlexR, Sep 22 '14 at 14:36

score 1 · Answer 1 · answered Sep 22 '14 at 14:41

Fixing the little mistake that $b=s-3r$, you can just conclude that $$(r,s) \in S_{s-3r}$$ by seeing that $$s = 3\cdot r + (s-3r)$$ This is already the direct proof:
Since you gave a specific $b$ for wich $(r,s)\in S_b$, you know that for all $(r,s) \in\mathbb R^2$ there exists some $b\in\mathbb R$ (given by $s-3r$) such that $(r,s)\in S_b$.

Proof with Cartesian coordinates.

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