Define function J : Q×Q → R by the rule J(r, s) = r+ sqrt(2)s for all (r, s) ∈ Q×Q
I have no real idea how to read this. My thoughts are: For every pair of rationals in QxQ, or the pair of any two rationals, r and s, J = the function listed above.
Now, typical strategy for 1-1 is to make r1, s1, r2, s2, and see if you can derive an equality; now, I've decided that this function is not 1-1, but I'm not exactly sure how to show a counter example being as sqrt(2) doesn't have a value I can really work with....
Either way, I'm expected to give a counterexample and I don't entirely know where to start.
Thank you for any help!
This function actually is one-to-one.
A function $f$ is $1-1$ if whenever $f(y) = f(z)$, it follows that $y = z$. Ok, so let's look at $J(r_{1}, r_{2})$ and $J(s_{1}, s_{2})$. Suppose they are equal. We want to show $(r_{1}, r_{2}) = (s_{1},s_{2})$.
Well, if $J(r_{1}, r_{2}) = J(s_{1}, s_{2})$, then $r_{1} + \sqrt{2}r_{2} = s_{1} + \sqrt{2}s_{2}$. That means $$r_{1} - s_{1} + \sqrt{2}(r_{2} - s_{2}) = 0 \tag{1}.$$ Since $r_{1}, r_{2}, s_{1}, s_{2}$ are rational numbers, the only way this can happen is if $r_{1} = s_{1}$ and $r_{2} = s_{2}$.
To see why this is, note that if $r_{1} \neq s_{1}$ and $r_{2} \neq s_{2}$, then $r_{1} - s_{1} \neq 0$ and $r_{2} - s_{2} \neq 0$.
By (1), we then get $\sqrt{2} = \dfrac{r_{1} - s_{1}}{r_{2}-s_{2}}$. But each piece of the right hand side is rational, so the whole right hand side is rational, while the left hand side is irrational! Contradiction! Thus, the differences have to equal $0$, and so $(r_{1}, r_{2}) = (s_{1}, s_{2})$.