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I have the following homework question:

$$\begin{split} f: \mathbb I \times \mathbb I &\to \mathbb R\times \mathbb R\\ f(x, y) &=(x+y, xy)\end{split}$$

  • Does there exist $(x, y) \in \mathbb I \times \mathbb I$ such that $f(x, y) \in \mathbb R\times \{0\}$?

  • Does there exist $(x, y) \in \mathbb I \times \mathbb I$ such that $f(x, y)=(1, 1)$?

So I know that $\mathbb R\times \mathbb R$ is the two-dimensional Cartesian product of all real numbers. So, $\mathbb I \times \mathbb I$ must be the two-dimensional Cartesian product of all irrational numbers.

And I know that $f$ maps a coordinate $(x,y)$ from one Cartesian grid to another. I think.

Anyways, I'm stumped after that. This problem is actually infuriating. Can someone help me out?

Jason
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  • Can't embed images so that's a problem. Sorry. – Jason Jul 10 '15 at 04:27
  • What did you tried? –  Jul 10 '15 at 04:28
  • Note that the edit that I made is a bit different as your question. I guess there is a typo in your question. –  Jul 10 '15 at 04:34
  • The first question - what two numbers multiplied together give you 0? Can both be irrational? The second question - you have x+y = 1 and xy=1, so if you were to substitute 1/x for y and try to solve x+y = 1 you'd get a quadratic equations - solve that and see if the solution can be irrational. – user247608 Jul 10 '15 at 04:36

2 Answers2

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For the first question: $\Bbb R \times \{ 0 \} = \{(a,0) | a \in \Bbb R\}$ I.e. the second coordinate of the tuple is always $0$.

Implying that the two irrationals $x,y$ have to multiply to $0$. So at least one of $x$ and $y$ must be $0$. But they're irrational, and $0$ is not irrational. By this contradiction, there cannot exist $(x,y) \in \Bbb I \times \Bbb I$ to satisfy this condition.

For the second question: again, we look at both coordinates of the tuple $(x+y,xy)$ individually. We want them to be $(1,1)$. (At this point, look at this thread for help.)

We want $xy = 1$ so $x = 1/y$. But we also want $x+y = 1$, so $x = 1-y$.

Substituting in variables takes us from $x=1/y$ to $1-y = 1/y$, i.e. $y - y^2 = 1$.

Rearranging: $0 = 1 - y + y^2$

Then use the quadratic formula to solve it. I'll leave the calculation to you, but you end up with two complex roots. So there's no real $y$ that satisfies this equation. Since the irrationals are a subset of the reals, there is no irrational $y$ to satisfy it either. So there can't exist $(x,y) \in \Bbb I \times \Bbb I$ to satisfy the condition.

Community
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Newb
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$\bullet$ No, $\nexists$ any such pair. because if $\exists$ such pair then $(x+y,xy)\in \mathbb{R}\times \{0\}$ So $xy=0\implies x=0~y=0$ which is a contradiction since $0$ is not a irrational number

brinkle martyn
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    Outside a formal sentence, $\not\exists$ is probably more confusing than helpful. Just write out the words "there does not exist." And be clear which question you are answering. – Thomas Andrews Jul 10 '15 at 13:21