0
$\begingroup$

I'd really appreciate your help with this one:

I got the following:

R = being real numbers

Q = rational numbers

N = natural numbers

$ R^2 $ being the pair of $ (a,b)$ with $a,b$ being real numbers

$$ 〖[(R - Q )∩ R^2 ] 〗^c ∪ N $$

My reasoning tell me R - Q would give me all the irrational numbers, the intersection between the irrational numbers and $R^2$ would be empty.

After that this is where it gets fuzzy.

Could someone clarify what would the answer be?

$\endgroup$
4
  • 2
    $\begingroup$ The universe set $U$ since the complement of an empty set is the universe. Since $\mathbb N \subset U$, $U \cup \mathbb N = U$. $\endgroup$
    – Andrew Li
    Aug 17, 2019 at 3:47
  • $\begingroup$ Are you sure your version of question is right? [Seems peculiar to set up a problem where a one dimensional set gets intersected with a two dimensional one, inside the inner brackets] $\endgroup$
    – coffeemath
    Aug 17, 2019 at 4:00
  • $\begingroup$ Yes I found it weird too. I also find weird the fact that they don't define the universe at all. $\endgroup$
    – Gerardo Hernández
    Aug 17, 2019 at 4:04
  • 1
    $\begingroup$ It's possible that they are considering $\mathbb R = \mathbb R\times \{0\} = \{(x,0)|x \in \mathbb R\}$ and that the universe is $\mathbb R^2$ of which $\mathbb R, \mathbb Q,\mathbb N$ are subsets. In that case the answer is $\{(x,y)|$ if $y=0$ then $x$ is rational but if $y\ne 0$ then $x$ can be anything$\}$ $\endgroup$
    – fleablood
    Aug 17, 2019 at 4:11

2 Answers 2

Reset to default
1
$\begingroup$

When you take a complement, you have to specify what universe you are taking the complement with respect to. You are correct that the intersection of $\Bbb {R-Q}$, which consists of single numbers (they happen to be irrational, but that doesn't matter) and $\Bbb R^2$, which consists of ordered pairs is the empty set. The complement is whatever universe you are working in. Assuming your universe is a superset of $\Bbb N$, the final answer is the universe.

$\endgroup$
1
  • $\begingroup$ I agree with you both. Thank you very much for your valuable insight. $\endgroup$
    – Gerardo Hernández
    Aug 17, 2019 at 4:03
1
$\begingroup$

$R - Q$ is the irrationals. This is a subset of the reals.

$R^2$ is the set of ordered pairs. These are not real numbers; they are pairs of numbers..

$(R-Q) \cap R^2$ is the intersection of the irrational real numbers and the ordered pairs, or in other words all the irrational numbers that are also an ordered pair of reals.

Well, no number is also a pair of numbers. Those are entirely different types of things.

So $(R-Q)\cap R^2$ is the empty set.

The complement of the empty set is everything that isn't in the empty set. Nothing is in the empty set so everything is in the complement of the empty set. $[(R-Q)]\cap R^2]^c$ is the entire universe of things. (We haven't actually been told what our actual universe is.... )

And UNIVERSE $\cup N$ is everything that is either something in the universe or a natural number. Well, everything is something in the universe so this set is everything in the universe.

So $[(R-Q)\cap R^2]^c \cup N = $UNIVERSE.

======

Alternative answer. $\mathbb R\times \{0\} = \{(x,0)|x\in \mathbb R\} \cong \mathbb R$ and so maybe the question is taking $\mathbb R$ to be equal to $\mathbb R\times \{0\}$.

If so the universe is $\mathbb R^2$ and $\mathbb N\cong \mathbb N\times\{0\} = \{(n,0)|n\in \mathbb N\}$ etc are all subsets of $\mathbb R^2$.

In which case:

$(\mathbb R-\mathbb Q)\cap \mathbb R^2 = \{(x,0)|x\not \in \mathbb Q\}$.

And $[(\mathbb R-\mathbb Q)\cap \mathbb R^2]^c = \{(x,y)|y \in \mathbb R, x\in \mathbb R$ and if $y=0$ then $x\in \mathbb Q\}$.

And $[(\mathbb R-\mathbb Q)\cap \mathbb R^2]^c\cup \mathbb N = \{(x,y)|y \in \mathbb R, x\in \mathbb R$ and if $y=0$ then $x\in \mathbb Q$ or $x\in \mathbb N\}$.

But as $\mathbb N\subset \mathbb Q$ that is $[(\mathbb R-\mathbb Q)\cap \mathbb R^2]^c\cup \mathbb N = \{(x,y)|y \in \mathbb R, x\in \mathbb R$ and if $y=0$ then $x\in \mathbb Q\}$

$\endgroup$
3
  • $\begingroup$ Great answer! The tutor could had done that asumption of R x (0). Thank you very much! $\endgroup$
    – Gerardo Hernández
    Aug 17, 2019 at 4:31
  • $\begingroup$ IMO that isn't a valid assumption unless it is pointed out. And then you have to state whether the universe is $\mathbb R^{|\mathbb N|}$ or $\mathbb R^2$. ... The third possibility is the are defining ordered pairs as a form of set so $\mathbb R^2 \subset P(\mathbb R)$ but I don't think that is common (if it is ever done) and somehow in some ZFC many defining $\mathbb R$ as sets of sets so that $\mathbb R\subset P(\mathbb R)$ (but somehow without the contradiction) but... that seems like a real abuse. I'm inclined to think the question is just plain wrong. $\endgroup$
    – fleablood
    Aug 17, 2019 at 15:18
  • $\begingroup$ The more I think about it i, the more I incline towards thinking the same thing. Thank you for help $\endgroup$
    – Gerardo Hernández
    Aug 17, 2019 at 19:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .