# Uncountable Subset

#### Lalitha183

Which of the following is an uncountable subset of $R^2$ ?
A) $$\displaystyle {(x,y) \epsilon R^2 : {x \epsilon Q} { or} {(x+y ) \epsilon Q} }$$
B)$$\displaystyle {(x,y) \epsilon R^2 : {x \epsilon Q} and {y \epsilon Q} }$$
C) $$\displaystyle {(x,y) \epsilon R^2 : {x \epsilon Q} { or} {y \epsilon Q} }$$
D) $$\displaystyle {(x,y) \epsilon R^2 : {x \epsilon Q} { or} {y^2 \epsilon Q} }$$

In my view Option D contains all irrational and rational values as y can be an irrational whose square gives us a rational number which belongs to $Q$.

Please correct me If I'm wrong.
Thanks #### romsek

Math Team
It will be (C)

if $x \in \mathbb{Q}$ then $y$ is only restricted in that it is real.

the same holds if $y \in \mathbb{Q}$ for $x$

The fact that $x$ or $y$ can be reals makes the pair uncountable.

The reason it is not (D) is that $x \in \mathbb{Q}$ is the rationals in $x$

and $y^2 \in \mathbb{Q}$ is a subset of the algebraic numbers.

Both are countable.

#### Lalitha183

It will be (C)

if $x \in \mathbb{Q}$ then $y$ is only restricted in that it is real.

the same holds if $y \in \mathbb{Q}$ for $x$

The fact that $x$ or $y$ can be reals makes the pair uncountable.

The reason it is not (D) is that $x \in \mathbb{Q}$ is the rationals in $x$

and $y^2 \in \mathbb{Q}$ is a subset of the algebraic numbers.

Both are countable.
So option D restricts the use of negatives for $y$ ?
and either $x$ or $y$ in option C covers the real numbers ??

#### romsek

Math Team
So option D restricts the use of negatives for $y$ ?
and either $x$ or $y$ in option C covers the real numbers ??
no

do you know what an algebraic number is?

#### Country Boy

Math Team
No. An "algebraic number" is a real number that is a solution to some polynomial equation with integer coefficients. A real number that is not "algebraic" is called "transcendental". All rational numbers are algebraic (since a/b satisfies bx= a). $$\displaystyle \sqrt{2}$$ is algebraic since it satisfies $$\displaystyle x^2= 2$$. e and $$\displaystyle \pi$$ are transcendental.

#### Country Boy

Math Team
No. An "algebraic number" is a real number that is a solution to some polynomial equation with integer coefficients. A real number that is not "algebraic" is called "transcendental". All rational numbers are algebraic (since a/b satisfies bx= a). $$\displaystyle \sqrt{2}$$ is algebraic since it satisfies $$\displaystyle x^2= 2$$. e and $$\displaystyle \pi$$ are transcendental.
The point, that I should have mentioned before, is that, while the set of all irrational numbers is uncountable, the set of all "algebraic numbers" is itself countable.

#### v8archie

Math Team
$\{(x,y) \in \mathbb R^2: x \in \mathbb Q\}$ is uncountable. It contains a "copy" of $\mathbb R$ for every $x \in \mathbb Q\}$. That makes both A, C and D uncountable I think.

B is countable because it is exactly $\mathbb Q^2$.