Uncountable Subset

Nov 2015
246
4
hyderabad
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
Sep 2015
2,959
1,673
USA
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.
 
Nov 2015
246
4
hyderabad
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
Sep 2015
2,959
1,673
USA
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
Jan 2015
3,261
899
Alabama
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
Jan 2015
3,261
899
Alabama
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
Dec 2013
7,712
2,682
Colombia
$\{(x,y) \in \mathbb R^2: x \in \mathbb Q\}$ 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 $\mathbb Q^2$.