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 May 17th, 2017, 11:01 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2 Uncountable Subset 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
 May 18th, 2017, 01:47 AM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,601 Thanks: 816 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.
May 18th, 2017, 01:59 AM   #3
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 Originally Posted by romsek 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 ??

May 18th, 2017, 03:10 AM   #4
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 Originally Posted by Lalitha183 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?

May 18th, 2017, 03:55 AM   #5
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 Originally Posted by romsek no do you know what an algebraic number is?
Complex numbers ?

 May 18th, 2017, 05:06 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,820 Thanks: 750 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). $\sqrt{2}$ is algebraic since it satisfies $x^2= 2$. e and $\pi$ are transcendental.
May 19th, 2017, 04:34 AM   #7
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 Originally Posted by Country Boy 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). $\sqrt{2}$ is algebraic since it satisfies $x^2= 2$. e and $\pi$ 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.

 May 19th, 2017, 07:00 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra $\{(x,y) \in \mathbb R^2: x \in \mathbb Q\}$ 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 $\mathbb Q^2$.

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