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May 17th, 2017, 11:01 PM   #1
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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.
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May 18th, 2017, 01:47 AM   #2
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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.
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May 18th, 2017, 01:59 AM   #3
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Quote:
Originally Posted by romsek View Post
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 ??
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May 18th, 2017, 03:10 AM   #4
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Quote:
Originally Posted by Lalitha183 View Post
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?
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May 18th, 2017, 03:55 AM   #5
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Originally Posted by romsek View Post
no

do you know what an algebraic number is?
Complex numbers ?
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May 18th, 2017, 05:06 AM   #6
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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). is algebraic since it satisfies . e and are transcendental.
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May 19th, 2017, 04:34 AM   #7
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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). is algebraic since it satisfies . e and 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.
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May 19th, 2017, 07:00 AM   #8
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$\{(x,y) \in \mathbb R^2: x \in \mathbb Q\}$ 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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