Mixing
logic and set theory
$begingroup$
Let $S$ be a set of real numbers that is closed under multiplication. Let $Tsubset S$ and $Usubset S$ such that $Tcup U= S$. Since the product of any three (not necessarily different) elements of $T$ is in $T$ and the product of any three elements of $U$ is in $U$, prove that at least one of the two subsets $T$, $U$ is closed under multiplication.
real-analysis
edited Jan 9 at 4:55
Lee
328111
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
$endgroup$
|
show 7 more comments
$begingroup$
Let $S$ be a set of real numbers that is closed under multiplication. Let $Tsubset S$ and $Usubset S$ such that $Tcup U= S$. Since the product of any three (not necessarily different) elements of $T$ is in $T$ and the product of any three elements of $U$ is in $U$, prove that at least one of the two subsets $T$, $U$ is closed under multiplication.
real-analysis
edited Jan 9 at 4:55
Lee
328111
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
$endgroup$
2
$begingroup$
What have you tried ?
$endgroup$
– T. Fo
Jan 9 at 0:44
2
$begingroup$
Shouldn't "since the product..." be "if the product..."? Certainly I can split $Bbb R$ into two subsets such that neither is closed under threeway products.
$endgroup$
– Ross Millikan
Jan 9 at 1:47
$begingroup$
How is that related to group theory? Is some other information given?
$endgroup$
– Martund
Jan 9 at 2:44
1
$begingroup$
okay, I'm sorry for saying such stupidity, it is clear that the products of two irrational can be a rational, for example the root of 2 by itself or the root of six by the root of twenty-four.
$endgroup$
– Luis guamushig
Jan 9 at 9:09
1
$begingroup$
besides that the product of two rationals is always rational
$endgroup$
– Luis guamushig
Jan 9 at 9:15
|
show 7 more comments
$begingroup$
Let $S$ be a set of real numbers that is closed under multiplication. Let $Tsubset S$ and $Usubset S$ such that $Tcup U= S$. Since the product of any three (not necessarily different) elements of $T$ is in $T$ and the product of any three elements of $U$ is in $U$, prove that at least one of the two subsets $T$, $U$ is closed under multiplication.
real-analysis
edited Jan 9 at 4:55
Lee
328111
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
$endgroup$
Let $S$ be a set of real numbers that is closed under multiplication. Let $Tsubset S$ and $Usubset S$ such that $Tcup U= S$. Since the product of any three (not necessarily different) elements of $T$ is in $T$ and the product of any three elements of $U$ is in $U$, prove that at least one of the two subsets $T$, $U$ is closed under multiplication.
real-analysis
real-analysis
edited Jan 9 at 4:55
Lee
328111
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
edited Jan 9 at 4:55
Lee
328111
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
edited Jan 9 at 4:55
Lee
328111
edited Jan 9 at 4:55
Lee
328111
edited Jan 9 at 4:55
Lee
328111
328111
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
asked Jan 9 at 0:43
Luis guamushigLuis guamushig
106
106
2
$begingroup$
What have you tried ?
$endgroup$
– T. Fo
Jan 9 at 0:44
2
$begingroup$
Shouldn't "since the product..." be "if the product..."? Certainly I can split $Bbb R$ into two subsets such that neither is closed under threeway products.
$endgroup$
– Ross Millikan
Jan 9 at 1:47
$begingroup$
How is that related to group theory? Is some other information given?
$endgroup$
– Martund
Jan 9 at 2:44
1
$begingroup$
okay, I'm sorry for saying such stupidity, it is clear that the products of two irrational can be a rational, for example the root of 2 by itself or the root of six by the root of twenty-four.
$endgroup$
– Luis guamushig
Jan 9 at 9:09
1
$begingroup$
besides that the product of two rationals is always rational
$endgroup$
– Luis guamushig
Jan 9 at 9:15
|
show 7 more comments
2
$begingroup$
What have you tried ?
$endgroup$
– T. Fo
Jan 9 at 0:44
2
$begingroup$
Shouldn't "since the product..." be "if the product..."? Certainly I can split $Bbb R$ into two subsets such that neither is closed under threeway products.
$endgroup$
– Ross Millikan
Jan 9 at 1:47
$begingroup$
How is that related to group theory? Is some other information given?
$endgroup$
– Martund
Jan 9 at 2:44
1
$begingroup$
okay, I'm sorry for saying such stupidity, it is clear that the products of two irrational can be a rational, for example the root of 2 by itself or the root of six by the root of twenty-four.
$endgroup$
– Luis guamushig
Jan 9 at 9:09
1
$begingroup$
besides that the product of two rationals is always rational
$endgroup$
– Luis guamushig
Jan 9 at 9:15
2
2
$begingroup$
What have you tried ?
$endgroup$
– T. Fo
Jan 9 at 0:44
$begingroup$
What have you tried ?
$endgroup$
– T. Fo
Jan 9 at 0:44
2
2
$begingroup$
Shouldn't "since the product..." be "if the product..."? Certainly I can split $Bbb R$ into two subsets such that neither is closed under threeway products.
$endgroup$
– Ross Millikan
Jan 9 at 1:47
$begingroup$
Shouldn't "since the product..." be "if the product..."? Certainly I can split $Bbb R$ into two subsets such that neither is closed under threeway products.
$endgroup$
– Ross Millikan
Jan 9 at 1:47
$begingroup$
How is that related to group theory? Is some other information given?
$endgroup$
– Martund
Jan 9 at 2:44
$begingroup$
How is that related to group theory? Is some other information given?
$endgroup$
– Martund
Jan 9 at 2:44
1
1
$begingroup$
okay, I'm sorry for saying such stupidity, it is clear that the products of two irrational can be a rational, for example the root of 2 by itself or the root of six by the root of twenty-four.
$endgroup$
– Luis guamushig
Jan 9 at 9:09
$begingroup$
okay, I'm sorry for saying such stupidity, it is clear that the products of two irrational can be a rational, for example the root of 2 by itself or the root of six by the root of twenty-four.
$endgroup$
– Luis guamushig
Jan 9 at 9:09
1
1
$begingroup$
besides that the product of two rationals is always rational
$endgroup$
– Luis guamushig
Jan 9 at 9:15
$begingroup$
besides that the product of two rationals is always rational
$endgroup$
– Luis guamushig
Jan 9 at 9:15
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let S be all even numbers, T = {8}, U = S - T.
Clearly, neither T nor S is closed under product of three elements.
To rescue the problem, change "since" to "assume".
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
$endgroup$
add a comment |
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$begingroup$
Let S be all even numbers, T = {8}, U = S - T.
Clearly, neither T nor S is closed under product of three elements.
To rescue the problem, change "since" to "assume".
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
$endgroup$
add a comment |
$begingroup$
Let S be all even numbers, T = {8}, U = S - T.
Clearly, neither T nor S is closed under product of three elements.
To rescue the problem, change "since" to "assume".
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
$endgroup$
add a comment |
$begingroup$
Let S be all even numbers, T = {8}, U = S - T.
Clearly, neither T nor S is closed under product of three elements.
To rescue the problem, change "since" to "assume".
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
$endgroup$
Let S be all even numbers, T = {8}, U = S - T.
Clearly, neither T nor S is closed under product of three elements.
To rescue the problem, change "since" to "assume".
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
answered Jan 9 at 3:53
William ElliotWilliam Elliot
7,9152720
7,9152720
add a comment |
add a comment |
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2
$begingroup$
What have you tried ?
$endgroup$
– T. Fo
Jan 9 at 0:44
2
$begingroup$
Shouldn't "since the product..." be "if the product..."? Certainly I can split $Bbb R$ into two subsets such that neither is closed under threeway products.
$endgroup$
– Ross Millikan
Jan 9 at 1:47
$begingroup$
How is that related to group theory? Is some other information given?
$endgroup$
– Martund
Jan 9 at 2:44
1
$begingroup$
okay, I'm sorry for saying such stupidity, it is clear that the products of two irrational can be a rational, for example the root of 2 by itself or the root of six by the root of twenty-four.
$endgroup$
– Luis guamushig
Jan 9 at 9:09
1
$begingroup$
besides that the product of two rationals is always rational
$endgroup$
– Luis guamushig
Jan 9 at 9:15