Mixing
Partitioning an Infinite Set Using Natural Numbers
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Can we partition all positive real numbers that are greater or equal to one, i.e. $[1, infty)$ into sets of $[n, n+1)$ such that $n$ is a natural number?
Intuitively it is obviously correct but is there a way to prove it by using archimedean property?
real-analysis elementary-number-theory elementary-set-theory
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
asked Jan 31 at 1:21
james blackjames black
424114
$endgroup$
add a comment |
$begingroup$
Can we partition all positive real numbers that are greater or equal to one, i.e. $[1, infty)$ into sets of $[n, n+1)$ such that $n$ is a natural number?
Intuitively it is obviously correct but is there a way to prove it by using archimedean property?
real-analysis elementary-number-theory elementary-set-theory
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
asked Jan 31 at 1:21
james blackjames black
424114
$endgroup$
add a comment |
$begingroup$
Can we partition all positive real numbers that are greater or equal to one, i.e. $[1, infty)$ into sets of $[n, n+1)$ such that $n$ is a natural number?
Intuitively it is obviously correct but is there a way to prove it by using archimedean property?
real-analysis elementary-number-theory elementary-set-theory
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
asked Jan 31 at 1:21
james blackjames black
424114
$endgroup$
Can we partition all positive real numbers that are greater or equal to one, i.e. $[1, infty)$ into sets of $[n, n+1)$ such that $n$ is a natural number?
Intuitively it is obviously correct but is there a way to prove it by using archimedean property?
real-analysis elementary-number-theory elementary-set-theory
real-analysis elementary-number-theory elementary-set-theory
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
asked Jan 31 at 1:21
james blackjames black
424114
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
asked Jan 31 at 1:21
james blackjames black
424114
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
edited Jan 31 at 1:35
Theo Bendit
20.3k12353
20.3k12353
asked Jan 31 at 1:21
james blackjames black
424114
asked Jan 31 at 1:21
james blackjames black
424114
asked Jan 31 at 1:21
james blackjames black
424114
424114
add a comment |
add a comment |
1 Answer
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$begingroup$
Assume 1 <= r. Prove B = { n in N : r < n } is non empty.
Next show B has minimum element. Call that integer n.
As 1 <= r < n, conclude 1 <= n-1 <= r < n.
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
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add a comment |
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1 Answer
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1 Answer
1
active
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$begingroup$
Assume 1 <= r. Prove B = { n in N : r < n } is non empty.
Next show B has minimum element. Call that integer n.
As 1 <= r < n, conclude 1 <= n-1 <= r < n.
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
$endgroup$
add a comment |
$begingroup$
Assume 1 <= r. Prove B = { n in N : r < n } is non empty.
Next show B has minimum element. Call that integer n.
As 1 <= r < n, conclude 1 <= n-1 <= r < n.
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
$endgroup$
add a comment |
$begingroup$
Assume 1 <= r. Prove B = { n in N : r < n } is non empty.
Next show B has minimum element. Call that integer n.
As 1 <= r < n, conclude 1 <= n-1 <= r < n.
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
$endgroup$
Assume 1 <= r. Prove B = { n in N : r < n } is non empty.
Next show B has minimum element. Call that integer n.
As 1 <= r < n, conclude 1 <= n-1 <= r < n.
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
answered Jan 31 at 2:05
William ElliotWilliam Elliot
8,9662820
8,9662820
add a comment |
add a comment |
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