How to prove #R + #P = #R
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I have already started this. I redefined the Reals as the Reals minus the Positive Integers (to make the two sets disjoint) so that I could prove that #(R - P) + #P = #R. I know that to prove this I will have to prove that #(R - P) = #(R). I would have to define a bijection, f:(R-P) -> R. I am having trouble coming up with a formula that will map the (R-P) to R. If anyone could point me in the right direction I would really appreciate it. Thanks
cardinals integers
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
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add a comment |
$begingroup$
I have already started this. I redefined the Reals as the Reals minus the Positive Integers (to make the two sets disjoint) so that I could prove that #(R - P) + #P = #R. I know that to prove this I will have to prove that #(R - P) = #(R). I would have to define a bijection, f:(R-P) -> R. I am having trouble coming up with a formula that will map the (R-P) to R. If anyone could point me in the right direction I would really appreciate it. Thanks
cardinals integers
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
$endgroup$
$begingroup$
Wait, so you're trying to prove $#(Bbb RbackslashBbb P)+#Bbb P=#Bbb R$ with $Bbb P:={xinBbb Z|x>0}$? Well, $#(Abackslash B)+#B=#A$ works for any sets $A,,B$ with $Bsubseteq A$, so perhaps I've misunderstood; perhaps you're trying to prove something less trivial.
$endgroup$
– J.G.
Dec 8 '18 at 21:49
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I am trying to prove #(reals) + #(positive integers) = #(reals) by defining a bijection f: (R-P) --> R. I'm really sorry if my question is confusing.
$endgroup$
– JaneDoe1094
Dec 8 '18 at 22:09
add a comment |
$begingroup$
I have already started this. I redefined the Reals as the Reals minus the Positive Integers (to make the two sets disjoint) so that I could prove that #(R - P) + #P = #R. I know that to prove this I will have to prove that #(R - P) = #(R). I would have to define a bijection, f:(R-P) -> R. I am having trouble coming up with a formula that will map the (R-P) to R. If anyone could point me in the right direction I would really appreciate it. Thanks
cardinals integers
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
$endgroup$
I have already started this. I redefined the Reals as the Reals minus the Positive Integers (to make the two sets disjoint) so that I could prove that #(R - P) + #P = #R. I know that to prove this I will have to prove that #(R - P) = #(R). I would have to define a bijection, f:(R-P) -> R. I am having trouble coming up with a formula that will map the (R-P) to R. If anyone could point me in the right direction I would really appreciate it. Thanks
cardinals integers
cardinals integers
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
asked Dec 8 '18 at 21:36
JaneDoe1094JaneDoe1094
1
1
$begingroup$
Wait, so you're trying to prove $#(Bbb RbackslashBbb P)+#Bbb P=#Bbb R$ with $Bbb P:={xinBbb Z|x>0}$? Well, $#(Abackslash B)+#B=#A$ works for any sets $A,,B$ with $Bsubseteq A$, so perhaps I've misunderstood; perhaps you're trying to prove something less trivial.
$endgroup$
– J.G.
Dec 8 '18 at 21:49
$begingroup$
I am trying to prove #(reals) + #(positive integers) = #(reals) by defining a bijection f: (R-P) --> R. I'm really sorry if my question is confusing.
$endgroup$
– JaneDoe1094
Dec 8 '18 at 22:09
add a comment |
$begingroup$
Wait, so you're trying to prove $#(Bbb RbackslashBbb P)+#Bbb P=#Bbb R$ with $Bbb P:={xinBbb Z|x>0}$? Well, $#(Abackslash B)+#B=#A$ works for any sets $A,,B$ with $Bsubseteq A$, so perhaps I've misunderstood; perhaps you're trying to prove something less trivial.
$endgroup$
– J.G.
Dec 8 '18 at 21:49
$begingroup$
I am trying to prove #(reals) + #(positive integers) = #(reals) by defining a bijection f: (R-P) --> R. I'm really sorry if my question is confusing.
$endgroup$
– JaneDoe1094
Dec 8 '18 at 22:09
$begingroup$
Wait, so you're trying to prove $#(Bbb RbackslashBbb P)+#Bbb P=#Bbb R$ with $Bbb P:={xinBbb Z|x>0}$? Well, $#(Abackslash B)+#B=#A$ works for any sets $A,,B$ with $Bsubseteq A$, so perhaps I've misunderstood; perhaps you're trying to prove something less trivial.
$endgroup$
– J.G.
Dec 8 '18 at 21:49
$begingroup$
Wait, so you're trying to prove $#(Bbb RbackslashBbb P)+#Bbb P=#Bbb R$ with $Bbb P:={xinBbb Z|x>0}$? Well, $#(Abackslash B)+#B=#A$ works for any sets $A,,B$ with $Bsubseteq A$, so perhaps I've misunderstood; perhaps you're trying to prove something less trivial.
$endgroup$
– J.G.
Dec 8 '18 at 21:49
$begingroup$
I am trying to prove #(reals) + #(positive integers) = #(reals) by defining a bijection f: (R-P) --> R. I'm really sorry if my question is confusing.
$endgroup$
– JaneDoe1094
Dec 8 '18 at 22:09
$begingroup$
I am trying to prove #(reals) + #(positive integers) = #(reals) by defining a bijection f: (R-P) --> R. I'm really sorry if my question is confusing.
$endgroup$
– JaneDoe1094
Dec 8 '18 at 22:09
add a comment |
1 Answer
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Do you know how to biject $[0,1)$ with $(0,1)$? You can then use the identity on $(-infty,0)$ and map $[n,n+1)$ to $(n,n+1)$
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
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add a comment |
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$begingroup$
Do you know how to biject $[0,1)$ with $(0,1)$? You can then use the identity on $(-infty,0)$ and map $[n,n+1)$ to $(n,n+1)$
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
Do you know how to biject $[0,1)$ with $(0,1)$? You can then use the identity on $(-infty,0)$ and map $[n,n+1)$ to $(n,n+1)$
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
Do you know how to biject $[0,1)$ with $(0,1)$? You can then use the identity on $(-infty,0)$ and map $[n,n+1)$ to $(n,n+1)$
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
$endgroup$
Do you know how to biject $[0,1)$ with $(0,1)$? You can then use the identity on $(-infty,0)$ and map $[n,n+1)$ to $(n,n+1)$
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
answered Dec 8 '18 at 22:16
Ross MillikanRoss Millikan
296k23198371
296k23198371
add a comment |
add a comment |
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$begingroup$
Wait, so you're trying to prove $#(Bbb RbackslashBbb P)+#Bbb P=#Bbb R$ with $Bbb P:={xinBbb Z|x>0}$? Well, $#(Abackslash B)+#B=#A$ works for any sets $A,,B$ with $Bsubseteq A$, so perhaps I've misunderstood; perhaps you're trying to prove something less trivial.
$endgroup$
– J.G.
Dec 8 '18 at 21:49
$begingroup$
I am trying to prove #(reals) + #(positive integers) = #(reals) by defining a bijection f: (R-P) --> R. I'm really sorry if my question is confusing.
$endgroup$
– JaneDoe1094
Dec 8 '18 at 22:09