Is there an easy enough way to show that between two algebraic numbers there is an infinite number of...
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We know that between two different rational numbers there is an infinite number of irrational numbers and that between two different irrational numbers there is an infinite number of rational numbers.
With the help of the fact that all rational numbers are algebraic and of the fact that rational numbers are dense enough on the real line we could show that between two different transcendental numbers there is an infinite number of algebraic numbers.
But how to show that between two different algebraic numbers there is an infinite number of transcendental numbers?
I would like to see the proof which is as simple as possible.
Thank you.
real-analysis transcendental-numbers
asked Mar 9 '16 at 14:59
Farewell
3,31011024
add a comment |
up vote
2
down vote
favorite
We know that between two different rational numbers there is an infinite number of irrational numbers and that between two different irrational numbers there is an infinite number of rational numbers.
With the help of the fact that all rational numbers are algebraic and of the fact that rational numbers are dense enough on the real line we could show that between two different transcendental numbers there is an infinite number of algebraic numbers.
But how to show that between two different algebraic numbers there is an infinite number of transcendental numbers?
I would like to see the proof which is as simple as possible.
Thank you.
real-analysis transcendental-numbers
asked Mar 9 '16 at 14:59
Farewell
3,31011024
$a + pi^{-n},, n geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:01
@DanielFischer Could that argument be modified to show that there is an uncountable number of them?
– Farewell
Mar 9 '16 at 15:03
The second argument shows that there are $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:04
@DanielFischer That thing with $pi^{-n}$ is easy enough way, you can post it as an answer.
– Farewell
Mar 9 '16 at 15:05
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We know that between two different rational numbers there is an infinite number of irrational numbers and that between two different irrational numbers there is an infinite number of rational numbers.
With the help of the fact that all rational numbers are algebraic and of the fact that rational numbers are dense enough on the real line we could show that between two different transcendental numbers there is an infinite number of algebraic numbers.
But how to show that between two different algebraic numbers there is an infinite number of transcendental numbers?
I would like to see the proof which is as simple as possible.
Thank you.
real-analysis transcendental-numbers
asked Mar 9 '16 at 14:59
Farewell
3,31011024
We know that between two different rational numbers there is an infinite number of irrational numbers and that between two different irrational numbers there is an infinite number of rational numbers.
With the help of the fact that all rational numbers are algebraic and of the fact that rational numbers are dense enough on the real line we could show that between two different transcendental numbers there is an infinite number of algebraic numbers.
But how to show that between two different algebraic numbers there is an infinite number of transcendental numbers?
I would like to see the proof which is as simple as possible.
Thank you.
real-analysis transcendental-numbers
real-analysis transcendental-numbers
asked Mar 9 '16 at 14:59
Farewell
3,31011024
asked Mar 9 '16 at 14:59
Farewell
3,31011024
asked Mar 9 '16 at 14:59
Farewell
3,31011024
asked Mar 9 '16 at 14:59
Farewell
3,31011024
asked Mar 9 '16 at 14:59
Farewell
3,31011024
3,31011024
$a + pi^{-n},, n geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:01
@DanielFischer Could that argument be modified to show that there is an uncountable number of them?
– Farewell
Mar 9 '16 at 15:03
The second argument shows that there are $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:04
@DanielFischer That thing with $pi^{-n}$ is easy enough way, you can post it as an answer.
– Farewell
Mar 9 '16 at 15:05
add a comment |
$a + pi^{-n},, n geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:01
@DanielFischer Could that argument be modified to show that there is an uncountable number of them?
– Farewell
Mar 9 '16 at 15:03
The second argument shows that there are $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:04
@DanielFischer That thing with $pi^{-n}$ is easy enough way, you can post it as an answer.
– Farewell
Mar 9 '16 at 15:05
$a + pi^{-n},, n geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:01
$a + pi^{-n},, n geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:01
@DanielFischer Could that argument be modified to show that there is an uncountable number of them?
– Farewell
Mar 9 '16 at 15:03
@DanielFischer Could that argument be modified to show that there is an uncountable number of them?
– Farewell
Mar 9 '16 at 15:03
The second argument shows that there are $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:04
The second argument shows that there are $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:04
@DanielFischer That thing with $pi^{-n}$ is easy enough way, you can post it as an answer.
– Farewell
Mar 9 '16 at 15:05
@DanielFischer That thing with $pi^{-n}$ is easy enough way, you can post it as an answer.
– Farewell
Mar 9 '16 at 15:05
add a comment |
2 Answers
2
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up vote
2
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accepted
We can explicitly give infinitely many transcendental numbers between any two algebraic numbers. Let the algebraic numbers be $a,b$ and suppose $a < b$. Choose $n_0 in mathbb{N}$ so large that $pi^{-n_0} < b-a$, then ${a + pi^{-n} : n geqslant n_0}$ is an infinite family of transcendental numbers between $a$ and $b$.
Since there are only countably many algebraic numbers, and the interval $(a,b)$ has cardinality $2^{aleph_0}$, there are in fact $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
add a comment |
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0
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I don't think this needs a very deep explanation. If you agree there are infinitely many transcendental numbers in any finite interval (say between 0 and 1), then you can easily (linearly) scale this between any two other numbers, be they algebraic or not, resulting in an infinite number of remapped transcendentals in the new interval. Without getting too rigorous, it should be pretty obvious that the remapped numbers are still transcendentals because they are of the form (algebraic + (algebraic) scalar * algebraic * transcendental = transcendental). You can't get the transcendental out of the transcendental by multiplying by an algebraic.
answered 2 days ago
ThePopMachine
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We can explicitly give infinitely many transcendental numbers between any two algebraic numbers. Let the algebraic numbers be $a,b$ and suppose $a < b$. Choose $n_0 in mathbb{N}$ so large that $pi^{-n_0} < b-a$, then ${a + pi^{-n} : n geqslant n_0}$ is an infinite family of transcendental numbers between $a$ and $b$.
Since there are only countably many algebraic numbers, and the interval $(a,b)$ has cardinality $2^{aleph_0}$, there are in fact $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
add a comment |
up vote
2
down vote
accepted
We can explicitly give infinitely many transcendental numbers between any two algebraic numbers. Let the algebraic numbers be $a,b$ and suppose $a < b$. Choose $n_0 in mathbb{N}$ so large that $pi^{-n_0} < b-a$, then ${a + pi^{-n} : n geqslant n_0}$ is an infinite family of transcendental numbers between $a$ and $b$.
Since there are only countably many algebraic numbers, and the interval $(a,b)$ has cardinality $2^{aleph_0}$, there are in fact $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We can explicitly give infinitely many transcendental numbers between any two algebraic numbers. Let the algebraic numbers be $a,b$ and suppose $a < b$. Choose $n_0 in mathbb{N}$ so large that $pi^{-n_0} < b-a$, then ${a + pi^{-n} : n geqslant n_0}$ is an infinite family of transcendental numbers between $a$ and $b$.
Since there are only countably many algebraic numbers, and the interval $(a,b)$ has cardinality $2^{aleph_0}$, there are in fact $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
We can explicitly give infinitely many transcendental numbers between any two algebraic numbers. Let the algebraic numbers be $a,b$ and suppose $a < b$. Choose $n_0 in mathbb{N}$ so large that $pi^{-n_0} < b-a$, then ${a + pi^{-n} : n geqslant n_0}$ is an infinite family of transcendental numbers between $a$ and $b$.
Since there are only countably many algebraic numbers, and the interval $(a,b)$ has cardinality $2^{aleph_0}$, there are in fact $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
answered Mar 9 '16 at 15:09
Daniel Fischer♦
172k16158281
172k16158281
add a comment |
add a comment |
up vote
0
down vote
I don't think this needs a very deep explanation. If you agree there are infinitely many transcendental numbers in any finite interval (say between 0 and 1), then you can easily (linearly) scale this between any two other numbers, be they algebraic or not, resulting in an infinite number of remapped transcendentals in the new interval. Without getting too rigorous, it should be pretty obvious that the remapped numbers are still transcendentals because they are of the form (algebraic + (algebraic) scalar * algebraic * transcendental = transcendental). You can't get the transcendental out of the transcendental by multiplying by an algebraic.
answered 2 days ago
ThePopMachine
1314
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
I don't think this needs a very deep explanation. If you agree there are infinitely many transcendental numbers in any finite interval (say between 0 and 1), then you can easily (linearly) scale this between any two other numbers, be they algebraic or not, resulting in an infinite number of remapped transcendentals in the new interval. Without getting too rigorous, it should be pretty obvious that the remapped numbers are still transcendentals because they are of the form (algebraic + (algebraic) scalar * algebraic * transcendental = transcendental). You can't get the transcendental out of the transcendental by multiplying by an algebraic.
answered 2 days ago
ThePopMachine
1314
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
I don't think this needs a very deep explanation. If you agree there are infinitely many transcendental numbers in any finite interval (say between 0 and 1), then you can easily (linearly) scale this between any two other numbers, be they algebraic or not, resulting in an infinite number of remapped transcendentals in the new interval. Without getting too rigorous, it should be pretty obvious that the remapped numbers are still transcendentals because they are of the form (algebraic + (algebraic) scalar * algebraic * transcendental = transcendental). You can't get the transcendental out of the transcendental by multiplying by an algebraic.
answered 2 days ago
ThePopMachine
1314
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't think this needs a very deep explanation. If you agree there are infinitely many transcendental numbers in any finite interval (say between 0 and 1), then you can easily (linearly) scale this between any two other numbers, be they algebraic or not, resulting in an infinite number of remapped transcendentals in the new interval. Without getting too rigorous, it should be pretty obvious that the remapped numbers are still transcendentals because they are of the form (algebraic + (algebraic) scalar * algebraic * transcendental = transcendental). You can't get the transcendental out of the transcendental by multiplying by an algebraic.
answered 2 days ago
ThePopMachine
1314
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
ThePopMachine
1314
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
ThePopMachine
1314
answered 2 days ago
ThePopMachine
1314
1314
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$a + pi^{-n},, n geqslant n_0$. Alternatively, $(a,b)$ is uncountable, but there are only countably many algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:01
@DanielFischer Could that argument be modified to show that there is an uncountable number of them?
– Farewell
Mar 9 '16 at 15:03
The second argument shows that there are $2^{aleph_0}$ transcendental numbers between any two algebraic numbers.
– Daniel Fischer♦
Mar 9 '16 at 15:04
@DanielFischer That thing with $pi^{-n}$ is easy enough way, you can post it as an answer.
– Farewell
Mar 9 '16 at 15:05