Mixing
Proving that $m+nsqrt2$ is dense in R
Clash Royale CLAN TAG#URR8PPP
up vote
41
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I am having trouble proving the statement:
Let $S = m + nsqrt 2 : m, n inmathbb Z$. Prove for every $epsilon > 0$, The intersection of $S$ and $(0, epsilon)$ is nonempty.
real-analysis irrational-numbers diophantine-approximation
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
asked Oct 17 '11 at 5:40
user11135
27038
add a comment |Â
up vote
41
down vote
favorite
I am having trouble proving the statement:
Let $S = m + nsqrt 2 : m, n inmathbb Z$. Prove for every $epsilon > 0$, The intersection of $S$ and $(0, epsilon)$ is nonempty.
real-analysis irrational-numbers diophantine-approximation
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
asked Oct 17 '11 at 5:40
user11135
27038
4
Consider rational numbers really close to $sqrt2$.
– Ted
Oct 17 '11 at 5:45
5
$Scap (0,epsilon)neemptysetquadforallepsilon>0$ only says $S$ has a limit point at $0$, not at every $xinmathbbR$...
– anon
Oct 17 '11 at 5:55
1
@anon if a subgroup of $mathbb R$ contains arbitrarily small positive numbers, it is dense in $mathbb R$.
– fredgoodman
Apr 6 at 16:02
add a comment |Â
up vote
41
down vote
favorite
up vote
41
down vote
favorite
I am having trouble proving the statement:
Let $S = m + nsqrt 2 : m, n inmathbb Z$. Prove for every $epsilon > 0$, The intersection of $S$ and $(0, epsilon)$ is nonempty.
real-analysis irrational-numbers diophantine-approximation
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
asked Oct 17 '11 at 5:40
user11135
27038
I am having trouble proving the statement:
Let $S = m + nsqrt 2 : m, n inmathbb Z$. Prove for every $epsilon > 0$, The intersection of $S$ and $(0, epsilon)$ is nonempty.
real-analysis irrational-numbers diophantine-approximation
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
asked Oct 17 '11 at 5:40
user11135
27038
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
edited Jan 30 '12 at 3:58
Srivatsan
20.6k368121
20.6k368121
asked Oct 17 '11 at 5:40
user11135
27038
asked Oct 17 '11 at 5:40
user11135
27038
asked Oct 17 '11 at 5:40
user11135
27038
27038
4
Consider rational numbers really close to $sqrt2$.
– Ted
Oct 17 '11 at 5:45
5
$Scap (0,epsilon)neemptysetquadforallepsilon>0$ only says $S$ has a limit point at $0$, not at every $xinmathbbR$...
– anon
Oct 17 '11 at 5:55
1
@anon if a subgroup of $mathbb R$ contains arbitrarily small positive numbers, it is dense in $mathbb R$.
– fredgoodman
Apr 6 at 16:02
add a comment |Â
4
Consider rational numbers really close to $sqrt2$.
– Ted
Oct 17 '11 at 5:45
5
$Scap (0,epsilon)neemptysetquadforallepsilon>0$ only says $S$ has a limit point at $0$, not at every $xinmathbbR$...
– anon
Oct 17 '11 at 5:55
1
@anon if a subgroup of $mathbb R$ contains arbitrarily small positive numbers, it is dense in $mathbb R$.
– fredgoodman
Apr 6 at 16:02
4
4
Consider rational numbers really close to $sqrt2$.
– Ted
Oct 17 '11 at 5:45
Consider rational numbers really close to $sqrt2$.
– Ted
Oct 17 '11 at 5:45
5
5
$Scap (0,epsilon)neemptysetquadforallepsilon>0$ only says $S$ has a limit point at $0$, not at every $xinmathbbR$...
– anon
Oct 17 '11 at 5:55
$Scap (0,epsilon)neemptysetquadforallepsilon>0$ only says $S$ has a limit point at $0$, not at every $xinmathbbR$...
– anon
Oct 17 '11 at 5:55
1
1
@anon if a subgroup of $mathbb R$ contains arbitrarily small positive numbers, it is dense in $mathbb R$.
– fredgoodman
Apr 6 at 16:02
@anon if a subgroup of $mathbb R$ contains arbitrarily small positive numbers, it is dense in $mathbb R$.
– fredgoodman
Apr 6 at 16:02
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
45
down vote
accepted
Hint: $|sqrt2 -1|<1/2$, so as $ntoinfty$ we have that $(sqrt2-1)^nto ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
2
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
2
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
add a comment |Â
up vote
29
down vote
Suppose not, so that
there exists an $varepsilon>0$ such that $(0,varepsilon)cap S=emptyset$. $qquadqquadqquad(star)$
It follows that $alpha=inf Scap(0,+infty)$ is a positive number.
- The choice of $alpha$ and its positivity implies that
the one and only element of $S$ which is in $[0,alpha)$ is $0$.
I claim that $alphain S$. Indeed, suppose not and let $alpha=inf Scap(0,+infty)$. The hypothesis implies that $alpha>0$, and the choice of $alpha$ implies that there exists elements $s$, $tin Scap(0,+infty)$ such that $$alphaleq s<tleq(1+tfrac14)alpha.$$ Then $u=t-s$ is an element of $S$ (because $S$ is closed under addition) such that $0<uleqtfrac14alpha<alpha$. This is absurd so we must have $alphain S$, as I claimed.
Let $sin Scap(0,+infty)$ and let $n=lfloor s/alpharfloor$ be the largest integer which is less than $s/alpha$. Then $nalphaleq s<(n+1)alpha$, so that $0leq s-nalpha <alpha$. This tells us that $s-nalpha$, which is an element of $S$, is in $[0,alpha)$. The choice of $alpha$ implies that we must then have $s-nalpha=0$, that is, $s=nalpha$. We conclude that every positive element of $S$ is an integer multiple of $alpha$.
In particular, since $1in S$ and $sqrt2in S$, there exists integers $n$ and $m$ such that $1=nalpha$ and $sqrt2=malpha$. But then $sqrt2=fracsqrt21=fracmalphanalpha=frac mninmathbb Q.$ This is absurd, and we can thus conclude that $(star)$ is an untenable hypothesis.
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
2
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
1
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
1
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
1
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
add a comment |Â
up vote
10
down vote
Taking a step into generalization, it is true that every additive subgroup $G$ of $mathbb R$ is either discrete or dense. This can be proved by considering $alpha = inf x in G : x>0 $. Then $G$ is discrete iff $alpha >0$, in which case $G=alpha mathbb Z$. In your case, Jyrki's suggestion implies that $alpha=0$ and so $S$ is dense.
answered Oct 17 '11 at 10:25
lhf
155k9160366
add a comment |Â
up vote
1
down vote
Let$epsilon>0$,Let $xinmathbbR$
$x-m-epsilon<x-m+epsilon$
between any two reals there are infinitely many rationals.
$fracx-m-epsilon<fracpq<fracx-m+epsilon$
therefore $$(x-epsilon,x+epsilon)$$ contains the member of the given set.
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
45
down vote
accepted
Hint: $|sqrt2 -1|<1/2$, so as $ntoinfty$ we have that $(sqrt2-1)^nto ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
2
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
2
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
add a comment |Â
up vote
45
down vote
accepted
Hint: $|sqrt2 -1|<1/2$, so as $ntoinfty$ we have that $(sqrt2-1)^nto ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
2
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
2
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
add a comment |Â
up vote
45
down vote
accepted
up vote
45
down vote
accepted
Hint: $|sqrt2 -1|<1/2$, so as $ntoinfty$ we have that $(sqrt2-1)^nto ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
Hint: $|sqrt2 -1|<1/2$, so as $ntoinfty$ we have that $(sqrt2-1)^nto ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
answered Oct 17 '11 at 6:03
Jyrki Lahtonen
104k12161355
104k12161355
2
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
2
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
add a comment |Â
2
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
2
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
2
2
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
If you are unfamiliar with the language of rings, you can just compute $(sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$.
– Jyrki Lahtonen
Oct 17 '11 at 7:54
2
2
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
If $sqrt2$ is replaced with a generic irrational real number $alpha$, then we can use the pigeon hole principle to prove the same result as Dirichlet's approximation theorem. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-)
– Jyrki Lahtonen
Oct 17 '11 at 13:24
add a comment |Â
up vote
29
down vote
Suppose not, so that
there exists an $varepsilon>0$ such that $(0,varepsilon)cap S=emptyset$. $qquadqquadqquad(star)$
It follows that $alpha=inf Scap(0,+infty)$ is a positive number.
- The choice of $alpha$ and its positivity implies that
the one and only element of $S$ which is in $[0,alpha)$ is $0$.
I claim that $alphain S$. Indeed, suppose not and let $alpha=inf Scap(0,+infty)$. The hypothesis implies that $alpha>0$, and the choice of $alpha$ implies that there exists elements $s$, $tin Scap(0,+infty)$ such that $$alphaleq s<tleq(1+tfrac14)alpha.$$ Then $u=t-s$ is an element of $S$ (because $S$ is closed under addition) such that $0<uleqtfrac14alpha<alpha$. This is absurd so we must have $alphain S$, as I claimed.
Let $sin Scap(0,+infty)$ and let $n=lfloor s/alpharfloor$ be the largest integer which is less than $s/alpha$. Then $nalphaleq s<(n+1)alpha$, so that $0leq s-nalpha <alpha$. This tells us that $s-nalpha$, which is an element of $S$, is in $[0,alpha)$. The choice of $alpha$ implies that we must then have $s-nalpha=0$, that is, $s=nalpha$. We conclude that every positive element of $S$ is an integer multiple of $alpha$.
In particular, since $1in S$ and $sqrt2in S$, there exists integers $n$ and $m$ such that $1=nalpha$ and $sqrt2=malpha$. But then $sqrt2=fracsqrt21=fracmalphanalpha=frac mninmathbb Q.$ This is absurd, and we can thus conclude that $(star)$ is an untenable hypothesis.
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
2
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
1
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
1
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
1
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
add a comment |Â
up vote
29
down vote
Suppose not, so that
there exists an $varepsilon>0$ such that $(0,varepsilon)cap S=emptyset$. $qquadqquadqquad(star)$
It follows that $alpha=inf Scap(0,+infty)$ is a positive number.
- The choice of $alpha$ and its positivity implies that
the one and only element of $S$ which is in $[0,alpha)$ is $0$.
I claim that $alphain S$. Indeed, suppose not and let $alpha=inf Scap(0,+infty)$. The hypothesis implies that $alpha>0$, and the choice of $alpha$ implies that there exists elements $s$, $tin Scap(0,+infty)$ such that $$alphaleq s<tleq(1+tfrac14)alpha.$$ Then $u=t-s$ is an element of $S$ (because $S$ is closed under addition) such that $0<uleqtfrac14alpha<alpha$. This is absurd so we must have $alphain S$, as I claimed.
Let $sin Scap(0,+infty)$ and let $n=lfloor s/alpharfloor$ be the largest integer which is less than $s/alpha$. Then $nalphaleq s<(n+1)alpha$, so that $0leq s-nalpha <alpha$. This tells us that $s-nalpha$, which is an element of $S$, is in $[0,alpha)$. The choice of $alpha$ implies that we must then have $s-nalpha=0$, that is, $s=nalpha$. We conclude that every positive element of $S$ is an integer multiple of $alpha$.
In particular, since $1in S$ and $sqrt2in S$, there exists integers $n$ and $m$ such that $1=nalpha$ and $sqrt2=malpha$. But then $sqrt2=fracsqrt21=fracmalphanalpha=frac mninmathbb Q.$ This is absurd, and we can thus conclude that $(star)$ is an untenable hypothesis.
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
2
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
1
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
1
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
1
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
add a comment |Â
up vote
29
down vote
up vote
29
down vote
Suppose not, so that
there exists an $varepsilon>0$ such that $(0,varepsilon)cap S=emptyset$. $qquadqquadqquad(star)$
It follows that $alpha=inf Scap(0,+infty)$ is a positive number.
- The choice of $alpha$ and its positivity implies that
the one and only element of $S$ which is in $[0,alpha)$ is $0$.
I claim that $alphain S$. Indeed, suppose not and let $alpha=inf Scap(0,+infty)$. The hypothesis implies that $alpha>0$, and the choice of $alpha$ implies that there exists elements $s$, $tin Scap(0,+infty)$ such that $$alphaleq s<tleq(1+tfrac14)alpha.$$ Then $u=t-s$ is an element of $S$ (because $S$ is closed under addition) such that $0<uleqtfrac14alpha<alpha$. This is absurd so we must have $alphain S$, as I claimed.
Let $sin Scap(0,+infty)$ and let $n=lfloor s/alpharfloor$ be the largest integer which is less than $s/alpha$. Then $nalphaleq s<(n+1)alpha$, so that $0leq s-nalpha <alpha$. This tells us that $s-nalpha$, which is an element of $S$, is in $[0,alpha)$. The choice of $alpha$ implies that we must then have $s-nalpha=0$, that is, $s=nalpha$. We conclude that every positive element of $S$ is an integer multiple of $alpha$.
In particular, since $1in S$ and $sqrt2in S$, there exists integers $n$ and $m$ such that $1=nalpha$ and $sqrt2=malpha$. But then $sqrt2=fracsqrt21=fracmalphanalpha=frac mninmathbb Q.$ This is absurd, and we can thus conclude that $(star)$ is an untenable hypothesis.
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
Suppose not, so that
there exists an $varepsilon>0$ such that $(0,varepsilon)cap S=emptyset$. $qquadqquadqquad(star)$
It follows that $alpha=inf Scap(0,+infty)$ is a positive number.
- The choice of $alpha$ and its positivity implies that
the one and only element of $S$ which is in $[0,alpha)$ is $0$.
I claim that $alphain S$. Indeed, suppose not and let $alpha=inf Scap(0,+infty)$. The hypothesis implies that $alpha>0$, and the choice of $alpha$ implies that there exists elements $s$, $tin Scap(0,+infty)$ such that $$alphaleq s<tleq(1+tfrac14)alpha.$$ Then $u=t-s$ is an element of $S$ (because $S$ is closed under addition) such that $0<uleqtfrac14alpha<alpha$. This is absurd so we must have $alphain S$, as I claimed.
Let $sin Scap(0,+infty)$ and let $n=lfloor s/alpharfloor$ be the largest integer which is less than $s/alpha$. Then $nalphaleq s<(n+1)alpha$, so that $0leq s-nalpha <alpha$. This tells us that $s-nalpha$, which is an element of $S$, is in $[0,alpha)$. The choice of $alpha$ implies that we must then have $s-nalpha=0$, that is, $s=nalpha$. We conclude that every positive element of $S$ is an integer multiple of $alpha$.
In particular, since $1in S$ and $sqrt2in S$, there exists integers $n$ and $m$ such that $1=nalpha$ and $sqrt2=malpha$. But then $sqrt2=fracsqrt21=fracmalphanalpha=frac mninmathbb Q.$ This is absurd, and we can thus conclude that $(star)$ is an untenable hypothesis.
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
edited Mar 31 '16 at 17:36
user940
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
answered Oct 17 '11 at 6:28
Mariano Suárez-Ãlvarez♦
109k7153277
109k7153277
2
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
1
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
1
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
1
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
add a comment |Â
2
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
1
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
1
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
1
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
2
2
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
This answer should get more votes. It is more technical than Jyrki's, but it works for every irrational number and not only $sqrt2$.
– KotelKanim
Oct 17 '11 at 8:39
1
1
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $alpha$ and (1+1/4)$alpha$? I know one can always find one such number by the definition of inf., but what about the other?
– Yang
Jan 30 '12 at 3:51
1
1
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
@Yang, pick one, call it $t$, such that $alpha<tleq(1+tfrac14)alpha$ and then pick another, call it $s$, such that $alphaleq s<t$.
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 3:59
1
1
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
(By the way, just call me Mariano, please! :) )
– Mariano Suárez-Ãlvarez♦
Jan 30 '12 at 4:00
add a comment |Â
up vote
10
down vote
Taking a step into generalization, it is true that every additive subgroup $G$ of $mathbb R$ is either discrete or dense. This can be proved by considering $alpha = inf x in G : x>0 $. Then $G$ is discrete iff $alpha >0$, in which case $G=alpha mathbb Z$. In your case, Jyrki's suggestion implies that $alpha=0$ and so $S$ is dense.
answered Oct 17 '11 at 10:25
lhf
155k9160366
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10
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Taking a step into generalization, it is true that every additive subgroup $G$ of $mathbb R$ is either discrete or dense. This can be proved by considering $alpha = inf x in G : x>0 $. Then $G$ is discrete iff $alpha >0$, in which case $G=alpha mathbb Z$. In your case, Jyrki's suggestion implies that $alpha=0$ and so $S$ is dense.
answered Oct 17 '11 at 10:25
lhf
155k9160366
add a comment |Â
up vote
10
down vote
up vote
10
down vote
Taking a step into generalization, it is true that every additive subgroup $G$ of $mathbb R$ is either discrete or dense. This can be proved by considering $alpha = inf x in G : x>0 $. Then $G$ is discrete iff $alpha >0$, in which case $G=alpha mathbb Z$. In your case, Jyrki's suggestion implies that $alpha=0$ and so $S$ is dense.
answered Oct 17 '11 at 10:25
lhf
155k9160366
Taking a step into generalization, it is true that every additive subgroup $G$ of $mathbb R$ is either discrete or dense. This can be proved by considering $alpha = inf x in G : x>0 $. Then $G$ is discrete iff $alpha >0$, in which case $G=alpha mathbb Z$. In your case, Jyrki's suggestion implies that $alpha=0$ and so $S$ is dense.
answered Oct 17 '11 at 10:25
lhf
155k9160366
edited Oct 17 '11 at 10:50
answered Oct 17 '11 at 10:25
lhf
155k9160366
answered Oct 17 '11 at 10:25
lhf
155k9160366
answered Oct 17 '11 at 10:25
lhf
155k9160366
155k9160366
add a comment |Â
add a comment |Â
up vote
1
down vote
Let$epsilon>0$,Let $xinmathbbR$
$x-m-epsilon<x-m+epsilon$
between any two reals there are infinitely many rationals.
$fracx-m-epsilon<fracpq<fracx-m+epsilon$
therefore $$(x-epsilon,x+epsilon)$$ contains the member of the given set.
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
add a comment |Â
up vote
1
down vote
Let$epsilon>0$,Let $xinmathbbR$
$x-m-epsilon<x-m+epsilon$
between any two reals there are infinitely many rationals.
$fracx-m-epsilon<fracpq<fracx-m+epsilon$
therefore $$(x-epsilon,x+epsilon)$$ contains the member of the given set.
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let$epsilon>0$,Let $xinmathbbR$
$x-m-epsilon<x-m+epsilon$
between any two reals there are infinitely many rationals.
$fracx-m-epsilon<fracpq<fracx-m+epsilon$
therefore $$(x-epsilon,x+epsilon)$$ contains the member of the given set.
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
Let$epsilon>0$,Let $xinmathbbR$
$x-m-epsilon<x-m+epsilon$
between any two reals there are infinitely many rationals.
$fracx-m-epsilon<fracpq<fracx-m+epsilon$
therefore $$(x-epsilon,x+epsilon)$$ contains the member of the given set.
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
answered Mar 31 '16 at 18:13
Rayees Ahmad
834312
834312
add a comment |Â
add a comment |Â
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4
Consider rational numbers really close to $sqrt2$.
– Ted
Oct 17 '11 at 5:45
5
$Scap (0,epsilon)neemptysetquadforallepsilon>0$ only says $S$ has a limit point at $0$, not at every $xinmathbbR$...
– anon
Oct 17 '11 at 5:55
1
@anon if a subgroup of $mathbb R$ contains arbitrarily small positive numbers, it is dense in $mathbb R$.
– fredgoodman
Apr 6 at 16:02