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Prove there is a square of a rational number between any two positive real numbers
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I'm doing this problem and the solution is below. I don't understand why this solution proved $m^2 le x$ firstly? Any help, thanks.
calculus real-numbers irrational-numbers
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
asked Aug 6 at 12:05
Cathy
1207
add a comment |Â
up vote
2
down vote
favorite
I'm doing this problem and the solution is below. I don't understand why this solution proved $m^2 le x$ firstly? Any help, thanks.
calculus real-numbers irrational-numbers
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
asked Aug 6 at 12:05
Cathy
1207
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm doing this problem and the solution is below. I don't understand why this solution proved $m^2 le x$ firstly? Any help, thanks.
calculus real-numbers irrational-numbers
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
asked Aug 6 at 12:05
Cathy
1207
I'm doing this problem and the solution is below. I don't understand why this solution proved $m^2 le x$ firstly? Any help, thanks.
calculus real-numbers irrational-numbers
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
asked Aug 6 at 12:05
Cathy
1207
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
edited Aug 6 at 12:17
José Carlos Santos
115k1698177
115k1698177
asked Aug 6 at 12:05
Cathy
1207
asked Aug 6 at 12:05
Cathy
1207
asked Aug 6 at 12:05
Cathy
1207
1207
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
The idea is to prove that there is a real number $mgeqslant0$ such that $m^2leqslant x$ which is such that for any rational number $r>m$ you have $r^2>x$. So, if you add a small number to $m$ (the $delta$ of the proof), then $(m+delta)^2>x$. So, what you have to do next is to prove that you can find such a $delta$ so that
- $m+deltainmathbb Q$;
- $x<(m+delta)^2<y$.
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
add a comment |Â
up vote
0
down vote
This is too long for a comment, so I'm posting it as an answer.
Instead of addressing the quoted proof, I give a quite short proof of the more general result that for any positive integer $k$ there exists a rational number $r$ such that $x < r^k < y$, without assuming any knowledge of the existence of $k^textth$ roots. (Taking $k = 2$ doesn't make the proof any shorter.)
For all positive integers $k, p$,
$$
frac(p + 1)^kp^k = left(1 + frac1pright)^k = sum_j=0^kbinomkjfrac1p^j leqslant 1 + sum_j=1^kbinomkjfrac1p = 1 + frac2^k - 1p.
$$
Given $y > x > 0$, define
$$
h = frac(2^k - 1)xy - x.
$$
Then
$$
tag1labelineq:p
p > h implies frac(p + 1)^kp^k < fracyx.
$$
There exists a positive integer $q$ such that
$$
tag2labelineq:q
q^kx geqslant (1 + h)^k.
$$
For instance, it is enough to take any $q$ such that
$$
q geqslant (1 + h)left(1 + fracmax1 - x, 0kxright).
$$
The inequality just given is equivalent to the conjunction of $q geqslant 1 + h$ with
$$
1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
and therefore it implies
$$
fracq^k(1 + h)^k = left(fracq1 + hright)^k = left(1 + fracq1 + h - 1right)^k geqslant 1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
as required.
Take any value of $q$ satisfying eqrefineq:q.
The inequality
$$
tag3labelineq
p^k leqslant q^kx
$$
holds for some values of $p$ (for example, $p = 1$), but not for all (for example, not for $p = qleft(1 + lceilxrceilright)$).
Therefore, there exists a unique largest value of $p$ for which eqrefineq holds.
For this value of $p$, we have $(p + 1)^k > q^kx geqslant (1 + h)^k$, and therefore $p > h$.
Invoking eqrefineq:p and eqrefineq, we get
$$
q^kx < (p + 1)^k < p^kfracyx leqslant q^ky,
$$
and therefore finally
$$
x < left(fracp + 1qright)^k < y.
$$
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The idea is to prove that there is a real number $mgeqslant0$ such that $m^2leqslant x$ which is such that for any rational number $r>m$ you have $r^2>x$. So, if you add a small number to $m$ (the $delta$ of the proof), then $(m+delta)^2>x$. So, what you have to do next is to prove that you can find such a $delta$ so that
- $m+deltainmathbb Q$;
- $x<(m+delta)^2<y$.
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
add a comment |Â
up vote
0
down vote
The idea is to prove that there is a real number $mgeqslant0$ such that $m^2leqslant x$ which is such that for any rational number $r>m$ you have $r^2>x$. So, if you add a small number to $m$ (the $delta$ of the proof), then $(m+delta)^2>x$. So, what you have to do next is to prove that you can find such a $delta$ so that
- $m+deltainmathbb Q$;
- $x<(m+delta)^2<y$.
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The idea is to prove that there is a real number $mgeqslant0$ such that $m^2leqslant x$ which is such that for any rational number $r>m$ you have $r^2>x$. So, if you add a small number to $m$ (the $delta$ of the proof), then $(m+delta)^2>x$. So, what you have to do next is to prove that you can find such a $delta$ so that
- $m+deltainmathbb Q$;
- $x<(m+delta)^2<y$.
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
The idea is to prove that there is a real number $mgeqslant0$ such that $m^2leqslant x$ which is such that for any rational number $r>m$ you have $r^2>x$. So, if you add a small number to $m$ (the $delta$ of the proof), then $(m+delta)^2>x$. So, what you have to do next is to prove that you can find such a $delta$ so that
- $m+deltainmathbb Q$;
- $x<(m+delta)^2<y$.
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
answered Aug 6 at 12:16
José Carlos Santos
115k1698177
115k1698177
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
add a comment |Â
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
Thanks very much. I'm a little confused about how this proof show that $m + delta in mathbbQ$?
– Cathy
Aug 6 at 12:50
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@Cathy $m + delta $ is not assumed to be in $mathbbQ $. Rather $m$ is a real number which satisfies $m leq r quad forall r in A$ and $m^2 leq x$. Then $delta$ is chosen such that $m lt m + delta$ and $(m+ delta)^2 leq y$. The last part of the proof is then just the assumption that there must exist a rational $r$ such that $m lt r lt m + delta$ (since otherwise $m+ delta$ would be a lower bound of $A$ greater than $m$, which is a contradiction).
– Moed Pol Bollo
Aug 6 at 13:00
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@MoedPolBollo Thanks! I just find I still confused about why we need $m^2 le x$?
– Cathy
Aug 6 at 13:09
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy I made a mistake in my answer. It is not necessarily true that $m+deltainmathbb Q$. But it follows from the definition of $m$ that there's some rational $rin(x+x+delta)$ such that $x<r^2<y$.
– José Carlos Santos
Aug 6 at 13:11
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
@Cathy The defining property of elements of $A$, is that their square is greater than $x$. $m$ is the infimum of $A$, which means it is the largest real number less than any element of $A$. So, think of any interval: $(m, m + a)$, then the fact that $m= infA$ means that there is always a rational $r in A$ in this interval, and we can shrink this interval down as small as we like. This is essentially what the proof shows. So I guess you can think of the fact that $m^2 leq x$ as acting like an anchor, which pulls the rationals down to where we want them.
– Moed Pol Bollo
Aug 6 at 13:49
add a comment |Â
up vote
0
down vote
This is too long for a comment, so I'm posting it as an answer.
Instead of addressing the quoted proof, I give a quite short proof of the more general result that for any positive integer $k$ there exists a rational number $r$ such that $x < r^k < y$, without assuming any knowledge of the existence of $k^textth$ roots. (Taking $k = 2$ doesn't make the proof any shorter.)
For all positive integers $k, p$,
$$
frac(p + 1)^kp^k = left(1 + frac1pright)^k = sum_j=0^kbinomkjfrac1p^j leqslant 1 + sum_j=1^kbinomkjfrac1p = 1 + frac2^k - 1p.
$$
Given $y > x > 0$, define
$$
h = frac(2^k - 1)xy - x.
$$
Then
$$
tag1labelineq:p
p > h implies frac(p + 1)^kp^k < fracyx.
$$
There exists a positive integer $q$ such that
$$
tag2labelineq:q
q^kx geqslant (1 + h)^k.
$$
For instance, it is enough to take any $q$ such that
$$
q geqslant (1 + h)left(1 + fracmax1 - x, 0kxright).
$$
The inequality just given is equivalent to the conjunction of $q geqslant 1 + h$ with
$$
1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
and therefore it implies
$$
fracq^k(1 + h)^k = left(fracq1 + hright)^k = left(1 + fracq1 + h - 1right)^k geqslant 1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
as required.
Take any value of $q$ satisfying eqrefineq:q.
The inequality
$$
tag3labelineq
p^k leqslant q^kx
$$
holds for some values of $p$ (for example, $p = 1$), but not for all (for example, not for $p = qleft(1 + lceilxrceilright)$).
Therefore, there exists a unique largest value of $p$ for which eqrefineq holds.
For this value of $p$, we have $(p + 1)^k > q^kx geqslant (1 + h)^k$, and therefore $p > h$.
Invoking eqrefineq:p and eqrefineq, we get
$$
q^kx < (p + 1)^k < p^kfracyx leqslant q^ky,
$$
and therefore finally
$$
x < left(fracp + 1qright)^k < y.
$$
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
add a comment |Â
up vote
0
down vote
This is too long for a comment, so I'm posting it as an answer.
Instead of addressing the quoted proof, I give a quite short proof of the more general result that for any positive integer $k$ there exists a rational number $r$ such that $x < r^k < y$, without assuming any knowledge of the existence of $k^textth$ roots. (Taking $k = 2$ doesn't make the proof any shorter.)
For all positive integers $k, p$,
$$
frac(p + 1)^kp^k = left(1 + frac1pright)^k = sum_j=0^kbinomkjfrac1p^j leqslant 1 + sum_j=1^kbinomkjfrac1p = 1 + frac2^k - 1p.
$$
Given $y > x > 0$, define
$$
h = frac(2^k - 1)xy - x.
$$
Then
$$
tag1labelineq:p
p > h implies frac(p + 1)^kp^k < fracyx.
$$
There exists a positive integer $q$ such that
$$
tag2labelineq:q
q^kx geqslant (1 + h)^k.
$$
For instance, it is enough to take any $q$ such that
$$
q geqslant (1 + h)left(1 + fracmax1 - x, 0kxright).
$$
The inequality just given is equivalent to the conjunction of $q geqslant 1 + h$ with
$$
1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
and therefore it implies
$$
fracq^k(1 + h)^k = left(fracq1 + hright)^k = left(1 + fracq1 + h - 1right)^k geqslant 1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
as required.
Take any value of $q$ satisfying eqrefineq:q.
The inequality
$$
tag3labelineq
p^k leqslant q^kx
$$
holds for some values of $p$ (for example, $p = 1$), but not for all (for example, not for $p = qleft(1 + lceilxrceilright)$).
Therefore, there exists a unique largest value of $p$ for which eqrefineq holds.
For this value of $p$, we have $(p + 1)^k > q^kx geqslant (1 + h)^k$, and therefore $p > h$.
Invoking eqrefineq:p and eqrefineq, we get
$$
q^kx < (p + 1)^k < p^kfracyx leqslant q^ky,
$$
and therefore finally
$$
x < left(fracp + 1qright)^k < y.
$$
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
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This is too long for a comment, so I'm posting it as an answer.
Instead of addressing the quoted proof, I give a quite short proof of the more general result that for any positive integer $k$ there exists a rational number $r$ such that $x < r^k < y$, without assuming any knowledge of the existence of $k^textth$ roots. (Taking $k = 2$ doesn't make the proof any shorter.)
For all positive integers $k, p$,
$$
frac(p + 1)^kp^k = left(1 + frac1pright)^k = sum_j=0^kbinomkjfrac1p^j leqslant 1 + sum_j=1^kbinomkjfrac1p = 1 + frac2^k - 1p.
$$
Given $y > x > 0$, define
$$
h = frac(2^k - 1)xy - x.
$$
Then
$$
tag1labelineq:p
p > h implies frac(p + 1)^kp^k < fracyx.
$$
There exists a positive integer $q$ such that
$$
tag2labelineq:q
q^kx geqslant (1 + h)^k.
$$
For instance, it is enough to take any $q$ such that
$$
q geqslant (1 + h)left(1 + fracmax1 - x, 0kxright).
$$
The inequality just given is equivalent to the conjunction of $q geqslant 1 + h$ with
$$
1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
and therefore it implies
$$
fracq^k(1 + h)^k = left(fracq1 + hright)^k = left(1 + fracq1 + h - 1right)^k geqslant 1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
as required.
Take any value of $q$ satisfying eqrefineq:q.
The inequality
$$
tag3labelineq
p^k leqslant q^kx
$$
holds for some values of $p$ (for example, $p = 1$), but not for all (for example, not for $p = qleft(1 + lceilxrceilright)$).
Therefore, there exists a unique largest value of $p$ for which eqrefineq holds.
For this value of $p$, we have $(p + 1)^k > q^kx geqslant (1 + h)^k$, and therefore $p > h$.
Invoking eqrefineq:p and eqrefineq, we get
$$
q^kx < (p + 1)^k < p^kfracyx leqslant q^ky,
$$
and therefore finally
$$
x < left(fracp + 1qright)^k < y.
$$
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
This is too long for a comment, so I'm posting it as an answer.
Instead of addressing the quoted proof, I give a quite short proof of the more general result that for any positive integer $k$ there exists a rational number $r$ such that $x < r^k < y$, without assuming any knowledge of the existence of $k^textth$ roots. (Taking $k = 2$ doesn't make the proof any shorter.)
For all positive integers $k, p$,
$$
frac(p + 1)^kp^k = left(1 + frac1pright)^k = sum_j=0^kbinomkjfrac1p^j leqslant 1 + sum_j=1^kbinomkjfrac1p = 1 + frac2^k - 1p.
$$
Given $y > x > 0$, define
$$
h = frac(2^k - 1)xy - x.
$$
Then
$$
tag1labelineq:p
p > h implies frac(p + 1)^kp^k < fracyx.
$$
There exists a positive integer $q$ such that
$$
tag2labelineq:q
q^kx geqslant (1 + h)^k.
$$
For instance, it is enough to take any $q$ such that
$$
q geqslant (1 + h)left(1 + fracmax1 - x, 0kxright).
$$
The inequality just given is equivalent to the conjunction of $q geqslant 1 + h$ with
$$
1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
and therefore it implies
$$
fracq^k(1 + h)^k = left(fracq1 + hright)^k = left(1 + fracq1 + h - 1right)^k geqslant 1 + kleft(fracq1 + h - 1right) geqslant frac1x,
$$
as required.
Take any value of $q$ satisfying eqrefineq:q.
The inequality
$$
tag3labelineq
p^k leqslant q^kx
$$
holds for some values of $p$ (for example, $p = 1$), but not for all (for example, not for $p = qleft(1 + lceilxrceilright)$).
Therefore, there exists a unique largest value of $p$ for which eqrefineq holds.
For this value of $p$, we have $(p + 1)^k > q^kx geqslant (1 + h)^k$, and therefore $p > h$.
Invoking eqrefineq:p and eqrefineq, we get
$$
q^kx < (p + 1)^k < p^kfracyx leqslant q^ky,
$$
and therefore finally
$$
x < left(fracp + 1qright)^k < y.
$$
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
answered Aug 6 at 18:39
Calum Gilhooley
2,782526
2,782526
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
add a comment |Â
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
Is there an even simpler proof by contradiction possible? There exists a rational $x<m<y$. Assume $mgt 1$, fix some $nin mathbbN$ and let $A=rin mathbbQ: r^n lt m$, $B=rin mathbbQ: r^n gt m$. Let $a=supA, b=infB$, we claim either $a=m$ or $b=m$. Assume the opposite; that is $a lt m lt b$. There exist rationals $q_1, q_2$ such that (1) $a lt q_1 lt m lt q_2 lt b$, and such that (2) $q_1^n geq m$ and $q_2^n leq m$. The contradiction between (1) and (2) shows either $a=m$ or $b=m$ (or both), which easily gives the case for $mgt 1$. Is there a flaw here?
– Moed Pol Bollo
Aug 13 at 0:20
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
@MoedPolBollo I don't think there is any flaw in your argument, and I think what this shows is that it is simpler to prove the existence of $n^textth$ roots than to try to prove the result in the question without using the existence of $n^textth$ roots (at least by the method in my comment or the method in the textbook).
– Calum Gilhooley
Aug 13 at 11:18
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
P.S. I forgot to mention that one intended advantage of my argument is that it includes the case $k = 1$: indeed, it was designed to generalise an argument for the density of the rationals in the reals in the simplest way I could think of.
– Calum Gilhooley
Aug 13 at 12:06
add a comment |Â
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