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Problem with reading a proof: uniqueness of integer solution to $0<(za-1)a^q-1<1$
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If $m,ninmathbb N, m<n, z=fracmn, qge 2$, the inequality $0<(za-1)a^q-1<1$ has solution $ainmathbb N$.
Then the solution to the inequality above is unique.
The question is simple: how to prove the statement above?
I am currently studying
Daniel D. Bonar & Michael Khoury Jr. Real Infinite Series. Classroom Resource Materials. The Mathematical Association of America, $2006$. isbn: $0883857456$.
The problem above is here: P.1, P.2.
The book did give a proof to the statement at beginning, as a lemma to that problem in the book.
The book said (I use $q$ instead of $e$ in the book) "$fracnb<mle m(b-a)<fracna^q-1$ implies a contradiction", but how? And, how to prove $m(b-a)<fracna^q-1$?
What I have thought:
If $a^q-1inmathbb Z$ again and $q-1>1$, then there may be a contradiction since there are just finite many integer between $a$ and $b$, (similar to proof by infinite descent). But well... I think I thought too much. It just doesn't work.
And I back to the problem itself and try, if $a<b$ are natural numbers satisfying the inequality,
beginalign*
0<(za-1)a^q-1&<(zb-1)b^q-1<1\
0<(ma-n)a^q-1&<(mb-n)b^q-1<n\
0<ma^q-na^q-1&<mb^q-nb^q-1<n\
n(b^q-1-a^q-1)&<m(b^q-a^q)\
z&>fracb^q-1-a^q-1b^q-a^q
endalign*
But it doesn't seem to be constructive.
I would like a proof to the statement at beginning. New proofs are good, in case there are, but I would like, if possible, someone telling me how the arguments in the book flow.
real-analysis inequality integers
asked Jul 19 at 9:37
Tony Ma
1,133225
add a comment |Â
up vote
1
down vote
favorite
If $m,ninmathbb N, m<n, z=fracmn, qge 2$, the inequality $0<(za-1)a^q-1<1$ has solution $ainmathbb N$.
Then the solution to the inequality above is unique.
The question is simple: how to prove the statement above?
I am currently studying
Daniel D. Bonar & Michael Khoury Jr. Real Infinite Series. Classroom Resource Materials. The Mathematical Association of America, $2006$. isbn: $0883857456$.
The problem above is here: P.1, P.2.
The book did give a proof to the statement at beginning, as a lemma to that problem in the book.
The book said (I use $q$ instead of $e$ in the book) "$fracnb<mle m(b-a)<fracna^q-1$ implies a contradiction", but how? And, how to prove $m(b-a)<fracna^q-1$?
What I have thought:
If $a^q-1inmathbb Z$ again and $q-1>1$, then there may be a contradiction since there are just finite many integer between $a$ and $b$, (similar to proof by infinite descent). But well... I think I thought too much. It just doesn't work.
And I back to the problem itself and try, if $a<b$ are natural numbers satisfying the inequality,
beginalign*
0<(za-1)a^q-1&<(zb-1)b^q-1<1\
0<(ma-n)a^q-1&<(mb-n)b^q-1<n\
0<ma^q-na^q-1&<mb^q-nb^q-1<n\
n(b^q-1-a^q-1)&<m(b^q-a^q)\
z&>fracb^q-1-a^q-1b^q-a^q
endalign*
But it doesn't seem to be constructive.
I would like a proof to the statement at beginning. New proofs are good, in case there are, but I would like, if possible, someone telling me how the arguments in the book flow.
real-analysis inequality integers
asked Jul 19 at 9:37
Tony Ma
1,133225
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $m,ninmathbb N, m<n, z=fracmn, qge 2$, the inequality $0<(za-1)a^q-1<1$ has solution $ainmathbb N$.
Then the solution to the inequality above is unique.
The question is simple: how to prove the statement above?
I am currently studying
Daniel D. Bonar & Michael Khoury Jr. Real Infinite Series. Classroom Resource Materials. The Mathematical Association of America, $2006$. isbn: $0883857456$.
The problem above is here: P.1, P.2.
The book did give a proof to the statement at beginning, as a lemma to that problem in the book.
The book said (I use $q$ instead of $e$ in the book) "$fracnb<mle m(b-a)<fracna^q-1$ implies a contradiction", but how? And, how to prove $m(b-a)<fracna^q-1$?
What I have thought:
If $a^q-1inmathbb Z$ again and $q-1>1$, then there may be a contradiction since there are just finite many integer between $a$ and $b$, (similar to proof by infinite descent). But well... I think I thought too much. It just doesn't work.
And I back to the problem itself and try, if $a<b$ are natural numbers satisfying the inequality,
beginalign*
0<(za-1)a^q-1&<(zb-1)b^q-1<1\
0<(ma-n)a^q-1&<(mb-n)b^q-1<n\
0<ma^q-na^q-1&<mb^q-nb^q-1<n\
n(b^q-1-a^q-1)&<m(b^q-a^q)\
z&>fracb^q-1-a^q-1b^q-a^q
endalign*
But it doesn't seem to be constructive.
I would like a proof to the statement at beginning. New proofs are good, in case there are, but I would like, if possible, someone telling me how the arguments in the book flow.
real-analysis inequality integers
asked Jul 19 at 9:37
Tony Ma
1,133225
If $m,ninmathbb N, m<n, z=fracmn, qge 2$, the inequality $0<(za-1)a^q-1<1$ has solution $ainmathbb N$.
Then the solution to the inequality above is unique.
The question is simple: how to prove the statement above?
I am currently studying
Daniel D. Bonar & Michael Khoury Jr. Real Infinite Series. Classroom Resource Materials. The Mathematical Association of America, $2006$. isbn: $0883857456$.
The problem above is here: P.1, P.2.
The book did give a proof to the statement at beginning, as a lemma to that problem in the book.
The book said (I use $q$ instead of $e$ in the book) "$fracnb<mle m(b-a)<fracna^q-1$ implies a contradiction", but how? And, how to prove $m(b-a)<fracna^q-1$?
What I have thought:
If $a^q-1inmathbb Z$ again and $q-1>1$, then there may be a contradiction since there are just finite many integer between $a$ and $b$, (similar to proof by infinite descent). But well... I think I thought too much. It just doesn't work.
And I back to the problem itself and try, if $a<b$ are natural numbers satisfying the inequality,
beginalign*
0<(za-1)a^q-1&<(zb-1)b^q-1<1\
0<(ma-n)a^q-1&<(mb-n)b^q-1<n\
0<ma^q-na^q-1&<mb^q-nb^q-1<n\
n(b^q-1-a^q-1)&<m(b^q-a^q)\
z&>fracb^q-1-a^q-1b^q-a^q
endalign*
But it doesn't seem to be constructive.
I would like a proof to the statement at beginning. New proofs are good, in case there are, but I would like, if possible, someone telling me how the arguments in the book flow.
real-analysis inequality integers
asked Jul 19 at 9:37
Tony Ma
1,133225
edited Jul 19 at 12:19
asked Jul 19 at 9:37
Tony Ma
1,133225
asked Jul 19 at 9:37
Tony Ma
1,133225
asked Jul 19 at 9:37
Tony Ma
1,133225
1,133225
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
1
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If the inequality
$$0 < (za-1)a^q-1$$
holds, then, since $a^q-1 > 0$, we must have $za = fracmna > 1$, or equivalently
$$m > fracna,.$$
If $a < b$ are two integers, then $b-a geqslant 1$, whence $m leqslant m(b-a)$. If $0 < a < b$ are integer solutions of
$$0 < (zx-1)x^q-1 < 1,, tag$ast$$$
then, since the middle expression in $(ast)$ is monotonic in $x$ for $x > z^-1$, we have
$$0 < (zb-1)b^q-1 - (za-1)a^q-1 < 1,.$$
Since $b^q-1 > a^q-1$ and $zb > za$, it follows that
$$0 < bigl((zb-1) - (za-1)bigr)a^q-1 = fracmn(b-a)a^q-1 leqslant (zb-1)b^q-1 - (za-1)a^q-1 < 1,,$$
and from that we obtain
$$m(b-a) < fracna^q-1,.$$
But from
$$m < fracna^q-1$$
it follows that
$$za = fracmna < frac1a^q-2 leqslant 1,,$$
which shows that $a$ can't be a solution of $(ast)$ by the remark at the top.
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If the inequality
$$0 < (za-1)a^q-1$$
holds, then, since $a^q-1 > 0$, we must have $za = fracmna > 1$, or equivalently
$$m > fracna,.$$
If $a < b$ are two integers, then $b-a geqslant 1$, whence $m leqslant m(b-a)$. If $0 < a < b$ are integer solutions of
$$0 < (zx-1)x^q-1 < 1,, tag$ast$$$
then, since the middle expression in $(ast)$ is monotonic in $x$ for $x > z^-1$, we have
$$0 < (zb-1)b^q-1 - (za-1)a^q-1 < 1,.$$
Since $b^q-1 > a^q-1$ and $zb > za$, it follows that
$$0 < bigl((zb-1) - (za-1)bigr)a^q-1 = fracmn(b-a)a^q-1 leqslant (zb-1)b^q-1 - (za-1)a^q-1 < 1,,$$
and from that we obtain
$$m(b-a) < fracna^q-1,.$$
But from
$$m < fracna^q-1$$
it follows that
$$za = fracmna < frac1a^q-2 leqslant 1,,$$
which shows that $a$ can't be a solution of $(ast)$ by the remark at the top.
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
1
down vote
accepted
If the inequality
$$0 < (za-1)a^q-1$$
holds, then, since $a^q-1 > 0$, we must have $za = fracmna > 1$, or equivalently
$$m > fracna,.$$
If $a < b$ are two integers, then $b-a geqslant 1$, whence $m leqslant m(b-a)$. If $0 < a < b$ are integer solutions of
$$0 < (zx-1)x^q-1 < 1,, tag$ast$$$
then, since the middle expression in $(ast)$ is monotonic in $x$ for $x > z^-1$, we have
$$0 < (zb-1)b^q-1 - (za-1)a^q-1 < 1,.$$
Since $b^q-1 > a^q-1$ and $zb > za$, it follows that
$$0 < bigl((zb-1) - (za-1)bigr)a^q-1 = fracmn(b-a)a^q-1 leqslant (zb-1)b^q-1 - (za-1)a^q-1 < 1,,$$
and from that we obtain
$$m(b-a) < fracna^q-1,.$$
But from
$$m < fracna^q-1$$
it follows that
$$za = fracmna < frac1a^q-2 leqslant 1,,$$
which shows that $a$ can't be a solution of $(ast)$ by the remark at the top.
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If the inequality
$$0 < (za-1)a^q-1$$
holds, then, since $a^q-1 > 0$, we must have $za = fracmna > 1$, or equivalently
$$m > fracna,.$$
If $a < b$ are two integers, then $b-a geqslant 1$, whence $m leqslant m(b-a)$. If $0 < a < b$ are integer solutions of
$$0 < (zx-1)x^q-1 < 1,, tag$ast$$$
then, since the middle expression in $(ast)$ is monotonic in $x$ for $x > z^-1$, we have
$$0 < (zb-1)b^q-1 - (za-1)a^q-1 < 1,.$$
Since $b^q-1 > a^q-1$ and $zb > za$, it follows that
$$0 < bigl((zb-1) - (za-1)bigr)a^q-1 = fracmn(b-a)a^q-1 leqslant (zb-1)b^q-1 - (za-1)a^q-1 < 1,,$$
and from that we obtain
$$m(b-a) < fracna^q-1,.$$
But from
$$m < fracna^q-1$$
it follows that
$$za = fracmna < frac1a^q-2 leqslant 1,,$$
which shows that $a$ can't be a solution of $(ast)$ by the remark at the top.
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
If the inequality
$$0 < (za-1)a^q-1$$
holds, then, since $a^q-1 > 0$, we must have $za = fracmna > 1$, or equivalently
$$m > fracna,.$$
If $a < b$ are two integers, then $b-a geqslant 1$, whence $m leqslant m(b-a)$. If $0 < a < b$ are integer solutions of
$$0 < (zx-1)x^q-1 < 1,, tag$ast$$$
then, since the middle expression in $(ast)$ is monotonic in $x$ for $x > z^-1$, we have
$$0 < (zb-1)b^q-1 - (za-1)a^q-1 < 1,.$$
Since $b^q-1 > a^q-1$ and $zb > za$, it follows that
$$0 < bigl((zb-1) - (za-1)bigr)a^q-1 = fracmn(b-a)a^q-1 leqslant (zb-1)b^q-1 - (za-1)a^q-1 < 1,,$$
and from that we obtain
$$m(b-a) < fracna^q-1,.$$
But from
$$m < fracna^q-1$$
it follows that
$$za = fracmna < frac1a^q-2 leqslant 1,,$$
which shows that $a$ can't be a solution of $(ast)$ by the remark at the top.
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
answered Jul 19 at 11:51
Daniel Fischer♦
171k16154274
171k16154274
add a comment |Â
add a comment |Â
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