Prove that there are no two integers $a$ and $b$, such that $fracab$ is non-terminating and non-recurring number
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I understand that all non-terminating and recurring real numbers can be expressed as $fracab$ where $a, b$ are integers. But I am unable to prove converse of this statement.
number-theory elementary-number-theory
edited Aug 28 at 17:28
tarit goswami
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asked Aug 28 at 16:42
meet112
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I understand that all non-terminating and recurring real numbers can be expressed as $fracab$ where $a, b$ are integers. But I am unable to prove converse of this statement.
number-theory elementary-number-theory
edited Aug 28 at 17:28
tarit goswami
1,139219
asked Aug 28 at 16:42
meet112
265
1
Consider the algorithm of long division by $b$: if it is non-terminating, as there are only $b$ possible partial remainders at each step, one of them will appear again in a further step. All the steps between the first and the second appearance of this remainder will be reproduced in the following step.
– Bernard
Aug 28 at 16:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I understand that all non-terminating and recurring real numbers can be expressed as $fracab$ where $a, b$ are integers. But I am unable to prove converse of this statement.
number-theory elementary-number-theory
edited Aug 28 at 17:28
tarit goswami
1,139219
asked Aug 28 at 16:42
meet112
265
I understand that all non-terminating and recurring real numbers can be expressed as $fracab$ where $a, b$ are integers. But I am unable to prove converse of this statement.
number-theory elementary-number-theory
edited Aug 28 at 17:28
tarit goswami
1,139219
asked Aug 28 at 16:42
meet112
265
edited Aug 28 at 17:28
tarit goswami
1,139219
edited Aug 28 at 17:28
tarit goswami
1,139219
edited Aug 28 at 17:28
tarit goswami
1,139219
1,139219
asked Aug 28 at 16:42
meet112
265
asked Aug 28 at 16:42
meet112
265
asked Aug 28 at 16:42
meet112
265
265
1
Consider the algorithm of long division by $b$: if it is non-terminating, as there are only $b$ possible partial remainders at each step, one of them will appear again in a further step. All the steps between the first and the second appearance of this remainder will be reproduced in the following step.
– Bernard
Aug 28 at 16:56
add a comment |Â
1
Consider the algorithm of long division by $b$: if it is non-terminating, as there are only $b$ possible partial remainders at each step, one of them will appear again in a further step. All the steps between the first and the second appearance of this remainder will be reproduced in the following step.
– Bernard
Aug 28 at 16:56
1
1
Consider the algorithm of long division by $b$: if it is non-terminating, as there are only $b$ possible partial remainders at each step, one of them will appear again in a further step. All the steps between the first and the second appearance of this remainder will be reproduced in the following step.
– Bernard
Aug 28 at 16:56
Consider the algorithm of long division by $b$: if it is non-terminating, as there are only $b$ possible partial remainders at each step, one of them will appear again in a further step. All the steps between the first and the second appearance of this remainder will be reproduced in the following step.
– Bernard
Aug 28 at 16:56
add a comment |Â
2 Answers
2
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2
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Suppose there exist two integers $p$ and $q$ where $qneq0$
When you divide $fracpq$ using long division, there are only $q$ remainders that are possible. If $0$ appears as a remainder, the decimal expansion terminates. If $0$ never occurs, then the algorithm can run at most $q − 1$ steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.
For more, read about irrational numbers
answered Aug 28 at 17:01
Dashi
615311
add a comment |Â
up vote
1
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Factor $b$ as $b=2^h5^kc$, where $c$ is divisible neither for $2$ nor for $5$. Then
$$
fracab=frac2^k5^ha10^h+kc
$$
and we can reduce to the case of $a/b$ where $b$ is divisible neither for $2$ nor for $5$ (because the original number is the same with the decimal point suitably shifted) and $b>1$. It is also not restrictive to assume $0<a<b$.
We then know that the decimal number we get is not terminating. Can you prove it?
The process of long division (adding zeros as we need them) will never leave a zero remainder, so the remainder $r$ at each stage satisfies $0<r<b$.
Since only $b-1$ remainders are available, at some stage we must get a remainder we already got before. At that point, the divisions will repeat.
answered Aug 28 at 17:05
egreg
166k1180187
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Suppose there exist two integers $p$ and $q$ where $qneq0$
When you divide $fracpq$ using long division, there are only $q$ remainders that are possible. If $0$ appears as a remainder, the decimal expansion terminates. If $0$ never occurs, then the algorithm can run at most $q − 1$ steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.
For more, read about irrational numbers
answered Aug 28 at 17:01
Dashi
615311
add a comment |Â
up vote
2
down vote
Suppose there exist two integers $p$ and $q$ where $qneq0$
When you divide $fracpq$ using long division, there are only $q$ remainders that are possible. If $0$ appears as a remainder, the decimal expansion terminates. If $0$ never occurs, then the algorithm can run at most $q − 1$ steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.
For more, read about irrational numbers
answered Aug 28 at 17:01
Dashi
615311
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Suppose there exist two integers $p$ and $q$ where $qneq0$
When you divide $fracpq$ using long division, there are only $q$ remainders that are possible. If $0$ appears as a remainder, the decimal expansion terminates. If $0$ never occurs, then the algorithm can run at most $q − 1$ steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.
For more, read about irrational numbers
answered Aug 28 at 17:01
Dashi
615311
Suppose there exist two integers $p$ and $q$ where $qneq0$
When you divide $fracpq$ using long division, there are only $q$ remainders that are possible. If $0$ appears as a remainder, the decimal expansion terminates. If $0$ never occurs, then the algorithm can run at most $q − 1$ steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.
For more, read about irrational numbers
answered Aug 28 at 17:01
Dashi
615311
edited Aug 28 at 17:43
answered Aug 28 at 17:01
Dashi
615311
answered Aug 28 at 17:01
Dashi
615311
answered Aug 28 at 17:01
Dashi
615311
615311
add a comment |Â
add a comment |Â
up vote
1
down vote
Factor $b$ as $b=2^h5^kc$, where $c$ is divisible neither for $2$ nor for $5$. Then
$$
fracab=frac2^k5^ha10^h+kc
$$
and we can reduce to the case of $a/b$ where $b$ is divisible neither for $2$ nor for $5$ (because the original number is the same with the decimal point suitably shifted) and $b>1$. It is also not restrictive to assume $0<a<b$.
We then know that the decimal number we get is not terminating. Can you prove it?
The process of long division (adding zeros as we need them) will never leave a zero remainder, so the remainder $r$ at each stage satisfies $0<r<b$.
Since only $b-1$ remainders are available, at some stage we must get a remainder we already got before. At that point, the divisions will repeat.
answered Aug 28 at 17:05
egreg
166k1180187
add a comment |Â
up vote
1
down vote
Factor $b$ as $b=2^h5^kc$, where $c$ is divisible neither for $2$ nor for $5$. Then
$$
fracab=frac2^k5^ha10^h+kc
$$
and we can reduce to the case of $a/b$ where $b$ is divisible neither for $2$ nor for $5$ (because the original number is the same with the decimal point suitably shifted) and $b>1$. It is also not restrictive to assume $0<a<b$.
We then know that the decimal number we get is not terminating. Can you prove it?
The process of long division (adding zeros as we need them) will never leave a zero remainder, so the remainder $r$ at each stage satisfies $0<r<b$.
Since only $b-1$ remainders are available, at some stage we must get a remainder we already got before. At that point, the divisions will repeat.
answered Aug 28 at 17:05
egreg
166k1180187
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Factor $b$ as $b=2^h5^kc$, where $c$ is divisible neither for $2$ nor for $5$. Then
$$
fracab=frac2^k5^ha10^h+kc
$$
and we can reduce to the case of $a/b$ where $b$ is divisible neither for $2$ nor for $5$ (because the original number is the same with the decimal point suitably shifted) and $b>1$. It is also not restrictive to assume $0<a<b$.
We then know that the decimal number we get is not terminating. Can you prove it?
The process of long division (adding zeros as we need them) will never leave a zero remainder, so the remainder $r$ at each stage satisfies $0<r<b$.
Since only $b-1$ remainders are available, at some stage we must get a remainder we already got before. At that point, the divisions will repeat.
answered Aug 28 at 17:05
egreg
166k1180187
Factor $b$ as $b=2^h5^kc$, where $c$ is divisible neither for $2$ nor for $5$. Then
$$
fracab=frac2^k5^ha10^h+kc
$$
and we can reduce to the case of $a/b$ where $b$ is divisible neither for $2$ nor for $5$ (because the original number is the same with the decimal point suitably shifted) and $b>1$. It is also not restrictive to assume $0<a<b$.
We then know that the decimal number we get is not terminating. Can you prove it?
The process of long division (adding zeros as we need them) will never leave a zero remainder, so the remainder $r$ at each stage satisfies $0<r<b$.
Since only $b-1$ remainders are available, at some stage we must get a remainder we already got before. At that point, the divisions will repeat.
answered Aug 28 at 17:05
egreg
166k1180187
answered Aug 28 at 17:05
egreg
166k1180187
answered Aug 28 at 17:05
egreg
166k1180187
answered Aug 28 at 17:05
egreg
166k1180187
166k1180187
add a comment |Â
add a comment |Â
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1
Consider the algorithm of long division by $b$: if it is non-terminating, as there are only $b$ possible partial remainders at each step, one of them will appear again in a further step. All the steps between the first and the second appearance of this remainder will be reproduced in the following step.
– Bernard
Aug 28 at 16:56