Decimal expansions of rational numbers.
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One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).
Using the same method I have run into a couple of problems.
- $x = frac{1}{6}$
In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$
so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.
$x = frac{1}{37}$.
In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.
Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?
modular-arithmetic rational-numbers decimal-expansion
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One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).
Using the same method I have run into a couple of problems.
- $x = frac{1}{6}$
In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$
so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.
$x = frac{1}{37}$.
In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.
Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?
modular-arithmetic rational-numbers decimal-expansion
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).
Using the same method I have run into a couple of problems.
- $x = frac{1}{6}$
In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$
so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.
$x = frac{1}{37}$.
In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.
Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?
modular-arithmetic rational-numbers decimal-expansion
One can use modular arithmetic to find the decimal expansion of a rational number. (see 1: https://i.stack.imgur.com/kw4Gk.png).
Using the same method I have run into a couple of problems.
- $x = frac{1}{6}$
In this case $m =1$ and $n=6$. Then the remainders are $1, 10 equiv 4, 40 equiv 4, ...$
so therefore $10^2 cdot 1 equiv 10^1 cdot 1 pmod{6}$. Now $10^2-10^1 = 90 = 6 cdot 15$ and so $left(10^2 -10^1right)x = 15$, but doing long division $x = 0.1overline{6}$ and not $x = 0.1overline{15}$. I am not really understanding the final step to determine $x$.
$x = frac{1}{37}$.
In this case $m =1$ and $n =37$. Then the remainders are $1, 10 equiv 10, 100 equiv 26, 260 equiv 1.$ So $10^3 cdot 1 equiv 1 pmod{37}$. Now $10^3 -1 = 999 = 27 cdot 37$. Therefore so $(10^3-1)x =27$, and by method above $x = 0.overline{27}$, but doing long division $x = 0.overline{027}$.
Clearly I am doing something wrong, the method should work for both of these cases. However I am not sure what it is?
modular-arithmetic rational-numbers decimal-expansion
modular-arithmetic rational-numbers decimal-expansion
asked Nov 3 at 17:10
xAly
192
192
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1 Answer
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1
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For $dfrac {1}{6} dots$
$begin {align}
1operatorname {mod} 6 equiv 1 \
10operatorname {mod} 6 equiv 4 \
40operatorname {mod} 6 equiv 4 \
end {align}$
$4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.
For $dfrac {1}{37} dots$
$begin {align}
1operatorname {mod} 37 equiv 1 \
10operatorname {mod} 37 equiv 10 \
100operatorname {mod} 37 equiv 26 \
260operatorname {mod} 37 equiv 1 \
end {align}$
$1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.
It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For $dfrac {1}{6} dots$
$begin {align}
1operatorname {mod} 6 equiv 1 \
10operatorname {mod} 6 equiv 4 \
40operatorname {mod} 6 equiv 4 \
end {align}$
$4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.
For $dfrac {1}{37} dots$
$begin {align}
1operatorname {mod} 37 equiv 1 \
10operatorname {mod} 37 equiv 10 \
100operatorname {mod} 37 equiv 26 \
260operatorname {mod} 37 equiv 1 \
end {align}$
$1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.
It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
add a comment |
up vote
1
down vote
accepted
For $dfrac {1}{6} dots$
$begin {align}
1operatorname {mod} 6 equiv 1 \
10operatorname {mod} 6 equiv 4 \
40operatorname {mod} 6 equiv 4 \
end {align}$
$4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.
For $dfrac {1}{37} dots$
$begin {align}
1operatorname {mod} 37 equiv 1 \
10operatorname {mod} 37 equiv 10 \
100operatorname {mod} 37 equiv 26 \
260operatorname {mod} 37 equiv 1 \
end {align}$
$1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.
It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For $dfrac {1}{6} dots$
$begin {align}
1operatorname {mod} 6 equiv 1 \
10operatorname {mod} 6 equiv 4 \
40operatorname {mod} 6 equiv 4 \
end {align}$
$4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.
For $dfrac {1}{37} dots$
$begin {align}
1operatorname {mod} 37 equiv 1 \
10operatorname {mod} 37 equiv 10 \
100operatorname {mod} 37 equiv 26 \
260operatorname {mod} 37 equiv 1 \
end {align}$
$1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.
It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.
For $dfrac {1}{6} dots$
$begin {align}
1operatorname {mod} 6 equiv 1 \
10operatorname {mod} 6 equiv 4 \
40operatorname {mod} 6 equiv 4 \
end {align}$
$4$ is the repeating number, so we say $color red {10^2 - 4} = 96 = 6 cdot 16$, and $(10^2 - 4)x = (16) rightarrow x = dfrac {16}{10^2-4} rightarrow dfrac {16}{96} rightarrow bbox [2px, border: 2px solid black]{x = .1overline{6}}$.
For $dfrac {1}{37} dots$
$begin {align}
1operatorname {mod} 37 equiv 1 \
10operatorname {mod} 37 equiv 10 \
100operatorname {mod} 37 equiv 26 \
260operatorname {mod} 37 equiv 1 \
end {align}$
$1$ is the repeating number, so we say $10^3 - 1 = 999 = 37 cdot 27$, and $color red {(10^3 - 1)}x = 27 rightarrow x = dfrac {27}{10^3-1} rightarrow x = dfrac {27}{999} rightarrow bbox [2px, border: 2px solid black] {x = overline{.027}}$.
It's equivalent to what we do when we write out the fraction in long division, because in each case we're multiplying by 10 when we reduce the modulo, then stop when we've found a repeating number. The highest power of $10$ in which a remainder repeates is where we find out the equivalent fraction.
edited Nov 13 at 1:10
answered Nov 3 at 18:07
bjcolby15
1,0961916
1,0961916
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
add a comment |
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
But then what about the example. 1 is subtracted, not 3.
– xAly
Nov 3 at 19:01
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
Divide $3$ from $3 cdot 10^6 equiv 3 operatorname {mod} 7$ and use Fermat's Little Theorem ($a^{p-1} = 1 operatorname {mod} p$). If you tried to subtract $3$ from $10^6$ you would have two primes that are not divisible by $7$.
– bjcolby15
Nov 3 at 20:20
add a comment |
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