What's the notation for writing a number as its digits
$begingroup$
Hoping this is a simple question, I'm pretty certain this is covered in number theory, but I haven't had much time to pour through my number theory book. I'm wondering what the notation for writing an integer as its digits is.
For example, given $435$ is it something like $4|3|5$?
Thanks
number-theory elementary-number-theory notation decimal-expansion number-systems
$endgroup$
|
show 5 more comments
$begingroup$
Hoping this is a simple question, I'm pretty certain this is covered in number theory, but I haven't had much time to pour through my number theory book. I'm wondering what the notation for writing an integer as its digits is.
For example, given $435$ is it something like $4|3|5$?
Thanks
number-theory elementary-number-theory notation decimal-expansion number-systems
$endgroup$
2
$begingroup$
I would say $435$ is writing $435$ 'as its digits'
$endgroup$
– Bram28
Apr 22 '18 at 20:24
$begingroup$
I've seen $(435)_{10}$.
$endgroup$
– Clement C.
Apr 22 '18 at 20:25
1
$begingroup$
Ah, so perhaps you're more interested in the symbolic case? Such as a way to write $a cdot 10^2 + b cdot 10 + c$ where $a, b, c in {0, dots, 9}$?
$endgroup$
– Aaron Montgomery
Apr 22 '18 at 20:29
1
$begingroup$
Sometimes you might write $d_nd_{n-1}...d_0$, say $435 = d_2d_1d_0$, $d_2 = 4, d_1 = 3, d_0 = 5$. Perhaps that's what you mean?
$endgroup$
– Countingstuff
Apr 22 '18 at 20:32
5
$begingroup$
I don't think there is any standard. Whatever you do, define and state what you are doing. It's enough to say, "let's write $a_na_{n-1}..a_1a_0$ to be the expression if the number in terms of its digits". Others prefer $a_n.a_{n-1}.....a_1.a_0$ as the "$.$" can resemble the standard of concatination. But it doesn't matter your notation as long as you specify what you are doing.
$endgroup$
– fleablood
Apr 22 '18 at 20:39
|
show 5 more comments
$begingroup$
Hoping this is a simple question, I'm pretty certain this is covered in number theory, but I haven't had much time to pour through my number theory book. I'm wondering what the notation for writing an integer as its digits is.
For example, given $435$ is it something like $4|3|5$?
Thanks
number-theory elementary-number-theory notation decimal-expansion number-systems
$endgroup$
Hoping this is a simple question, I'm pretty certain this is covered in number theory, but I haven't had much time to pour through my number theory book. I'm wondering what the notation for writing an integer as its digits is.
For example, given $435$ is it something like $4|3|5$?
Thanks
number-theory elementary-number-theory notation decimal-expansion number-systems
number-theory elementary-number-theory notation decimal-expansion number-systems
edited Jan 2 at 4:18
MJD
47.7k29215397
47.7k29215397
asked Apr 22 '18 at 20:22
Joseph EckJoseph Eck
569413
569413
2
$begingroup$
I would say $435$ is writing $435$ 'as its digits'
$endgroup$
– Bram28
Apr 22 '18 at 20:24
$begingroup$
I've seen $(435)_{10}$.
$endgroup$
– Clement C.
Apr 22 '18 at 20:25
1
$begingroup$
Ah, so perhaps you're more interested in the symbolic case? Such as a way to write $a cdot 10^2 + b cdot 10 + c$ where $a, b, c in {0, dots, 9}$?
$endgroup$
– Aaron Montgomery
Apr 22 '18 at 20:29
1
$begingroup$
Sometimes you might write $d_nd_{n-1}...d_0$, say $435 = d_2d_1d_0$, $d_2 = 4, d_1 = 3, d_0 = 5$. Perhaps that's what you mean?
$endgroup$
– Countingstuff
Apr 22 '18 at 20:32
5
$begingroup$
I don't think there is any standard. Whatever you do, define and state what you are doing. It's enough to say, "let's write $a_na_{n-1}..a_1a_0$ to be the expression if the number in terms of its digits". Others prefer $a_n.a_{n-1}.....a_1.a_0$ as the "$.$" can resemble the standard of concatination. But it doesn't matter your notation as long as you specify what you are doing.
$endgroup$
– fleablood
Apr 22 '18 at 20:39
|
show 5 more comments
2
$begingroup$
I would say $435$ is writing $435$ 'as its digits'
$endgroup$
– Bram28
Apr 22 '18 at 20:24
$begingroup$
I've seen $(435)_{10}$.
$endgroup$
– Clement C.
Apr 22 '18 at 20:25
1
$begingroup$
Ah, so perhaps you're more interested in the symbolic case? Such as a way to write $a cdot 10^2 + b cdot 10 + c$ where $a, b, c in {0, dots, 9}$?
$endgroup$
– Aaron Montgomery
Apr 22 '18 at 20:29
1
$begingroup$
Sometimes you might write $d_nd_{n-1}...d_0$, say $435 = d_2d_1d_0$, $d_2 = 4, d_1 = 3, d_0 = 5$. Perhaps that's what you mean?
$endgroup$
– Countingstuff
Apr 22 '18 at 20:32
5
$begingroup$
I don't think there is any standard. Whatever you do, define and state what you are doing. It's enough to say, "let's write $a_na_{n-1}..a_1a_0$ to be the expression if the number in terms of its digits". Others prefer $a_n.a_{n-1}.....a_1.a_0$ as the "$.$" can resemble the standard of concatination. But it doesn't matter your notation as long as you specify what you are doing.
$endgroup$
– fleablood
Apr 22 '18 at 20:39
2
2
$begingroup$
I would say $435$ is writing $435$ 'as its digits'
$endgroup$
– Bram28
Apr 22 '18 at 20:24
$begingroup$
I would say $435$ is writing $435$ 'as its digits'
$endgroup$
– Bram28
Apr 22 '18 at 20:24
$begingroup$
I've seen $(435)_{10}$.
$endgroup$
– Clement C.
Apr 22 '18 at 20:25
$begingroup$
I've seen $(435)_{10}$.
$endgroup$
– Clement C.
Apr 22 '18 at 20:25
1
1
$begingroup$
Ah, so perhaps you're more interested in the symbolic case? Such as a way to write $a cdot 10^2 + b cdot 10 + c$ where $a, b, c in {0, dots, 9}$?
$endgroup$
– Aaron Montgomery
Apr 22 '18 at 20:29
$begingroup$
Ah, so perhaps you're more interested in the symbolic case? Such as a way to write $a cdot 10^2 + b cdot 10 + c$ where $a, b, c in {0, dots, 9}$?
$endgroup$
– Aaron Montgomery
Apr 22 '18 at 20:29
1
1
$begingroup$
Sometimes you might write $d_nd_{n-1}...d_0$, say $435 = d_2d_1d_0$, $d_2 = 4, d_1 = 3, d_0 = 5$. Perhaps that's what you mean?
$endgroup$
– Countingstuff
Apr 22 '18 at 20:32
$begingroup$
Sometimes you might write $d_nd_{n-1}...d_0$, say $435 = d_2d_1d_0$, $d_2 = 4, d_1 = 3, d_0 = 5$. Perhaps that's what you mean?
$endgroup$
– Countingstuff
Apr 22 '18 at 20:32
5
5
$begingroup$
I don't think there is any standard. Whatever you do, define and state what you are doing. It's enough to say, "let's write $a_na_{n-1}..a_1a_0$ to be the expression if the number in terms of its digits". Others prefer $a_n.a_{n-1}.....a_1.a_0$ as the "$.$" can resemble the standard of concatination. But it doesn't matter your notation as long as you specify what you are doing.
$endgroup$
– fleablood
Apr 22 '18 at 20:39
$begingroup$
I don't think there is any standard. Whatever you do, define and state what you are doing. It's enough to say, "let's write $a_na_{n-1}..a_1a_0$ to be the expression if the number in terms of its digits". Others prefer $a_n.a_{n-1}.....a_1.a_0$ as the "$.$" can resemble the standard of concatination. But it doesn't matter your notation as long as you specify what you are doing.
$endgroup$
– fleablood
Apr 22 '18 at 20:39
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I think you're looking for something like $overline{ABCD}$ as a shorthand for $1000A+100B+10C+D$. I see this notation used sometimes in problems dealing with a number's digits.
$endgroup$
add a comment |
$begingroup$
I always used
$$[a,b,c,dots,z]_B$$
where $a,b,c,dots,z$ are base-$10$ numbersto represent a number in base $B$. The advantage is that $a,b,c, dots$ don't have to be single-digit integers.
For example, $[10, 9, 8]_{16} = 10cdot 16^2 + 9 cdot 16 + 8$. If it is clear what I'm doing, I don't use the brackets when doing arithmetic in base $B$.
For example, to compute $3 times [10, 9, 8]_{16}$:
begin{array}{c}
& 10 & 9 & 8 \
times & & & 3 \
-- & -- & -- & -- \
& 30 & 27 & 24 \
end{array}
and $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.
$endgroup$
add a comment |
$begingroup$
If you want to do something, just do it.
Say: For purpose of notation am going to indicate a number use expression seperated by $|$ to mean an integer whose digits are the values between bars. What I mean for example $|a|9-a|2$ will mean a three digit number where the first digit is $a$, the second $9-a$, and the third digit is $2$. i.e. $a|9-a|2 = a*10^2 + (9-a)*10 + 2$.
It doesn't matter if there is or isn't any standard notation (there isn't). You have expressed what you intend.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you're looking for something like $overline{ABCD}$ as a shorthand for $1000A+100B+10C+D$. I see this notation used sometimes in problems dealing with a number's digits.
$endgroup$
add a comment |
$begingroup$
I think you're looking for something like $overline{ABCD}$ as a shorthand for $1000A+100B+10C+D$. I see this notation used sometimes in problems dealing with a number's digits.
$endgroup$
add a comment |
$begingroup$
I think you're looking for something like $overline{ABCD}$ as a shorthand for $1000A+100B+10C+D$. I see this notation used sometimes in problems dealing with a number's digits.
$endgroup$
I think you're looking for something like $overline{ABCD}$ as a shorthand for $1000A+100B+10C+D$. I see this notation used sometimes in problems dealing with a number's digits.
answered Apr 22 '18 at 20:49
ericw31415ericw31415
452212
452212
add a comment |
add a comment |
$begingroup$
I always used
$$[a,b,c,dots,z]_B$$
where $a,b,c,dots,z$ are base-$10$ numbersto represent a number in base $B$. The advantage is that $a,b,c, dots$ don't have to be single-digit integers.
For example, $[10, 9, 8]_{16} = 10cdot 16^2 + 9 cdot 16 + 8$. If it is clear what I'm doing, I don't use the brackets when doing arithmetic in base $B$.
For example, to compute $3 times [10, 9, 8]_{16}$:
begin{array}{c}
& 10 & 9 & 8 \
times & & & 3 \
-- & -- & -- & -- \
& 30 & 27 & 24 \
end{array}
and $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.
$endgroup$
add a comment |
$begingroup$
I always used
$$[a,b,c,dots,z]_B$$
where $a,b,c,dots,z$ are base-$10$ numbersto represent a number in base $B$. The advantage is that $a,b,c, dots$ don't have to be single-digit integers.
For example, $[10, 9, 8]_{16} = 10cdot 16^2 + 9 cdot 16 + 8$. If it is clear what I'm doing, I don't use the brackets when doing arithmetic in base $B$.
For example, to compute $3 times [10, 9, 8]_{16}$:
begin{array}{c}
& 10 & 9 & 8 \
times & & & 3 \
-- & -- & -- & -- \
& 30 & 27 & 24 \
end{array}
and $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.
$endgroup$
add a comment |
$begingroup$
I always used
$$[a,b,c,dots,z]_B$$
where $a,b,c,dots,z$ are base-$10$ numbersto represent a number in base $B$. The advantage is that $a,b,c, dots$ don't have to be single-digit integers.
For example, $[10, 9, 8]_{16} = 10cdot 16^2 + 9 cdot 16 + 8$. If it is clear what I'm doing, I don't use the brackets when doing arithmetic in base $B$.
For example, to compute $3 times [10, 9, 8]_{16}$:
begin{array}{c}
& 10 & 9 & 8 \
times & & & 3 \
-- & -- & -- & -- \
& 30 & 27 & 24 \
end{array}
and $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.
$endgroup$
I always used
$$[a,b,c,dots,z]_B$$
where $a,b,c,dots,z$ are base-$10$ numbersto represent a number in base $B$. The advantage is that $a,b,c, dots$ don't have to be single-digit integers.
For example, $[10, 9, 8]_{16} = 10cdot 16^2 + 9 cdot 16 + 8$. If it is clear what I'm doing, I don't use the brackets when doing arithmetic in base $B$.
For example, to compute $3 times [10, 9, 8]_{16}$:
begin{array}{c}
& 10 & 9 & 8 \
times & & & 3 \
-- & -- & -- & -- \
& 30 & 27 & 24 \
end{array}
and $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.
answered Jan 2 at 2:17
steven gregorysteven gregory
18.3k32358
18.3k32358
add a comment |
add a comment |
$begingroup$
If you want to do something, just do it.
Say: For purpose of notation am going to indicate a number use expression seperated by $|$ to mean an integer whose digits are the values between bars. What I mean for example $|a|9-a|2$ will mean a three digit number where the first digit is $a$, the second $9-a$, and the third digit is $2$. i.e. $a|9-a|2 = a*10^2 + (9-a)*10 + 2$.
It doesn't matter if there is or isn't any standard notation (there isn't). You have expressed what you intend.
$endgroup$
add a comment |
$begingroup$
If you want to do something, just do it.
Say: For purpose of notation am going to indicate a number use expression seperated by $|$ to mean an integer whose digits are the values between bars. What I mean for example $|a|9-a|2$ will mean a three digit number where the first digit is $a$, the second $9-a$, and the third digit is $2$. i.e. $a|9-a|2 = a*10^2 + (9-a)*10 + 2$.
It doesn't matter if there is or isn't any standard notation (there isn't). You have expressed what you intend.
$endgroup$
add a comment |
$begingroup$
If you want to do something, just do it.
Say: For purpose of notation am going to indicate a number use expression seperated by $|$ to mean an integer whose digits are the values between bars. What I mean for example $|a|9-a|2$ will mean a three digit number where the first digit is $a$, the second $9-a$, and the third digit is $2$. i.e. $a|9-a|2 = a*10^2 + (9-a)*10 + 2$.
It doesn't matter if there is or isn't any standard notation (there isn't). You have expressed what you intend.
$endgroup$
If you want to do something, just do it.
Say: For purpose of notation am going to indicate a number use expression seperated by $|$ to mean an integer whose digits are the values between bars. What I mean for example $|a|9-a|2$ will mean a three digit number where the first digit is $a$, the second $9-a$, and the third digit is $2$. i.e. $a|9-a|2 = a*10^2 + (9-a)*10 + 2$.
It doesn't matter if there is or isn't any standard notation (there isn't). You have expressed what you intend.
answered Jan 2 at 2:27
fleabloodfleablood
73.6k22891
73.6k22891
add a comment |
add a comment |
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2
$begingroup$
I would say $435$ is writing $435$ 'as its digits'
$endgroup$
– Bram28
Apr 22 '18 at 20:24
$begingroup$
I've seen $(435)_{10}$.
$endgroup$
– Clement C.
Apr 22 '18 at 20:25
1
$begingroup$
Ah, so perhaps you're more interested in the symbolic case? Such as a way to write $a cdot 10^2 + b cdot 10 + c$ where $a, b, c in {0, dots, 9}$?
$endgroup$
– Aaron Montgomery
Apr 22 '18 at 20:29
1
$begingroup$
Sometimes you might write $d_nd_{n-1}...d_0$, say $435 = d_2d_1d_0$, $d_2 = 4, d_1 = 3, d_0 = 5$. Perhaps that's what you mean?
$endgroup$
– Countingstuff
Apr 22 '18 at 20:32
5
$begingroup$
I don't think there is any standard. Whatever you do, define and state what you are doing. It's enough to say, "let's write $a_na_{n-1}..a_1a_0$ to be the expression if the number in terms of its digits". Others prefer $a_n.a_{n-1}.....a_1.a_0$ as the "$.$" can resemble the standard of concatination. But it doesn't matter your notation as long as you specify what you are doing.
$endgroup$
– fleablood
Apr 22 '18 at 20:39