What's the notation for writing a number as its digits












2












$begingroup$


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










share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


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










share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 4:18









MJD

47.9k29217398




47.9k29217398










asked Apr 22 '18 at 20:22









Joseph EckJoseph Eck

571413




571413








  • 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




    $begingroup$
    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










3 Answers
3






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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 $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      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.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 22 '18 at 20:49









            ericw31415ericw31415

            462212




            462212























                0












                $begingroup$

                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 $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  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 $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    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 $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.






                    share|cite|improve this answer









                    $endgroup$



                    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 $[30, 27, 24]_{16} = [30, 28, 8]_{16} = [31, 12, 8]_{16} = [15, 1, 12, 8]_{16}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 2 at 2:17









                    steven gregorysteven gregory

                    18.5k32359




                    18.5k32359























                        0












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            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.






                            share|cite|improve this answer









                            $endgroup$



                            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.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 2 at 2:27









                            fleabloodfleablood

                            1




                            1






























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