Multiply $1111_2$ with $1111_2$












1












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I tried to multiply $1111_2$ with $1111_2$ but I came across a problem, namely adding $1+1+1+1$ and $1+1+1+1+1$. Here is my method:



enter image description here



I can't figure out how to add the $3rd$ and $4th$ column. I thought that $1+1+1+1 = 4 equiv 100$ but how can I implement it here?










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  • 1




    $begingroup$
    Add enough columns to the left...
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 5 '18 at 12:24










  • $begingroup$
    If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
    $endgroup$
    – Damien
    Dec 5 '18 at 12:26












  • $begingroup$
    You can write $0$ and carry over the $10$. The rest is similar.
    $endgroup$
    – AryanSonwatikar
    Dec 5 '18 at 12:31










  • $begingroup$
    @MauroALLEGRANZA And don't forget leaving some space on top,
    $endgroup$
    – CopyPasteIt
    Dec 5 '18 at 15:39
















1












$begingroup$


I tried to multiply $1111_2$ with $1111_2$ but I came across a problem, namely adding $1+1+1+1$ and $1+1+1+1+1$. Here is my method:



enter image description here



I can't figure out how to add the $3rd$ and $4th$ column. I thought that $1+1+1+1 = 4 equiv 100$ but how can I implement it here?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Add enough columns to the left...
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 5 '18 at 12:24










  • $begingroup$
    If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
    $endgroup$
    – Damien
    Dec 5 '18 at 12:26












  • $begingroup$
    You can write $0$ and carry over the $10$. The rest is similar.
    $endgroup$
    – AryanSonwatikar
    Dec 5 '18 at 12:31










  • $begingroup$
    @MauroALLEGRANZA And don't forget leaving some space on top,
    $endgroup$
    – CopyPasteIt
    Dec 5 '18 at 15:39














1












1








1





$begingroup$


I tried to multiply $1111_2$ with $1111_2$ but I came across a problem, namely adding $1+1+1+1$ and $1+1+1+1+1$. Here is my method:



enter image description here



I can't figure out how to add the $3rd$ and $4th$ column. I thought that $1+1+1+1 = 4 equiv 100$ but how can I implement it here?










share|cite|improve this question











$endgroup$




I tried to multiply $1111_2$ with $1111_2$ but I came across a problem, namely adding $1+1+1+1$ and $1+1+1+1+1$. Here is my method:



enter image description here



I can't figure out how to add the $3rd$ and $4th$ column. I thought that $1+1+1+1 = 4 equiv 100$ but how can I implement it here?







arithmetic binary






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share|cite|improve this question













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edited Jan 11 at 5:57









Eric Wofsey

188k14216345




188k14216345










asked Dec 5 '18 at 12:17









xxxtentacionxxxtentacion

324113




324113








  • 1




    $begingroup$
    Add enough columns to the left...
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 5 '18 at 12:24










  • $begingroup$
    If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
    $endgroup$
    – Damien
    Dec 5 '18 at 12:26












  • $begingroup$
    You can write $0$ and carry over the $10$. The rest is similar.
    $endgroup$
    – AryanSonwatikar
    Dec 5 '18 at 12:31










  • $begingroup$
    @MauroALLEGRANZA And don't forget leaving some space on top,
    $endgroup$
    – CopyPasteIt
    Dec 5 '18 at 15:39














  • 1




    $begingroup$
    Add enough columns to the left...
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 5 '18 at 12:24










  • $begingroup$
    If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
    $endgroup$
    – Damien
    Dec 5 '18 at 12:26












  • $begingroup$
    You can write $0$ and carry over the $10$. The rest is similar.
    $endgroup$
    – AryanSonwatikar
    Dec 5 '18 at 12:31










  • $begingroup$
    @MauroALLEGRANZA And don't forget leaving some space on top,
    $endgroup$
    – CopyPasteIt
    Dec 5 '18 at 15:39








1




1




$begingroup$
Add enough columns to the left...
$endgroup$
– Mauro ALLEGRANZA
Dec 5 '18 at 12:24




$begingroup$
Add enough columns to the left...
$endgroup$
– Mauro ALLEGRANZA
Dec 5 '18 at 12:24












$begingroup$
If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
$endgroup$
– Damien
Dec 5 '18 at 12:26






$begingroup$
If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
$endgroup$
– Damien
Dec 5 '18 at 12:26














$begingroup$
You can write $0$ and carry over the $10$. The rest is similar.
$endgroup$
– AryanSonwatikar
Dec 5 '18 at 12:31




$begingroup$
You can write $0$ and carry over the $10$. The rest is similar.
$endgroup$
– AryanSonwatikar
Dec 5 '18 at 12:31












$begingroup$
@MauroALLEGRANZA And don't forget leaving some space on top,
$endgroup$
– CopyPasteIt
Dec 5 '18 at 15:39




$begingroup$
@MauroALLEGRANZA And don't forget leaving some space on top,
$endgroup$
– CopyPasteIt
Dec 5 '18 at 15:39










5 Answers
5






active

oldest

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1












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From the end, first digit: $color{red}1$.



2nd digit: $1+1=color{blue}1color{red}0$.



3rd digit: $1+1+1+color{blue}1=10+10=color{blue}{10}color{red}0$.



4th digit: $1+1+1+1+color{blue}{10}=10+10+10=100+10=color{blue}{11}color{red}0$.



5th digit: $1+1+1+color{blue}{11}=10+100=color{blue}{11}color{red}0$.



6th digit: $1+1+color{blue}{11}=1+100=color{blue}{10}color{red}1$.



7th digit: $1+color{blue}{10}=color{red}{11}$.



Collecting backwards: $1111_2times1111_2=11100001$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    You carry over $color{blue}{2=10_2}$, and then continue addition. It is that simple.



    Remember, even if we are adding single-digit numbers while moving down a column, there is no guarantee that the carry over will be a single digit number. For example, if you perform , say:
    $$
    29 \ + \ 09 \ +\ 09 \+ \ vdots \ + \ 09
    $$



    where the number of $09$s is $12$, then the sum of the first column will be $117$, so your sum will be:
    $$
    2^{11}9 \ + \ 09 \+ \09 \ + \ + \ vdots \ + \ 09 \ ----- \ 2^{11}7 = 137 \ -----
    $$



    In this case, for example, we would have :
    $$
    begin{matrix}
    color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
    end{matrix}
    $$






    share|cite|improve this answer











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      2












      $begingroup$

      $$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
      =2^5cdot111_2+1_2=11100000+1=11100001
      $$






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        I used google sheets to organize the manual work, allowing for plenty of space on the left and top. If you need more workspace once you start you can insert rows and columns.



        So the 'carry stuff' is above the red line and has to be included when you add the numbers between the two black lines. Each 'carry' goes in the next higher row.



        Here is the work:



        enter image description here



        This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
          $endgroup$
          – астон вілла олоф мэллбэрг
          Dec 5 '18 at 15:26






        • 1




          $begingroup$
          My update came at exact time as your comment!
          $endgroup$
          – CopyPasteIt
          Dec 5 '18 at 15:30






        • 1




          $begingroup$
          Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
          $endgroup$
          – астон вілла олоф мэллбэрг
          Dec 5 '18 at 15:33






        • 1




          $begingroup$
          @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
          $endgroup$
          – CopyPasteIt
          Dec 5 '18 at 15:36





















        1












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        As you appear to have done, you can start the calculation by splitting up one of the numbers into its constituent powers of $2$ and then distributing the multiplication:



        $$begin{aligned}color{blue}{1111_2}times1111_2&=(color{blue}{1000_2}+color{blue}{100_2}+color{blue}{10_2}+color{blue}{1_2})times1111_2\
        &=(1000_2times1111_2)+(100_2times1111_2)+(10_2times1111_2)+(1_2times1111_2)\
        &=1111000_2+111100_2+11110_2+1111_2quad(*)\
        end{aligned}$$



        In your third column from the right, where you get stuck, you have:



        $$begin{aligned}underbrace{1}_{text{carried}}+1+1+1&=100_2
        end{aligned}$$



        For $100$, your bit is $0$ and you can carry the rest ($10_2$). Then, in the next column, you'd have:



        $$begin{aligned}underbrace{10_2}_{text{carried}}+1+1+1+1&=110_2
        end{aligned}$$



        And just like before, your bit is $0$ but you carry the rest ($11_2$).



        If you keep on with that method, you'd get:



        $$begin{aligned}1111_2times1111_2&=color{red}{11100001_2}\
        15times15,color{green}{✔}&=225,color{green}{✔}end{aligned}$$





        (*) When multiplying any binary number by a power of $2$, you can just append the $0$'s of the power of $2$ onto the end of the other number. For example, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
        &=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
        &=10000000_2+10000_2+1000_2\&=10011000_2\
        8times38,color{green}{✔}&=152,color{green}{✔}
        end{aligned}$$






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          5 Answers
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          active

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          5 Answers
          5






          active

          oldest

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          1












          $begingroup$

          From the end, first digit: $color{red}1$.



          2nd digit: $1+1=color{blue}1color{red}0$.



          3rd digit: $1+1+1+color{blue}1=10+10=color{blue}{10}color{red}0$.



          4th digit: $1+1+1+1+color{blue}{10}=10+10+10=100+10=color{blue}{11}color{red}0$.



          5th digit: $1+1+1+color{blue}{11}=10+100=color{blue}{11}color{red}0$.



          6th digit: $1+1+color{blue}{11}=1+100=color{blue}{10}color{red}1$.



          7th digit: $1+color{blue}{10}=color{red}{11}$.



          Collecting backwards: $1111_2times1111_2=11100001$.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            From the end, first digit: $color{red}1$.



            2nd digit: $1+1=color{blue}1color{red}0$.



            3rd digit: $1+1+1+color{blue}1=10+10=color{blue}{10}color{red}0$.



            4th digit: $1+1+1+1+color{blue}{10}=10+10+10=100+10=color{blue}{11}color{red}0$.



            5th digit: $1+1+1+color{blue}{11}=10+100=color{blue}{11}color{red}0$.



            6th digit: $1+1+color{blue}{11}=1+100=color{blue}{10}color{red}1$.



            7th digit: $1+color{blue}{10}=color{red}{11}$.



            Collecting backwards: $1111_2times1111_2=11100001$.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              From the end, first digit: $color{red}1$.



              2nd digit: $1+1=color{blue}1color{red}0$.



              3rd digit: $1+1+1+color{blue}1=10+10=color{blue}{10}color{red}0$.



              4th digit: $1+1+1+1+color{blue}{10}=10+10+10=100+10=color{blue}{11}color{red}0$.



              5th digit: $1+1+1+color{blue}{11}=10+100=color{blue}{11}color{red}0$.



              6th digit: $1+1+color{blue}{11}=1+100=color{blue}{10}color{red}1$.



              7th digit: $1+color{blue}{10}=color{red}{11}$.



              Collecting backwards: $1111_2times1111_2=11100001$.






              share|cite|improve this answer









              $endgroup$



              From the end, first digit: $color{red}1$.



              2nd digit: $1+1=color{blue}1color{red}0$.



              3rd digit: $1+1+1+color{blue}1=10+10=color{blue}{10}color{red}0$.



              4th digit: $1+1+1+1+color{blue}{10}=10+10+10=100+10=color{blue}{11}color{red}0$.



              5th digit: $1+1+1+color{blue}{11}=10+100=color{blue}{11}color{red}0$.



              6th digit: $1+1+color{blue}{11}=1+100=color{blue}{10}color{red}1$.



              7th digit: $1+color{blue}{10}=color{red}{11}$.



              Collecting backwards: $1111_2times1111_2=11100001$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 5 '18 at 12:46









              farruhotafarruhota

              20.6k2740




              20.6k2740























                  4












                  $begingroup$

                  You carry over $color{blue}{2=10_2}$, and then continue addition. It is that simple.



                  Remember, even if we are adding single-digit numbers while moving down a column, there is no guarantee that the carry over will be a single digit number. For example, if you perform , say:
                  $$
                  29 \ + \ 09 \ +\ 09 \+ \ vdots \ + \ 09
                  $$



                  where the number of $09$s is $12$, then the sum of the first column will be $117$, so your sum will be:
                  $$
                  2^{11}9 \ + \ 09 \+ \09 \ + \ + \ vdots \ + \ 09 \ ----- \ 2^{11}7 = 137 \ -----
                  $$



                  In this case, for example, we would have :
                  $$
                  begin{matrix}
                  color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
                  end{matrix}
                  $$






                  share|cite|improve this answer











                  $endgroup$


















                    4












                    $begingroup$

                    You carry over $color{blue}{2=10_2}$, and then continue addition. It is that simple.



                    Remember, even if we are adding single-digit numbers while moving down a column, there is no guarantee that the carry over will be a single digit number. For example, if you perform , say:
                    $$
                    29 \ + \ 09 \ +\ 09 \+ \ vdots \ + \ 09
                    $$



                    where the number of $09$s is $12$, then the sum of the first column will be $117$, so your sum will be:
                    $$
                    2^{11}9 \ + \ 09 \+ \09 \ + \ + \ vdots \ + \ 09 \ ----- \ 2^{11}7 = 137 \ -----
                    $$



                    In this case, for example, we would have :
                    $$
                    begin{matrix}
                    color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
                    end{matrix}
                    $$






                    share|cite|improve this answer











                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      You carry over $color{blue}{2=10_2}$, and then continue addition. It is that simple.



                      Remember, even if we are adding single-digit numbers while moving down a column, there is no guarantee that the carry over will be a single digit number. For example, if you perform , say:
                      $$
                      29 \ + \ 09 \ +\ 09 \+ \ vdots \ + \ 09
                      $$



                      where the number of $09$s is $12$, then the sum of the first column will be $117$, so your sum will be:
                      $$
                      2^{11}9 \ + \ 09 \+ \09 \ + \ + \ vdots \ + \ 09 \ ----- \ 2^{11}7 = 137 \ -----
                      $$



                      In this case, for example, we would have :
                      $$
                      begin{matrix}
                      color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
                      end{matrix}
                      $$






                      share|cite|improve this answer











                      $endgroup$



                      You carry over $color{blue}{2=10_2}$, and then continue addition. It is that simple.



                      Remember, even if we are adding single-digit numbers while moving down a column, there is no guarantee that the carry over will be a single digit number. For example, if you perform , say:
                      $$
                      29 \ + \ 09 \ +\ 09 \+ \ vdots \ + \ 09
                      $$



                      where the number of $09$s is $12$, then the sum of the first column will be $117$, so your sum will be:
                      $$
                      2^{11}9 \ + \ 09 \+ \09 \ + \ + \ vdots \ + \ 09 \ ----- \ 2^{11}7 = 137 \ -----
                      $$



                      In this case, for example, we would have :
                      $$
                      begin{matrix}
                      color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
                      end{matrix}
                      $$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 5 '18 at 12:49

























                      answered Dec 5 '18 at 12:30









                      астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                      38.9k33477




                      38.9k33477























                          2












                          $begingroup$

                          $$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
                          =2^5cdot111_2+1_2=11100000+1=11100001
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            $$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
                            =2^5cdot111_2+1_2=11100000+1=11100001
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              $$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
                              =2^5cdot111_2+1_2=11100000+1=11100001
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              $$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
                              =2^5cdot111_2+1_2=11100000+1=11100001
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 5 '18 at 12:27









                              Yiorgos S. SmyrlisYiorgos S. Smyrlis

                              63.4k1385163




                              63.4k1385163























                                  2












                                  $begingroup$

                                  I used google sheets to organize the manual work, allowing for plenty of space on the left and top. If you need more workspace once you start you can insert rows and columns.



                                  So the 'carry stuff' is above the red line and has to be included when you add the numbers between the two black lines. Each 'carry' goes in the next higher row.



                                  Here is the work:



                                  enter image description here



                                  This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:26






                                  • 1




                                    $begingroup$
                                    My update came at exact time as your comment!
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:30






                                  • 1




                                    $begingroup$
                                    Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:33






                                  • 1




                                    $begingroup$
                                    @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:36


















                                  2












                                  $begingroup$

                                  I used google sheets to organize the manual work, allowing for plenty of space on the left and top. If you need more workspace once you start you can insert rows and columns.



                                  So the 'carry stuff' is above the red line and has to be included when you add the numbers between the two black lines. Each 'carry' goes in the next higher row.



                                  Here is the work:



                                  enter image description here



                                  This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:26






                                  • 1




                                    $begingroup$
                                    My update came at exact time as your comment!
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:30






                                  • 1




                                    $begingroup$
                                    Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:33






                                  • 1




                                    $begingroup$
                                    @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:36
















                                  2












                                  2








                                  2





                                  $begingroup$

                                  I used google sheets to organize the manual work, allowing for plenty of space on the left and top. If you need more workspace once you start you can insert rows and columns.



                                  So the 'carry stuff' is above the red line and has to be included when you add the numbers between the two black lines. Each 'carry' goes in the next higher row.



                                  Here is the work:



                                  enter image description here



                                  This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.






                                  share|cite|improve this answer











                                  $endgroup$



                                  I used google sheets to organize the manual work, allowing for plenty of space on the left and top. If you need more workspace once you start you can insert rows and columns.



                                  So the 'carry stuff' is above the red line and has to be included when you add the numbers between the two black lines. Each 'carry' goes in the next higher row.



                                  Here is the work:



                                  enter image description here



                                  This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited Dec 5 '18 at 15:26

























                                  answered Dec 5 '18 at 15:08









                                  CopyPasteItCopyPasteIt

                                  4,2031628




                                  4,2031628












                                  • $begingroup$
                                    I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:26






                                  • 1




                                    $begingroup$
                                    My update came at exact time as your comment!
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:30






                                  • 1




                                    $begingroup$
                                    Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:33






                                  • 1




                                    $begingroup$
                                    @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:36




















                                  • $begingroup$
                                    I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:26






                                  • 1




                                    $begingroup$
                                    My update came at exact time as your comment!
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:30






                                  • 1




                                    $begingroup$
                                    Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
                                    $endgroup$
                                    – астон вілла олоф мэллбэрг
                                    Dec 5 '18 at 15:33






                                  • 1




                                    $begingroup$
                                    @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
                                    $endgroup$
                                    – CopyPasteIt
                                    Dec 5 '18 at 15:36


















                                  $begingroup$
                                  I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
                                  $endgroup$
                                  – астон вілла олоф мэллбэрг
                                  Dec 5 '18 at 15:26




                                  $begingroup$
                                  I like the fact that you spaced out your sheet. I had to make do with the matrix construct from mathjax. Nice one, +1.
                                  $endgroup$
                                  – астон вілла олоф мэллбэрг
                                  Dec 5 '18 at 15:26




                                  1




                                  1




                                  $begingroup$
                                  My update came at exact time as your comment!
                                  $endgroup$
                                  – CopyPasteIt
                                  Dec 5 '18 at 15:30




                                  $begingroup$
                                  My update came at exact time as your comment!
                                  $endgroup$
                                  – CopyPasteIt
                                  Dec 5 '18 at 15:30




                                  1




                                  1




                                  $begingroup$
                                  Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
                                  $endgroup$
                                  – астон вілла олоф мэллбэрг
                                  Dec 5 '18 at 15:33




                                  $begingroup$
                                  Yes, a very interesting coincidence. Interestingly enough, just because we don't perform , say, addition of ten twelve numbers at a time usually in school, one doesn't move beyond having single digit carries. I think that is the crux of this problem : something that should be clarified in school itself.
                                  $endgroup$
                                  – астон вілла олоф мэллбэрг
                                  Dec 5 '18 at 15:33




                                  1




                                  1




                                  $begingroup$
                                  @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
                                  $endgroup$
                                  – CopyPasteIt
                                  Dec 5 '18 at 15:36






                                  $begingroup$
                                  @астонвіллаолофмэллбэрг As I worked it out it was like a puzzle and only one way to go to make sense of it all. And I was thinking, "Gee, I don;t remember doing this as a kid with base 10". Your comment clarifies the situation...
                                  $endgroup$
                                  – CopyPasteIt
                                  Dec 5 '18 at 15:36













                                  1












                                  $begingroup$

                                  As you appear to have done, you can start the calculation by splitting up one of the numbers into its constituent powers of $2$ and then distributing the multiplication:



                                  $$begin{aligned}color{blue}{1111_2}times1111_2&=(color{blue}{1000_2}+color{blue}{100_2}+color{blue}{10_2}+color{blue}{1_2})times1111_2\
                                  &=(1000_2times1111_2)+(100_2times1111_2)+(10_2times1111_2)+(1_2times1111_2)\
                                  &=1111000_2+111100_2+11110_2+1111_2quad(*)\
                                  end{aligned}$$



                                  In your third column from the right, where you get stuck, you have:



                                  $$begin{aligned}underbrace{1}_{text{carried}}+1+1+1&=100_2
                                  end{aligned}$$



                                  For $100$, your bit is $0$ and you can carry the rest ($10_2$). Then, in the next column, you'd have:



                                  $$begin{aligned}underbrace{10_2}_{text{carried}}+1+1+1+1&=110_2
                                  end{aligned}$$



                                  And just like before, your bit is $0$ but you carry the rest ($11_2$).



                                  If you keep on with that method, you'd get:



                                  $$begin{aligned}1111_2times1111_2&=color{red}{11100001_2}\
                                  15times15,color{green}{✔}&=225,color{green}{✔}end{aligned}$$





                                  (*) When multiplying any binary number by a power of $2$, you can just append the $0$'s of the power of $2$ onto the end of the other number. For example, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
                                  &=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
                                  &=10000000_2+10000_2+1000_2\&=10011000_2\
                                  8times38,color{green}{✔}&=152,color{green}{✔}
                                  end{aligned}$$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    As you appear to have done, you can start the calculation by splitting up one of the numbers into its constituent powers of $2$ and then distributing the multiplication:



                                    $$begin{aligned}color{blue}{1111_2}times1111_2&=(color{blue}{1000_2}+color{blue}{100_2}+color{blue}{10_2}+color{blue}{1_2})times1111_2\
                                    &=(1000_2times1111_2)+(100_2times1111_2)+(10_2times1111_2)+(1_2times1111_2)\
                                    &=1111000_2+111100_2+11110_2+1111_2quad(*)\
                                    end{aligned}$$



                                    In your third column from the right, where you get stuck, you have:



                                    $$begin{aligned}underbrace{1}_{text{carried}}+1+1+1&=100_2
                                    end{aligned}$$



                                    For $100$, your bit is $0$ and you can carry the rest ($10_2$). Then, in the next column, you'd have:



                                    $$begin{aligned}underbrace{10_2}_{text{carried}}+1+1+1+1&=110_2
                                    end{aligned}$$



                                    And just like before, your bit is $0$ but you carry the rest ($11_2$).



                                    If you keep on with that method, you'd get:



                                    $$begin{aligned}1111_2times1111_2&=color{red}{11100001_2}\
                                    15times15,color{green}{✔}&=225,color{green}{✔}end{aligned}$$





                                    (*) When multiplying any binary number by a power of $2$, you can just append the $0$'s of the power of $2$ onto the end of the other number. For example, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
                                    &=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
                                    &=10000000_2+10000_2+1000_2\&=10011000_2\
                                    8times38,color{green}{✔}&=152,color{green}{✔}
                                    end{aligned}$$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      As you appear to have done, you can start the calculation by splitting up one of the numbers into its constituent powers of $2$ and then distributing the multiplication:



                                      $$begin{aligned}color{blue}{1111_2}times1111_2&=(color{blue}{1000_2}+color{blue}{100_2}+color{blue}{10_2}+color{blue}{1_2})times1111_2\
                                      &=(1000_2times1111_2)+(100_2times1111_2)+(10_2times1111_2)+(1_2times1111_2)\
                                      &=1111000_2+111100_2+11110_2+1111_2quad(*)\
                                      end{aligned}$$



                                      In your third column from the right, where you get stuck, you have:



                                      $$begin{aligned}underbrace{1}_{text{carried}}+1+1+1&=100_2
                                      end{aligned}$$



                                      For $100$, your bit is $0$ and you can carry the rest ($10_2$). Then, in the next column, you'd have:



                                      $$begin{aligned}underbrace{10_2}_{text{carried}}+1+1+1+1&=110_2
                                      end{aligned}$$



                                      And just like before, your bit is $0$ but you carry the rest ($11_2$).



                                      If you keep on with that method, you'd get:



                                      $$begin{aligned}1111_2times1111_2&=color{red}{11100001_2}\
                                      15times15,color{green}{✔}&=225,color{green}{✔}end{aligned}$$





                                      (*) When multiplying any binary number by a power of $2$, you can just append the $0$'s of the power of $2$ onto the end of the other number. For example, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
                                      &=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
                                      &=10000000_2+10000_2+1000_2\&=10011000_2\
                                      8times38,color{green}{✔}&=152,color{green}{✔}
                                      end{aligned}$$






                                      share|cite|improve this answer











                                      $endgroup$



                                      As you appear to have done, you can start the calculation by splitting up one of the numbers into its constituent powers of $2$ and then distributing the multiplication:



                                      $$begin{aligned}color{blue}{1111_2}times1111_2&=(color{blue}{1000_2}+color{blue}{100_2}+color{blue}{10_2}+color{blue}{1_2})times1111_2\
                                      &=(1000_2times1111_2)+(100_2times1111_2)+(10_2times1111_2)+(1_2times1111_2)\
                                      &=1111000_2+111100_2+11110_2+1111_2quad(*)\
                                      end{aligned}$$



                                      In your third column from the right, where you get stuck, you have:



                                      $$begin{aligned}underbrace{1}_{text{carried}}+1+1+1&=100_2
                                      end{aligned}$$



                                      For $100$, your bit is $0$ and you can carry the rest ($10_2$). Then, in the next column, you'd have:



                                      $$begin{aligned}underbrace{10_2}_{text{carried}}+1+1+1+1&=110_2
                                      end{aligned}$$



                                      And just like before, your bit is $0$ but you carry the rest ($11_2$).



                                      If you keep on with that method, you'd get:



                                      $$begin{aligned}1111_2times1111_2&=color{red}{11100001_2}\
                                      15times15,color{green}{✔}&=225,color{green}{✔}end{aligned}$$





                                      (*) When multiplying any binary number by a power of $2$, you can just append the $0$'s of the power of $2$ onto the end of the other number. For example, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
                                      &=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
                                      &=10000000_2+10000_2+1000_2\&=10011000_2\
                                      8times38,color{green}{✔}&=152,color{green}{✔}
                                      end{aligned}$$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 5 '18 at 19:54

























                                      answered Dec 5 '18 at 13:49









                                      JamJam

                                      4,98221431




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