Multiply $1111_2$ with $1111_2$
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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:
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
$endgroup$
add a comment |
$begingroup$
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:
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
$endgroup$
1
$begingroup$
Add enough columns to the left...
$endgroup$
– Mauro ALLEGRANZA
Dec 5 '18 at 12:24
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If you get 4, write 4 mod 2 (0) and report 4/2 (2, integer division). Basically same procedure as in base 10
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– Damien
Dec 5 '18 at 12:26
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You can write $0$ and carry over the $10$. The rest is similar.
$endgroup$
– AryanSonwatikar
Dec 5 '18 at 12:31
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@MauroALLEGRANZA And don't forget leaving some space on top,
$endgroup$
– CopyPasteIt
Dec 5 '18 at 15:39
add a comment |
$begingroup$
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:
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
$endgroup$
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:
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
arithmetic binary
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
add a comment |
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
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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: $1111_2times1111_2=11100001$.
$endgroup$
add a comment |
$begingroup$
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 :
$$
begin{matrix}
color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
end{matrix}
$$
$endgroup$
add a comment |
$begingroup$
$$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
=2^5cdot111_2+1_2=11100000+1=11100001
$$
$endgroup$
add a comment |
$begingroup$
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:
This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.
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$begingroup$
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
add a comment |
$begingroup$
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, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
&=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
&=10000000_2+10000_2+1000_2\&=10011000_2\
8times38,color{green}{✔}&=152,color{green}{✔}
end{aligned}$$
$endgroup$
add a comment |
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5 Answers
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active
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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: $1111_2times1111_2=11100001$.
$endgroup$
add a comment |
$begingroup$
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: $1111_2times1111_2=11100001$.
$endgroup$
add a comment |
$begingroup$
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: $1111_2times1111_2=11100001$.
$endgroup$
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: $1111_2times1111_2=11100001$.
answered Dec 5 '18 at 12:46
farruhotafarruhota
20.6k2740
20.6k2740
add a comment |
add a comment |
$begingroup$
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 :
$$
begin{matrix}
color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
end{matrix}
$$
$endgroup$
add a comment |
$begingroup$
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 :
$$
begin{matrix}
color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
end{matrix}
$$
$endgroup$
add a comment |
$begingroup$
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 :
$$
begin{matrix}
color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
end{matrix}
$$
$endgroup$
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 :
$$
begin{matrix}
color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
end{matrix}
$$
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
add a comment |
add a comment |
$begingroup$
$$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
=2^5cdot111_2+1_2=11100000+1=11100001
$$
$endgroup$
add a comment |
$begingroup$
$$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
=2^5cdot111_2+1_2=11100000+1=11100001
$$
$endgroup$
add a comment |
$begingroup$
$$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
=2^5cdot111_2+1_2=11100000+1=11100001
$$
$endgroup$
$$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
=2^5cdot111_2+1_2=11100000+1=11100001
$$
answered Dec 5 '18 at 12:27
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385163
63.4k1385163
add a comment |
add a comment |
$begingroup$
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:
This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.
$endgroup$
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:
This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.
edited Dec 5 '18 at 15:26
answered Dec 5 '18 at 15:08
CopyPasteItCopyPasteIt
4,2031628
4,2031628
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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
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My update came at exact time as your comment!
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– 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.
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– 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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
&=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
&=10000000_2+10000_2+1000_2\&=10011000_2\
8times38,color{green}{✔}&=152,color{green}{✔}
end{aligned}$$
$endgroup$
add a comment |
$begingroup$
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, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
&=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
&=10000000_2+10000_2+1000_2\&=10011000_2\
8times38,color{green}{✔}&=152,color{green}{✔}
end{aligned}$$
$endgroup$
add a comment |
$begingroup$
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, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
&=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
&=10000000_2+10000_2+1000_2\&=10011000_2\
8times38,color{green}{✔}&=152,color{green}{✔}
end{aligned}$$
$endgroup$
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, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
&=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
&=10000000_2+10000_2+1000_2\&=10011000_2\
8times38,color{green}{✔}&=152,color{green}{✔}
end{aligned}$$
edited Dec 5 '18 at 19:54
answered Dec 5 '18 at 13:49
JamJam
4,98221431
4,98221431
add a comment |
add a comment |
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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