Show me the very working to this question that even a child in basic one will understand. 1111011 base 2 *1101base2

To multiply binary numbers, we use a process similar to long multiplication with decimal numbers.

Here are the rules for binary multiplication:

• $0 \times 0 = 0$
• $0 \times 1 = 0$
• $1 \times 0 = 0$
• $1 \times 1 = 1$

And for binary addition:

• $0 + 0 = 0$
• $0 + 1 = 1$
• $1 + 0 = 1$
• $1 + 1 = 10_2$ (which means 0 with a carry of 1)
• $1 + 1 + 1 = 11_2$ (which means 1 with a carry of 1)

We need to calculate $1111011_2 \times 1101_2$.

Step 1: Write down the numbers for long multiplication.

Step 2: Multiply the top number ($1111011_2$) by each digit of the bottom number ($1101_2$), starting from the rightmost digit. Remember to shift each new product to the left.

• Multiply by the rightmost 1: $1111011_2 \times 1_2 = 1111011_2$
• Multiply by the next digit 0: $1111011_2 \times 0_2 = 0000000_2$. Shift this one place to the left: $00000000_2$
• Multiply by the next digit 1: $1111011_2 \times 1_2 = 1111011_2$. Shift this two places to the left: $111101100_2$
• Multiply by the leftmost digit 1: $1111011_2 \times 1_2 = 1111011_2$. Shift this three places to the left: $1111011000_2$

Now, arrange these partial products for addition:

Step 3: Add the partial products together, column by column, from right to left. Remember to carry over any $1$s when the sum is $10_2$ or $11_2$.

Let's add them carefully:

Let's break down the addition column by column:

• Rightmost column: $1+0+0+0 = 1$.
• Second column: $1+0+0+0 = 1$.
• Third column: $0+0+1+0 = 1$.
• Fourth column: $1+0+1+0 = 10_2$. Write 0, carry 1.
• Fifth column: $1+0+0+1 + (\text{carry } 1) = 1+0+0+1+1 = 11_2$. Write 1, carry 1.
• Sixth column: $1+0+1+1 + (\text{carry } 1) = 1+0+1+1+1 = 100_2$. Write 0, carry $10_2$ (which is $2_{10}$). This is $1+1+1+1 = 100_2$. Write 0, carry $10_2$. Let's re-evaluate this step. It's easier to add two numbers at a time.

Let's add the partial products step-by-step:

First, add the first two partial products:

Next, add the third partial product ($111101100_2$) to the result:

Let's verify this addition:
$1+0=1$
$1+0=1$
$0+1=1$
$1+1=0$ (carry 1)
$1+0+1 (\text{carry})=0$ (carry 1)
$1+1+1 (\text{carry})=1$ (carry 1)
$1+1 (\text{carry})=0$ (carry 1)
$0+1 (\text{carry})=1$
$0+0=0$
$0+0=0$
Result: $100110011_2$. (This is the sum of the first three lines)

Finally, add the fourth partial product ($1111011000_2$) to this new sum:

Let's verify this final addition:
$1+0=1$
$1+0=1$
$0+0=0$
$0+1=1$
$1+1=0$ (carry 1)
$1+0+1 (\text{carry})=0$ (carry 1)
$0+1+1 (\text{carry})=0$ (carry 1)
$0+1+1 (\text{carry})=0$ (carry 1)
$1+1 (\text{carry})=0$ (carry 1)
$0+1 (\text{carry})=1$
$0+0=0$
Result: $100000100111_2$.

Let's re-do the final addition one more time, very carefully.
Sum of first three: $100110011_2$
Fourth partial product: $1111011000_2$

Align them:

  000100110011  (Sum of first three, padded)
+ 111101100000  (Fourth partial product, padded)
--------------

Adding column by column from right to left:

1. $1+0 = 1$
2. $1+0 = 1$
3. $0+0 = 0$
4. $0+1 = 1$
5. $1+1 = 0$ (carry 1)
6. $0+0+1 (\text{carry}) = 1$
7. $0+1 = 1$
8. $1+1 = 0$ (carry 1)
9. $0+1+1 (\text{carry}) = 0$ (carry 1)
10. $0+1+1 (\text{carry}) = 0$ (carry 1)
11. $0+1 (\text{carry}) = 1$

The final result is $100001101011_2$.

Let's double check with decimal conversion:
$1111011_2 = 1 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 64+32+16+8+0+2+1 = 123_{10}$
$1101_2 = 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = 8+4+0+1 = 13_{10}$
$123_{10} \times 13_{10} = 1599_{10}$

Now convert $1599_{10}$ to binary:
$1599 \div 2 = 799$ R 1
$799 \div 2 = 399$ R 1
$399 \div 2 = 199$ R 1
$199 \div 2 = 99$ R 1
$99 \div 2 = 49$ R 1
$49 \div 2 = 24$ R 1
$24 \div 2 = 12$ R 0
$12 \div 2 = 6$ R 0
$6 \div 2 = 3$ R 0
$3 \div 2 = 1$ R 1
$1 \div 2 = 0$ R 1
Reading the remainders from bottom up: $11000111111_2$.

My binary multiplication result $100001101011_2$ is still not matching $11000111111_2$.
There is a consistent error in my binary addition. Let's re-do the addition of the four lines using a different method, or just be extremely careful.

Let's write the four lines again:
Line 1: 0000001111011
Line 2: 0000000000000 (shifted)
Line 3: 0000111101100 (shifted)
Line 4: 0111101100000 (shifted)

Let's add them column by column, keeping track of carries.

  Carries:  1 1 1 1 1 1 1 1 1 0 0 0 0
            0 0 0 0 0 0 1 1 1 1 0 1 1  (Line 1)
            0 0 0 0 0 0 0 0 0 0 0 0 0  (Line 2)
            0 0 0 0 1 1 1 1 0 1 1 0 0  (Line 3)
          + 0 1 1 1 1 0 1 1 0 0 0 0 0  (Line 4)
          ---------------------------
            1 0 0 1 1 0 0 1 1 1 1 1 1

Let's trace the carries and sums:

1. Rightmost column: $1+0+0+0 = 1$. Sum: 1. Carry: 0.
2. Next column: $1+0+0+0 = 1$. Sum: 1. Carry: 0.
3. Next column: $0+0+1+0 = 1$. Sum: 1. Carry: 0.
4. Next column: $1+0+1+0 = 10_2$. Sum: 0. Carry: 1.
5. Next column: $1+0+0+1 + (\text{carry } 1) = 1+0+0+1+1 = 11_2$. Sum: 1. Carry: 1.
6. Next column: $1+0+1+1 + (\text{carry } 1) = 1+0+1+1+1 = 100_2$. Sum: 0. Carry: $10_2$ (which is $2_{10}$). This is where the "basic one" explanation gets tricky. $1+1+1+1 = 100_2$. So, write 0, carry $10_2$. Let's write the carries as $1$s and $0$s. $1+1+1+1 = 100_2$. So, 0 in the current column, and $10_2$ (which is $2_{10}$) is carried over. This means two $1$s are carried to the next column.

Let's use a simpler method for addition, adding two numbers at a time.

1. $1111011_2$
2. $00000000_2$
3. $111101100_2$
4. $1111011000_2$

Sum of (1) and (2): $1111011_2$

Sum of ($1111011_2$) and (3) ($111101100_2$):

  0001111011
+ 00111101100
--------------
  01011001111  (This was the result I got earlier, $687_{10}$, which was wrong)

Let's re-do this specific addition:
0001111011
00111101100

$1+0 = 1$
$1+0 = 1$
$0+1 = 1$
$1+1 = 0$ (carry 1)
$1+0+1 (\text{carry}) = 0$ (carry 1)
$1+1+1 (\text{carry}) = 1$ (carry 1)
$1+1 (\text{carry}) = 0$ (carry 1)
$0+1 (\text{carry}) = 1$
$0+0 = 0$
$0+0 = 0$
Result: $01010010111_2$. (This is $663_{10}$, still not $615_{10}$)

The error is in my binary addition. Let's use the standard long addition format with carries written above.

Adding $1111011_2$ and $111101100_2$:

  Carries: