# Convert the Signed Integer Number 271 362 650 271 888 to a Signed Binary. Converting and Writing the Base Ten Decimal System Signed Integer Number as a Signed Binary Code (Written in Base Two). Detailed Explanations

## Signed integer number 271 362 650 271 888(10)converted and written as a signed binary (base 2) = ?

### 1. Divide the number repeatedly by 2:

#### We stop when we get a quotient that is equal to zero.

• division = quotient + remainder;
• 271 362 650 271 888 ÷ 2 = 135 681 325 135 944 + 0;
• 135 681 325 135 944 ÷ 2 = 67 840 662 567 972 + 0;
• 67 840 662 567 972 ÷ 2 = 33 920 331 283 986 + 0;
• 33 920 331 283 986 ÷ 2 = 16 960 165 641 993 + 0;
• 16 960 165 641 993 ÷ 2 = 8 480 082 820 996 + 1;
• 8 480 082 820 996 ÷ 2 = 4 240 041 410 498 + 0;
• 4 240 041 410 498 ÷ 2 = 2 120 020 705 249 + 0;
• 2 120 020 705 249 ÷ 2 = 1 060 010 352 624 + 1;
• 1 060 010 352 624 ÷ 2 = 530 005 176 312 + 0;
• 530 005 176 312 ÷ 2 = 265 002 588 156 + 0;
• 265 002 588 156 ÷ 2 = 132 501 294 078 + 0;
• 132 501 294 078 ÷ 2 = 66 250 647 039 + 0;
• 66 250 647 039 ÷ 2 = 33 125 323 519 + 1;
• 33 125 323 519 ÷ 2 = 16 562 661 759 + 1;
• 16 562 661 759 ÷ 2 = 8 281 330 879 + 1;
• 8 281 330 879 ÷ 2 = 4 140 665 439 + 1;
• 4 140 665 439 ÷ 2 = 2 070 332 719 + 1;
• 2 070 332 719 ÷ 2 = 1 035 166 359 + 1;
• 1 035 166 359 ÷ 2 = 517 583 179 + 1;
• 517 583 179 ÷ 2 = 258 791 589 + 1;
• 258 791 589 ÷ 2 = 129 395 794 + 1;
• 129 395 794 ÷ 2 = 64 697 897 + 0;
• 64 697 897 ÷ 2 = 32 348 948 + 1;
• 32 348 948 ÷ 2 = 16 174 474 + 0;
• 16 174 474 ÷ 2 = 8 087 237 + 0;
• 8 087 237 ÷ 2 = 4 043 618 + 1;
• 4 043 618 ÷ 2 = 2 021 809 + 0;
• 2 021 809 ÷ 2 = 1 010 904 + 1;
• 1 010 904 ÷ 2 = 505 452 + 0;
• 505 452 ÷ 2 = 252 726 + 0;
• 252 726 ÷ 2 = 126 363 + 0;
• 126 363 ÷ 2 = 63 181 + 1;
• 63 181 ÷ 2 = 31 590 + 1;
• 31 590 ÷ 2 = 15 795 + 0;
• 15 795 ÷ 2 = 7 897 + 1;
• 7 897 ÷ 2 = 3 948 + 1;
• 3 948 ÷ 2 = 1 974 + 0;
• 1 974 ÷ 2 = 987 + 0;
• 987 ÷ 2 = 493 + 1;
• 493 ÷ 2 = 246 + 1;
• 246 ÷ 2 = 123 + 0;
• 123 ÷ 2 = 61 + 1;
• 61 ÷ 2 = 30 + 1;
• 30 ÷ 2 = 15 + 0;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

## How to convert signed integers from decimal system to binary code system

### Follow the steps below to convert a signed base ten integer number to signed binary:

• 1. In a signed binary, first bit (the leftmost) is reserved for sign: 0 = positive integer number, 1 = positive integer number. If the number to be converted is negative, start with its positive version.
• 2. Divide repeatedly by 2 the positive integer number keeping track of each remainder. STOP when we get a quotient that is ZERO.
• 3. Construct the base 2 representation of the positive number, by taking all the remainders starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
• 4. Binary numbers represented in computer language have a length of 4, 8, 16, 32, 64, ... bits (power of 2) - if needed, fill in extra '0' bits in front of the base 2 number (to the left), up to the right length; this way the first bit (the leftmost one) is always '0', as for a positive representation.
• 5. To get the negative reprezentation of the number, simply switch the first bit (the leftmost one), from '0' to '1'.

### Example: convert the negative number -63 from decimal system (base ten) to signed binary code system:

• 1. Start with the positive version of the number: |-63| = 63;
• 2. Divide repeatedly 63 by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
• division = quotient + remainder
• 63 ÷ 2 = 31 + 1
• 31 ÷ 2 = 15 + 1
• 15 ÷ 2 = 7 + 1
• 7 ÷ 2 = 3 + 1
• 3 ÷ 2 = 1 + 1
• 1 ÷ 2 = 0 + 1
• 3. Construct the base 2 representation of the positive number, by taking all the remainders starting from the bottom of the list constructed above:
63(10) = 11 1111(2)
• 4. The actual length of base 2 representation number is 6, so the positive binary computer representation length of the signed binary will take in this case 8 bits (the least power of 2 higher than 6) - add extra '0's in front (to the left), up to the required length; this way the first bit (the leftmost one) is to be '0', as for a positive number:
63(10) = 0011 1111(2)
• 5. To get the negative integer number representation simply change the first bit (the leftmost), from '0' to '1':
-63(10) = 1011 1111