-12 345.123 456 789 123 456 789 123 463 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -12 345.123 456 789 123 456 789 123 463 2(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-12 345.123 456 789 123 456 789 123 463 2(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-12 345.123 456 789 123 456 789 123 463 2| = 12 345.123 456 789 123 456 789 123 463 2


2. First, convert to binary (in base 2) the integer part: 12 345.
Divide the number repeatedly by 2.

Keep track of each remainder.

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


  • division = quotient + remainder;
  • 12 345 ÷ 2 = 6 172 + 1;
  • 6 172 ÷ 2 = 3 086 + 0;
  • 3 086 ÷ 2 = 1 543 + 0;
  • 1 543 ÷ 2 = 771 + 1;
  • 771 ÷ 2 = 385 + 1;
  • 385 ÷ 2 = 192 + 1;
  • 192 ÷ 2 = 96 + 0;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

12 345(10) =

11 0000 0011 1001(2)


4. Convert to binary (base 2) the fractional part: 0.123 456 789 123 456 789 123 463 2.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.123 456 789 123 456 789 123 463 2 × 2 = 0 + 0.246 913 578 246 913 578 246 926 4;
  • 2) 0.246 913 578 246 913 578 246 926 4 × 2 = 0 + 0.493 827 156 493 827 156 493 852 8;
  • 3) 0.493 827 156 493 827 156 493 852 8 × 2 = 0 + 0.987 654 312 987 654 312 987 705 6;
  • 4) 0.987 654 312 987 654 312 987 705 6 × 2 = 1 + 0.975 308 625 975 308 625 975 411 2;
  • 5) 0.975 308 625 975 308 625 975 411 2 × 2 = 1 + 0.950 617 251 950 617 251 950 822 4;
  • 6) 0.950 617 251 950 617 251 950 822 4 × 2 = 1 + 0.901 234 503 901 234 503 901 644 8;
  • 7) 0.901 234 503 901 234 503 901 644 8 × 2 = 1 + 0.802 469 007 802 469 007 803 289 6;
  • 8) 0.802 469 007 802 469 007 803 289 6 × 2 = 1 + 0.604 938 015 604 938 015 606 579 2;
  • 9) 0.604 938 015 604 938 015 606 579 2 × 2 = 1 + 0.209 876 031 209 876 031 213 158 4;
  • 10) 0.209 876 031 209 876 031 213 158 4 × 2 = 0 + 0.419 752 062 419 752 062 426 316 8;
  • 11) 0.419 752 062 419 752 062 426 316 8 × 2 = 0 + 0.839 504 124 839 504 124 852 633 6;
  • 12) 0.839 504 124 839 504 124 852 633 6 × 2 = 1 + 0.679 008 249 679 008 249 705 267 2;
  • 13) 0.679 008 249 679 008 249 705 267 2 × 2 = 1 + 0.358 016 499 358 016 499 410 534 4;
  • 14) 0.358 016 499 358 016 499 410 534 4 × 2 = 0 + 0.716 032 998 716 032 998 821 068 8;
  • 15) 0.716 032 998 716 032 998 821 068 8 × 2 = 1 + 0.432 065 997 432 065 997 642 137 6;
  • 16) 0.432 065 997 432 065 997 642 137 6 × 2 = 0 + 0.864 131 994 864 131 995 284 275 2;
  • 17) 0.864 131 994 864 131 995 284 275 2 × 2 = 1 + 0.728 263 989 728 263 990 568 550 4;
  • 18) 0.728 263 989 728 263 990 568 550 4 × 2 = 1 + 0.456 527 979 456 527 981 137 100 8;
  • 19) 0.456 527 979 456 527 981 137 100 8 × 2 = 0 + 0.913 055 958 913 055 962 274 201 6;
  • 20) 0.913 055 958 913 055 962 274 201 6 × 2 = 1 + 0.826 111 917 826 111 924 548 403 2;
  • 21) 0.826 111 917 826 111 924 548 403 2 × 2 = 1 + 0.652 223 835 652 223 849 096 806 4;
  • 22) 0.652 223 835 652 223 849 096 806 4 × 2 = 1 + 0.304 447 671 304 447 698 193 612 8;
  • 23) 0.304 447 671 304 447 698 193 612 8 × 2 = 0 + 0.608 895 342 608 895 396 387 225 6;
  • 24) 0.608 895 342 608 895 396 387 225 6 × 2 = 1 + 0.217 790 685 217 790 792 774 451 2;
  • 25) 0.217 790 685 217 790 792 774 451 2 × 2 = 0 + 0.435 581 370 435 581 585 548 902 4;
  • 26) 0.435 581 370 435 581 585 548 902 4 × 2 = 0 + 0.871 162 740 871 163 171 097 804 8;
  • 27) 0.871 162 740 871 163 171 097 804 8 × 2 = 1 + 0.742 325 481 742 326 342 195 609 6;
  • 28) 0.742 325 481 742 326 342 195 609 6 × 2 = 1 + 0.484 650 963 484 652 684 391 219 2;
  • 29) 0.484 650 963 484 652 684 391 219 2 × 2 = 0 + 0.969 301 926 969 305 368 782 438 4;
  • 30) 0.969 301 926 969 305 368 782 438 4 × 2 = 1 + 0.938 603 853 938 610 737 564 876 8;
  • 31) 0.938 603 853 938 610 737 564 876 8 × 2 = 1 + 0.877 207 707 877 221 475 129 753 6;
  • 32) 0.877 207 707 877 221 475 129 753 6 × 2 = 1 + 0.754 415 415 754 442 950 259 507 2;
  • 33) 0.754 415 415 754 442 950 259 507 2 × 2 = 1 + 0.508 830 831 508 885 900 519 014 4;
  • 34) 0.508 830 831 508 885 900 519 014 4 × 2 = 1 + 0.017 661 663 017 771 801 038 028 8;
  • 35) 0.017 661 663 017 771 801 038 028 8 × 2 = 0 + 0.035 323 326 035 543 602 076 057 6;
  • 36) 0.035 323 326 035 543 602 076 057 6 × 2 = 0 + 0.070 646 652 071 087 204 152 115 2;
  • 37) 0.070 646 652 071 087 204 152 115 2 × 2 = 0 + 0.141 293 304 142 174 408 304 230 4;
  • 38) 0.141 293 304 142 174 408 304 230 4 × 2 = 0 + 0.282 586 608 284 348 816 608 460 8;
  • 39) 0.282 586 608 284 348 816 608 460 8 × 2 = 0 + 0.565 173 216 568 697 633 216 921 6;
  • 40) 0.565 173 216 568 697 633 216 921 6 × 2 = 1 + 0.130 346 433 137 395 266 433 843 2;
  • 41) 0.130 346 433 137 395 266 433 843 2 × 2 = 0 + 0.260 692 866 274 790 532 867 686 4;
  • 42) 0.260 692 866 274 790 532 867 686 4 × 2 = 0 + 0.521 385 732 549 581 065 735 372 8;
  • 43) 0.521 385 732 549 581 065 735 372 8 × 2 = 1 + 0.042 771 465 099 162 131 470 745 6;
  • 44) 0.042 771 465 099 162 131 470 745 6 × 2 = 0 + 0.085 542 930 198 324 262 941 491 2;
  • 45) 0.085 542 930 198 324 262 941 491 2 × 2 = 0 + 0.171 085 860 396 648 525 882 982 4;
  • 46) 0.171 085 860 396 648 525 882 982 4 × 2 = 0 + 0.342 171 720 793 297 051 765 964 8;
  • 47) 0.342 171 720 793 297 051 765 964 8 × 2 = 0 + 0.684 343 441 586 594 103 531 929 6;
  • 48) 0.684 343 441 586 594 103 531 929 6 × 2 = 1 + 0.368 686 883 173 188 207 063 859 2;
  • 49) 0.368 686 883 173 188 207 063 859 2 × 2 = 0 + 0.737 373 766 346 376 414 127 718 4;
  • 50) 0.737 373 766 346 376 414 127 718 4 × 2 = 1 + 0.474 747 532 692 752 828 255 436 8;
  • 51) 0.474 747 532 692 752 828 255 436 8 × 2 = 0 + 0.949 495 065 385 505 656 510 873 6;
  • 52) 0.949 495 065 385 505 656 510 873 6 × 2 = 1 + 0.898 990 130 771 011 313 021 747 2;
  • 53) 0.898 990 130 771 011 313 021 747 2 × 2 = 1 + 0.797 980 261 542 022 626 043 494 4;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.123 456 789 123 456 789 123 463 2(10) =

0.0001 1111 1001 1010 1101 1101 0011 0111 1100 0001 0010 0001 0101 1(2)

6. Positive number before normalization:

12 345.123 456 789 123 456 789 123 463 2(10) =

11 0000 0011 1001.0001 1111 1001 1010 1101 1101 0011 0111 1100 0001 0010 0001 0101 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the left, so that only one non zero digit remains to the left of it:


12 345.123 456 789 123 456 789 123 463 2(10) =

11 0000 0011 1001.0001 1111 1001 1010 1101 1101 0011 0111 1100 0001 0010 0001 0101 1(2) =

11 0000 0011 1001.0001 1111 1001 1010 1101 1101 0011 0111 1100 0001 0010 0001 0101 1(2) × 20 =

1.1000 0001 1100 1000 1111 1100 1101 0110 1110 1001 1011 1110 0000 1001 0000 1010 11(2) × 213


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): 13

Mantissa (not normalized):
1.1000 0001 1100 1000 1111 1100 1101 0110 1110 1001 1011 1110 0000 1001 0000 1010 11


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

13 + 2(11-1) - 1 =

(13 + 1 023)(10) =

1 036(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 036 ÷ 2 = 518 + 0;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1036(10) =

100 0000 1100(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1000 0001 1100 1000 1111 1100 1101 0110 1110 1001 1011 1110 0000 10 0100 0010 1011 =

1000 0001 1100 1000 1111 1100 1101 0110 1110 1001 1011 1110 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 1100

Mantissa (52 bits) =
1000 0001 1100 1000 1111 1100 1101 0110 1110 1001 1011 1110 0000


Decimal number -12 345.123 456 789 123 456 789 123 463 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 1100 - 1000 0001 1100 1000 1111 1100 1101 0110 1110 1001 1011 1110 0000

» Convert decimal -12 345.123 456 789 123 456 789 123 453 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, 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. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100