-12 345.123 456 789 123 456 789 123 466 02 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -12 345.123 456 789 123 456 789 123 466 02(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
-12 345.123 456 789 123 456 789 123 466 02(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:

|-12 345.123 456 789 123 456 789 123 466 02| = 12 345.123 456 789 123 456 789 123 466 02


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: 0.123 456 789 123 456 789 123 466 02.

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) 0.123 456 789 123 456 789 123 466 02 × 2 = 0 + 0.246 913 578 246 913 578 246 932 04;
  • 2) 0.246 913 578 246 913 578 246 932 04 × 2 = 0 + 0.493 827 156 493 827 156 493 864 08;
  • 3) 0.493 827 156 493 827 156 493 864 08 × 2 = 0 + 0.987 654 312 987 654 312 987 728 16;
  • 4) 0.987 654 312 987 654 312 987 728 16 × 2 = 1 + 0.975 308 625 975 308 625 975 456 32;
  • 5) 0.975 308 625 975 308 625 975 456 32 × 2 = 1 + 0.950 617 251 950 617 251 950 912 64;
  • 6) 0.950 617 251 950 617 251 950 912 64 × 2 = 1 + 0.901 234 503 901 234 503 901 825 28;
  • 7) 0.901 234 503 901 234 503 901 825 28 × 2 = 1 + 0.802 469 007 802 469 007 803 650 56;
  • 8) 0.802 469 007 802 469 007 803 650 56 × 2 = 1 + 0.604 938 015 604 938 015 607 301 12;
  • 9) 0.604 938 015 604 938 015 607 301 12 × 2 = 1 + 0.209 876 031 209 876 031 214 602 24;
  • 10) 0.209 876 031 209 876 031 214 602 24 × 2 = 0 + 0.419 752 062 419 752 062 429 204 48;
  • 11) 0.419 752 062 419 752 062 429 204 48 × 2 = 0 + 0.839 504 124 839 504 124 858 408 96;
  • 12) 0.839 504 124 839 504 124 858 408 96 × 2 = 1 + 0.679 008 249 679 008 249 716 817 92;
  • 13) 0.679 008 249 679 008 249 716 817 92 × 2 = 1 + 0.358 016 499 358 016 499 433 635 84;
  • 14) 0.358 016 499 358 016 499 433 635 84 × 2 = 0 + 0.716 032 998 716 032 998 867 271 68;
  • 15) 0.716 032 998 716 032 998 867 271 68 × 2 = 1 + 0.432 065 997 432 065 997 734 543 36;
  • 16) 0.432 065 997 432 065 997 734 543 36 × 2 = 0 + 0.864 131 994 864 131 995 469 086 72;
  • 17) 0.864 131 994 864 131 995 469 086 72 × 2 = 1 + 0.728 263 989 728 263 990 938 173 44;
  • 18) 0.728 263 989 728 263 990 938 173 44 × 2 = 1 + 0.456 527 979 456 527 981 876 346 88;
  • 19) 0.456 527 979 456 527 981 876 346 88 × 2 = 0 + 0.913 055 958 913 055 963 752 693 76;
  • 20) 0.913 055 958 913 055 963 752 693 76 × 2 = 1 + 0.826 111 917 826 111 927 505 387 52;
  • 21) 0.826 111 917 826 111 927 505 387 52 × 2 = 1 + 0.652 223 835 652 223 855 010 775 04;
  • 22) 0.652 223 835 652 223 855 010 775 04 × 2 = 1 + 0.304 447 671 304 447 710 021 550 08;
  • 23) 0.304 447 671 304 447 710 021 550 08 × 2 = 0 + 0.608 895 342 608 895 420 043 100 16;
  • 24) 0.608 895 342 608 895 420 043 100 16 × 2 = 1 + 0.217 790 685 217 790 840 086 200 32;
  • 25) 0.217 790 685 217 790 840 086 200 32 × 2 = 0 + 0.435 581 370 435 581 680 172 400 64;
  • 26) 0.435 581 370 435 581 680 172 400 64 × 2 = 0 + 0.871 162 740 871 163 360 344 801 28;
  • 27) 0.871 162 740 871 163 360 344 801 28 × 2 = 1 + 0.742 325 481 742 326 720 689 602 56;
  • 28) 0.742 325 481 742 326 720 689 602 56 × 2 = 1 + 0.484 650 963 484 653 441 379 205 12;
  • 29) 0.484 650 963 484 653 441 379 205 12 × 2 = 0 + 0.969 301 926 969 306 882 758 410 24;
  • 30) 0.969 301 926 969 306 882 758 410 24 × 2 = 1 + 0.938 603 853 938 613 765 516 820 48;
  • 31) 0.938 603 853 938 613 765 516 820 48 × 2 = 1 + 0.877 207 707 877 227 531 033 640 96;
  • 32) 0.877 207 707 877 227 531 033 640 96 × 2 = 1 + 0.754 415 415 754 455 062 067 281 92;
  • 33) 0.754 415 415 754 455 062 067 281 92 × 2 = 1 + 0.508 830 831 508 910 124 134 563 84;
  • 34) 0.508 830 831 508 910 124 134 563 84 × 2 = 1 + 0.017 661 663 017 820 248 269 127 68;
  • 35) 0.017 661 663 017 820 248 269 127 68 × 2 = 0 + 0.035 323 326 035 640 496 538 255 36;
  • 36) 0.035 323 326 035 640 496 538 255 36 × 2 = 0 + 0.070 646 652 071 280 993 076 510 72;
  • 37) 0.070 646 652 071 280 993 076 510 72 × 2 = 0 + 0.141 293 304 142 561 986 153 021 44;
  • 38) 0.141 293 304 142 561 986 153 021 44 × 2 = 0 + 0.282 586 608 285 123 972 306 042 88;
  • 39) 0.282 586 608 285 123 972 306 042 88 × 2 = 0 + 0.565 173 216 570 247 944 612 085 76;
  • 40) 0.565 173 216 570 247 944 612 085 76 × 2 = 1 + 0.130 346 433 140 495 889 224 171 52;
  • 41) 0.130 346 433 140 495 889 224 171 52 × 2 = 0 + 0.260 692 866 280 991 778 448 343 04;
  • 42) 0.260 692 866 280 991 778 448 343 04 × 2 = 0 + 0.521 385 732 561 983 556 896 686 08;
  • 43) 0.521 385 732 561 983 556 896 686 08 × 2 = 1 + 0.042 771 465 123 967 113 793 372 16;
  • 44) 0.042 771 465 123 967 113 793 372 16 × 2 = 0 + 0.085 542 930 247 934 227 586 744 32;
  • 45) 0.085 542 930 247 934 227 586 744 32 × 2 = 0 + 0.171 085 860 495 868 455 173 488 64;
  • 46) 0.171 085 860 495 868 455 173 488 64 × 2 = 0 + 0.342 171 720 991 736 910 346 977 28;
  • 47) 0.342 171 720 991 736 910 346 977 28 × 2 = 0 + 0.684 343 441 983 473 820 693 954 56;
  • 48) 0.684 343 441 983 473 820 693 954 56 × 2 = 1 + 0.368 686 883 966 947 641 387 909 12;
  • 49) 0.368 686 883 966 947 641 387 909 12 × 2 = 0 + 0.737 373 767 933 895 282 775 818 24;
  • 50) 0.737 373 767 933 895 282 775 818 24 × 2 = 1 + 0.474 747 535 867 790 565 551 636 48;
  • 51) 0.474 747 535 867 790 565 551 636 48 × 2 = 0 + 0.949 495 071 735 581 131 103 272 96;
  • 52) 0.949 495 071 735 581 131 103 272 96 × 2 = 1 + 0.898 990 143 471 162 262 206 545 92;
  • 53) 0.898 990 143 471 162 262 206 545 92 × 2 = 1 + 0.797 980 286 942 324 524 413 091 84;

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:


0.123 456 789 123 456 789 123 466 02(10) =

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

6. Positive number before normalization:

12 345.123 456 789 123 456 789 123 466 02(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:


12 345.123 456 789 123 456 789 123 466 02(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 -12 345.123 456 789 123 456 789 123 466 02 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 -12 345.123 456 789 123 456 789 123 465 06 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