-12 345.123 456 789 123 456 789 123 453 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -12 345.123 456 789 123 456 789 123 453 4(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
-12 345.123 456 789 123 456 789 123 453 4(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:

|-12 345.123 456 789 123 456 789 123 453 4| = 12 345.123 456 789 123 456 789 123 453 4


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: 0.123 456 789 123 456 789 123 453 4.

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) 0.123 456 789 123 456 789 123 453 4 × 2 = 0 + 0.246 913 578 246 913 578 246 906 8;
  • 2) 0.246 913 578 246 913 578 246 906 8 × 2 = 0 + 0.493 827 156 493 827 156 493 813 6;
  • 3) 0.493 827 156 493 827 156 493 813 6 × 2 = 0 + 0.987 654 312 987 654 312 987 627 2;
  • 4) 0.987 654 312 987 654 312 987 627 2 × 2 = 1 + 0.975 308 625 975 308 625 975 254 4;
  • 5) 0.975 308 625 975 308 625 975 254 4 × 2 = 1 + 0.950 617 251 950 617 251 950 508 8;
  • 6) 0.950 617 251 950 617 251 950 508 8 × 2 = 1 + 0.901 234 503 901 234 503 901 017 6;
  • 7) 0.901 234 503 901 234 503 901 017 6 × 2 = 1 + 0.802 469 007 802 469 007 802 035 2;
  • 8) 0.802 469 007 802 469 007 802 035 2 × 2 = 1 + 0.604 938 015 604 938 015 604 070 4;
  • 9) 0.604 938 015 604 938 015 604 070 4 × 2 = 1 + 0.209 876 031 209 876 031 208 140 8;
  • 10) 0.209 876 031 209 876 031 208 140 8 × 2 = 0 + 0.419 752 062 419 752 062 416 281 6;
  • 11) 0.419 752 062 419 752 062 416 281 6 × 2 = 0 + 0.839 504 124 839 504 124 832 563 2;
  • 12) 0.839 504 124 839 504 124 832 563 2 × 2 = 1 + 0.679 008 249 679 008 249 665 126 4;
  • 13) 0.679 008 249 679 008 249 665 126 4 × 2 = 1 + 0.358 016 499 358 016 499 330 252 8;
  • 14) 0.358 016 499 358 016 499 330 252 8 × 2 = 0 + 0.716 032 998 716 032 998 660 505 6;
  • 15) 0.716 032 998 716 032 998 660 505 6 × 2 = 1 + 0.432 065 997 432 065 997 321 011 2;
  • 16) 0.432 065 997 432 065 997 321 011 2 × 2 = 0 + 0.864 131 994 864 131 994 642 022 4;
  • 17) 0.864 131 994 864 131 994 642 022 4 × 2 = 1 + 0.728 263 989 728 263 989 284 044 8;
  • 18) 0.728 263 989 728 263 989 284 044 8 × 2 = 1 + 0.456 527 979 456 527 978 568 089 6;
  • 19) 0.456 527 979 456 527 978 568 089 6 × 2 = 0 + 0.913 055 958 913 055 957 136 179 2;
  • 20) 0.913 055 958 913 055 957 136 179 2 × 2 = 1 + 0.826 111 917 826 111 914 272 358 4;
  • 21) 0.826 111 917 826 111 914 272 358 4 × 2 = 1 + 0.652 223 835 652 223 828 544 716 8;
  • 22) 0.652 223 835 652 223 828 544 716 8 × 2 = 1 + 0.304 447 671 304 447 657 089 433 6;
  • 23) 0.304 447 671 304 447 657 089 433 6 × 2 = 0 + 0.608 895 342 608 895 314 178 867 2;
  • 24) 0.608 895 342 608 895 314 178 867 2 × 2 = 1 + 0.217 790 685 217 790 628 357 734 4;
  • 25) 0.217 790 685 217 790 628 357 734 4 × 2 = 0 + 0.435 581 370 435 581 256 715 468 8;
  • 26) 0.435 581 370 435 581 256 715 468 8 × 2 = 0 + 0.871 162 740 871 162 513 430 937 6;
  • 27) 0.871 162 740 871 162 513 430 937 6 × 2 = 1 + 0.742 325 481 742 325 026 861 875 2;
  • 28) 0.742 325 481 742 325 026 861 875 2 × 2 = 1 + 0.484 650 963 484 650 053 723 750 4;
  • 29) 0.484 650 963 484 650 053 723 750 4 × 2 = 0 + 0.969 301 926 969 300 107 447 500 8;
  • 30) 0.969 301 926 969 300 107 447 500 8 × 2 = 1 + 0.938 603 853 938 600 214 895 001 6;
  • 31) 0.938 603 853 938 600 214 895 001 6 × 2 = 1 + 0.877 207 707 877 200 429 790 003 2;
  • 32) 0.877 207 707 877 200 429 790 003 2 × 2 = 1 + 0.754 415 415 754 400 859 580 006 4;
  • 33) 0.754 415 415 754 400 859 580 006 4 × 2 = 1 + 0.508 830 831 508 801 719 160 012 8;
  • 34) 0.508 830 831 508 801 719 160 012 8 × 2 = 1 + 0.017 661 663 017 603 438 320 025 6;
  • 35) 0.017 661 663 017 603 438 320 025 6 × 2 = 0 + 0.035 323 326 035 206 876 640 051 2;
  • 36) 0.035 323 326 035 206 876 640 051 2 × 2 = 0 + 0.070 646 652 070 413 753 280 102 4;
  • 37) 0.070 646 652 070 413 753 280 102 4 × 2 = 0 + 0.141 293 304 140 827 506 560 204 8;
  • 38) 0.141 293 304 140 827 506 560 204 8 × 2 = 0 + 0.282 586 608 281 655 013 120 409 6;
  • 39) 0.282 586 608 281 655 013 120 409 6 × 2 = 0 + 0.565 173 216 563 310 026 240 819 2;
  • 40) 0.565 173 216 563 310 026 240 819 2 × 2 = 1 + 0.130 346 433 126 620 052 481 638 4;
  • 41) 0.130 346 433 126 620 052 481 638 4 × 2 = 0 + 0.260 692 866 253 240 104 963 276 8;
  • 42) 0.260 692 866 253 240 104 963 276 8 × 2 = 0 + 0.521 385 732 506 480 209 926 553 6;
  • 43) 0.521 385 732 506 480 209 926 553 6 × 2 = 1 + 0.042 771 465 012 960 419 853 107 2;
  • 44) 0.042 771 465 012 960 419 853 107 2 × 2 = 0 + 0.085 542 930 025 920 839 706 214 4;
  • 45) 0.085 542 930 025 920 839 706 214 4 × 2 = 0 + 0.171 085 860 051 841 679 412 428 8;
  • 46) 0.171 085 860 051 841 679 412 428 8 × 2 = 0 + 0.342 171 720 103 683 358 824 857 6;
  • 47) 0.342 171 720 103 683 358 824 857 6 × 2 = 0 + 0.684 343 440 207 366 717 649 715 2;
  • 48) 0.684 343 440 207 366 717 649 715 2 × 2 = 1 + 0.368 686 880 414 733 435 299 430 4;
  • 49) 0.368 686 880 414 733 435 299 430 4 × 2 = 0 + 0.737 373 760 829 466 870 598 860 8;
  • 50) 0.737 373 760 829 466 870 598 860 8 × 2 = 1 + 0.474 747 521 658 933 741 197 721 6;
  • 51) 0.474 747 521 658 933 741 197 721 6 × 2 = 0 + 0.949 495 043 317 867 482 395 443 2;
  • 52) 0.949 495 043 317 867 482 395 443 2 × 2 = 1 + 0.898 990 086 635 734 964 790 886 4;
  • 53) 0.898 990 086 635 734 964 790 886 4 × 2 = 1 + 0.797 980 173 271 469 929 581 772 8;

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:


0.123 456 789 123 456 789 123 453 4(10) =

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

6. Positive number before normalization:

12 345.123 456 789 123 456 789 123 453 4(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:


12 345.123 456 789 123 456 789 123 453 4(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 -12 345.123 456 789 123 456 789 123 453 4 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 -12 345.123 456 789 123 456 789 123 446 9 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