-12 345.123 456 789 123 456 789 123 446 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -12 345.123 456 789 123 456 789 123 446 9(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
-12 345.123 456 789 123 456 789 123 446 9(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:

|-12 345.123 456 789 123 456 789 123 446 9| = 12 345.123 456 789 123 456 789 123 446 9


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: 0.123 456 789 123 456 789 123 446 9.

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) 0.123 456 789 123 456 789 123 446 9 × 2 = 0 + 0.246 913 578 246 913 578 246 893 8;
  • 2) 0.246 913 578 246 913 578 246 893 8 × 2 = 0 + 0.493 827 156 493 827 156 493 787 6;
  • 3) 0.493 827 156 493 827 156 493 787 6 × 2 = 0 + 0.987 654 312 987 654 312 987 575 2;
  • 4) 0.987 654 312 987 654 312 987 575 2 × 2 = 1 + 0.975 308 625 975 308 625 975 150 4;
  • 5) 0.975 308 625 975 308 625 975 150 4 × 2 = 1 + 0.950 617 251 950 617 251 950 300 8;
  • 6) 0.950 617 251 950 617 251 950 300 8 × 2 = 1 + 0.901 234 503 901 234 503 900 601 6;
  • 7) 0.901 234 503 901 234 503 900 601 6 × 2 = 1 + 0.802 469 007 802 469 007 801 203 2;
  • 8) 0.802 469 007 802 469 007 801 203 2 × 2 = 1 + 0.604 938 015 604 938 015 602 406 4;
  • 9) 0.604 938 015 604 938 015 602 406 4 × 2 = 1 + 0.209 876 031 209 876 031 204 812 8;
  • 10) 0.209 876 031 209 876 031 204 812 8 × 2 = 0 + 0.419 752 062 419 752 062 409 625 6;
  • 11) 0.419 752 062 419 752 062 409 625 6 × 2 = 0 + 0.839 504 124 839 504 124 819 251 2;
  • 12) 0.839 504 124 839 504 124 819 251 2 × 2 = 1 + 0.679 008 249 679 008 249 638 502 4;
  • 13) 0.679 008 249 679 008 249 638 502 4 × 2 = 1 + 0.358 016 499 358 016 499 277 004 8;
  • 14) 0.358 016 499 358 016 499 277 004 8 × 2 = 0 + 0.716 032 998 716 032 998 554 009 6;
  • 15) 0.716 032 998 716 032 998 554 009 6 × 2 = 1 + 0.432 065 997 432 065 997 108 019 2;
  • 16) 0.432 065 997 432 065 997 108 019 2 × 2 = 0 + 0.864 131 994 864 131 994 216 038 4;
  • 17) 0.864 131 994 864 131 994 216 038 4 × 2 = 1 + 0.728 263 989 728 263 988 432 076 8;
  • 18) 0.728 263 989 728 263 988 432 076 8 × 2 = 1 + 0.456 527 979 456 527 976 864 153 6;
  • 19) 0.456 527 979 456 527 976 864 153 6 × 2 = 0 + 0.913 055 958 913 055 953 728 307 2;
  • 20) 0.913 055 958 913 055 953 728 307 2 × 2 = 1 + 0.826 111 917 826 111 907 456 614 4;
  • 21) 0.826 111 917 826 111 907 456 614 4 × 2 = 1 + 0.652 223 835 652 223 814 913 228 8;
  • 22) 0.652 223 835 652 223 814 913 228 8 × 2 = 1 + 0.304 447 671 304 447 629 826 457 6;
  • 23) 0.304 447 671 304 447 629 826 457 6 × 2 = 0 + 0.608 895 342 608 895 259 652 915 2;
  • 24) 0.608 895 342 608 895 259 652 915 2 × 2 = 1 + 0.217 790 685 217 790 519 305 830 4;
  • 25) 0.217 790 685 217 790 519 305 830 4 × 2 = 0 + 0.435 581 370 435 581 038 611 660 8;
  • 26) 0.435 581 370 435 581 038 611 660 8 × 2 = 0 + 0.871 162 740 871 162 077 223 321 6;
  • 27) 0.871 162 740 871 162 077 223 321 6 × 2 = 1 + 0.742 325 481 742 324 154 446 643 2;
  • 28) 0.742 325 481 742 324 154 446 643 2 × 2 = 1 + 0.484 650 963 484 648 308 893 286 4;
  • 29) 0.484 650 963 484 648 308 893 286 4 × 2 = 0 + 0.969 301 926 969 296 617 786 572 8;
  • 30) 0.969 301 926 969 296 617 786 572 8 × 2 = 1 + 0.938 603 853 938 593 235 573 145 6;
  • 31) 0.938 603 853 938 593 235 573 145 6 × 2 = 1 + 0.877 207 707 877 186 471 146 291 2;
  • 32) 0.877 207 707 877 186 471 146 291 2 × 2 = 1 + 0.754 415 415 754 372 942 292 582 4;
  • 33) 0.754 415 415 754 372 942 292 582 4 × 2 = 1 + 0.508 830 831 508 745 884 585 164 8;
  • 34) 0.508 830 831 508 745 884 585 164 8 × 2 = 1 + 0.017 661 663 017 491 769 170 329 6;
  • 35) 0.017 661 663 017 491 769 170 329 6 × 2 = 0 + 0.035 323 326 034 983 538 340 659 2;
  • 36) 0.035 323 326 034 983 538 340 659 2 × 2 = 0 + 0.070 646 652 069 967 076 681 318 4;
  • 37) 0.070 646 652 069 967 076 681 318 4 × 2 = 0 + 0.141 293 304 139 934 153 362 636 8;
  • 38) 0.141 293 304 139 934 153 362 636 8 × 2 = 0 + 0.282 586 608 279 868 306 725 273 6;
  • 39) 0.282 586 608 279 868 306 725 273 6 × 2 = 0 + 0.565 173 216 559 736 613 450 547 2;
  • 40) 0.565 173 216 559 736 613 450 547 2 × 2 = 1 + 0.130 346 433 119 473 226 901 094 4;
  • 41) 0.130 346 433 119 473 226 901 094 4 × 2 = 0 + 0.260 692 866 238 946 453 802 188 8;
  • 42) 0.260 692 866 238 946 453 802 188 8 × 2 = 0 + 0.521 385 732 477 892 907 604 377 6;
  • 43) 0.521 385 732 477 892 907 604 377 6 × 2 = 1 + 0.042 771 464 955 785 815 208 755 2;
  • 44) 0.042 771 464 955 785 815 208 755 2 × 2 = 0 + 0.085 542 929 911 571 630 417 510 4;
  • 45) 0.085 542 929 911 571 630 417 510 4 × 2 = 0 + 0.171 085 859 823 143 260 835 020 8;
  • 46) 0.171 085 859 823 143 260 835 020 8 × 2 = 0 + 0.342 171 719 646 286 521 670 041 6;
  • 47) 0.342 171 719 646 286 521 670 041 6 × 2 = 0 + 0.684 343 439 292 573 043 340 083 2;
  • 48) 0.684 343 439 292 573 043 340 083 2 × 2 = 1 + 0.368 686 878 585 146 086 680 166 4;
  • 49) 0.368 686 878 585 146 086 680 166 4 × 2 = 0 + 0.737 373 757 170 292 173 360 332 8;
  • 50) 0.737 373 757 170 292 173 360 332 8 × 2 = 1 + 0.474 747 514 340 584 346 720 665 6;
  • 51) 0.474 747 514 340 584 346 720 665 6 × 2 = 0 + 0.949 495 028 681 168 693 441 331 2;
  • 52) 0.949 495 028 681 168 693 441 331 2 × 2 = 1 + 0.898 990 057 362 337 386 882 662 4;
  • 53) 0.898 990 057 362 337 386 882 662 4 × 2 = 1 + 0.797 980 114 724 674 773 765 324 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:


0.123 456 789 123 456 789 123 446 9(10) =

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

6. Positive number before normalization:

12 345.123 456 789 123 456 789 123 446 9(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:


12 345.123 456 789 123 456 789 123 446 9(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 -12 345.123 456 789 123 456 789 123 446 9 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 -12 345.123 456 789 123 456 789 123 436 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