-12 345.123 456 789 123 456 789 123 436 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -12 345.123 456 789 123 456 789 123 436 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
-12 345.123 456 789 123 456 789 123 436 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:

|-12 345.123 456 789 123 456 789 123 436 9| = 12 345.123 456 789 123 456 789 123 436 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: 0.123 456 789 123 456 789 123 436 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) 0.123 456 789 123 456 789 123 436 9 × 2 = 0 + 0.246 913 578 246 913 578 246 873 8;
  • 2) 0.246 913 578 246 913 578 246 873 8 × 2 = 0 + 0.493 827 156 493 827 156 493 747 6;
  • 3) 0.493 827 156 493 827 156 493 747 6 × 2 = 0 + 0.987 654 312 987 654 312 987 495 2;
  • 4) 0.987 654 312 987 654 312 987 495 2 × 2 = 1 + 0.975 308 625 975 308 625 974 990 4;
  • 5) 0.975 308 625 975 308 625 974 990 4 × 2 = 1 + 0.950 617 251 950 617 251 949 980 8;
  • 6) 0.950 617 251 950 617 251 949 980 8 × 2 = 1 + 0.901 234 503 901 234 503 899 961 6;
  • 7) 0.901 234 503 901 234 503 899 961 6 × 2 = 1 + 0.802 469 007 802 469 007 799 923 2;
  • 8) 0.802 469 007 802 469 007 799 923 2 × 2 = 1 + 0.604 938 015 604 938 015 599 846 4;
  • 9) 0.604 938 015 604 938 015 599 846 4 × 2 = 1 + 0.209 876 031 209 876 031 199 692 8;
  • 10) 0.209 876 031 209 876 031 199 692 8 × 2 = 0 + 0.419 752 062 419 752 062 399 385 6;
  • 11) 0.419 752 062 419 752 062 399 385 6 × 2 = 0 + 0.839 504 124 839 504 124 798 771 2;
  • 12) 0.839 504 124 839 504 124 798 771 2 × 2 = 1 + 0.679 008 249 679 008 249 597 542 4;
  • 13) 0.679 008 249 679 008 249 597 542 4 × 2 = 1 + 0.358 016 499 358 016 499 195 084 8;
  • 14) 0.358 016 499 358 016 499 195 084 8 × 2 = 0 + 0.716 032 998 716 032 998 390 169 6;
  • 15) 0.716 032 998 716 032 998 390 169 6 × 2 = 1 + 0.432 065 997 432 065 996 780 339 2;
  • 16) 0.432 065 997 432 065 996 780 339 2 × 2 = 0 + 0.864 131 994 864 131 993 560 678 4;
  • 17) 0.864 131 994 864 131 993 560 678 4 × 2 = 1 + 0.728 263 989 728 263 987 121 356 8;
  • 18) 0.728 263 989 728 263 987 121 356 8 × 2 = 1 + 0.456 527 979 456 527 974 242 713 6;
  • 19) 0.456 527 979 456 527 974 242 713 6 × 2 = 0 + 0.913 055 958 913 055 948 485 427 2;
  • 20) 0.913 055 958 913 055 948 485 427 2 × 2 = 1 + 0.826 111 917 826 111 896 970 854 4;
  • 21) 0.826 111 917 826 111 896 970 854 4 × 2 = 1 + 0.652 223 835 652 223 793 941 708 8;
  • 22) 0.652 223 835 652 223 793 941 708 8 × 2 = 1 + 0.304 447 671 304 447 587 883 417 6;
  • 23) 0.304 447 671 304 447 587 883 417 6 × 2 = 0 + 0.608 895 342 608 895 175 766 835 2;
  • 24) 0.608 895 342 608 895 175 766 835 2 × 2 = 1 + 0.217 790 685 217 790 351 533 670 4;
  • 25) 0.217 790 685 217 790 351 533 670 4 × 2 = 0 + 0.435 581 370 435 580 703 067 340 8;
  • 26) 0.435 581 370 435 580 703 067 340 8 × 2 = 0 + 0.871 162 740 871 161 406 134 681 6;
  • 27) 0.871 162 740 871 161 406 134 681 6 × 2 = 1 + 0.742 325 481 742 322 812 269 363 2;
  • 28) 0.742 325 481 742 322 812 269 363 2 × 2 = 1 + 0.484 650 963 484 645 624 538 726 4;
  • 29) 0.484 650 963 484 645 624 538 726 4 × 2 = 0 + 0.969 301 926 969 291 249 077 452 8;
  • 30) 0.969 301 926 969 291 249 077 452 8 × 2 = 1 + 0.938 603 853 938 582 498 154 905 6;
  • 31) 0.938 603 853 938 582 498 154 905 6 × 2 = 1 + 0.877 207 707 877 164 996 309 811 2;
  • 32) 0.877 207 707 877 164 996 309 811 2 × 2 = 1 + 0.754 415 415 754 329 992 619 622 4;
  • 33) 0.754 415 415 754 329 992 619 622 4 × 2 = 1 + 0.508 830 831 508 659 985 239 244 8;
  • 34) 0.508 830 831 508 659 985 239 244 8 × 2 = 1 + 0.017 661 663 017 319 970 478 489 6;
  • 35) 0.017 661 663 017 319 970 478 489 6 × 2 = 0 + 0.035 323 326 034 639 940 956 979 2;
  • 36) 0.035 323 326 034 639 940 956 979 2 × 2 = 0 + 0.070 646 652 069 279 881 913 958 4;
  • 37) 0.070 646 652 069 279 881 913 958 4 × 2 = 0 + 0.141 293 304 138 559 763 827 916 8;
  • 38) 0.141 293 304 138 559 763 827 916 8 × 2 = 0 + 0.282 586 608 277 119 527 655 833 6;
  • 39) 0.282 586 608 277 119 527 655 833 6 × 2 = 0 + 0.565 173 216 554 239 055 311 667 2;
  • 40) 0.565 173 216 554 239 055 311 667 2 × 2 = 1 + 0.130 346 433 108 478 110 623 334 4;
  • 41) 0.130 346 433 108 478 110 623 334 4 × 2 = 0 + 0.260 692 866 216 956 221 246 668 8;
  • 42) 0.260 692 866 216 956 221 246 668 8 × 2 = 0 + 0.521 385 732 433 912 442 493 337 6;
  • 43) 0.521 385 732 433 912 442 493 337 6 × 2 = 1 + 0.042 771 464 867 824 884 986 675 2;
  • 44) 0.042 771 464 867 824 884 986 675 2 × 2 = 0 + 0.085 542 929 735 649 769 973 350 4;
  • 45) 0.085 542 929 735 649 769 973 350 4 × 2 = 0 + 0.171 085 859 471 299 539 946 700 8;
  • 46) 0.171 085 859 471 299 539 946 700 8 × 2 = 0 + 0.342 171 718 942 599 079 893 401 6;
  • 47) 0.342 171 718 942 599 079 893 401 6 × 2 = 0 + 0.684 343 437 885 198 159 786 803 2;
  • 48) 0.684 343 437 885 198 159 786 803 2 × 2 = 1 + 0.368 686 875 770 396 319 573 606 4;
  • 49) 0.368 686 875 770 396 319 573 606 4 × 2 = 0 + 0.737 373 751 540 792 639 147 212 8;
  • 50) 0.737 373 751 540 792 639 147 212 8 × 2 = 1 + 0.474 747 503 081 585 278 294 425 6;
  • 51) 0.474 747 503 081 585 278 294 425 6 × 2 = 0 + 0.949 495 006 163 170 556 588 851 2;
  • 52) 0.949 495 006 163 170 556 588 851 2 × 2 = 1 + 0.898 990 012 326 341 113 177 702 4;
  • 53) 0.898 990 012 326 341 113 177 702 4 × 2 = 1 + 0.797 980 024 652 682 226 355 404 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:


0.123 456 789 123 456 789 123 436 9(10) =

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

6. Positive number before normalization:

12 345.123 456 789 123 456 789 123 436 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:


12 345.123 456 789 123 456 789 123 436 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 -12 345.123 456 789 123 456 789 123 436 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 -12 345.123 456 789 123 456 789 123 433 2 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