654.599 999 999 999 909 050 514 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 514(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
654.599 999 999 999 909 050 514(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. First, convert to binary (in base 2) the integer part: 654.
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;
  • 654 ÷ 2 = 327 + 0;
  • 327 ÷ 2 = 163 + 1;
  • 163 ÷ 2 = 81 + 1;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

654(10) =

10 1000 1110(2)


3. Convert to binary (base 2) the fractional part: 0.599 999 999 999 909 050 514.

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) 0.599 999 999 999 909 050 514 × 2 = 1 + 0.199 999 999 999 818 101 028;
  • 2) 0.199 999 999 999 818 101 028 × 2 = 0 + 0.399 999 999 999 636 202 056;
  • 3) 0.399 999 999 999 636 202 056 × 2 = 0 + 0.799 999 999 999 272 404 112;
  • 4) 0.799 999 999 999 272 404 112 × 2 = 1 + 0.599 999 999 998 544 808 224;
  • 5) 0.599 999 999 998 544 808 224 × 2 = 1 + 0.199 999 999 997 089 616 448;
  • 6) 0.199 999 999 997 089 616 448 × 2 = 0 + 0.399 999 999 994 179 232 896;
  • 7) 0.399 999 999 994 179 232 896 × 2 = 0 + 0.799 999 999 988 358 465 792;
  • 8) 0.799 999 999 988 358 465 792 × 2 = 1 + 0.599 999 999 976 716 931 584;
  • 9) 0.599 999 999 976 716 931 584 × 2 = 1 + 0.199 999 999 953 433 863 168;
  • 10) 0.199 999 999 953 433 863 168 × 2 = 0 + 0.399 999 999 906 867 726 336;
  • 11) 0.399 999 999 906 867 726 336 × 2 = 0 + 0.799 999 999 813 735 452 672;
  • 12) 0.799 999 999 813 735 452 672 × 2 = 1 + 0.599 999 999 627 470 905 344;
  • 13) 0.599 999 999 627 470 905 344 × 2 = 1 + 0.199 999 999 254 941 810 688;
  • 14) 0.199 999 999 254 941 810 688 × 2 = 0 + 0.399 999 998 509 883 621 376;
  • 15) 0.399 999 998 509 883 621 376 × 2 = 0 + 0.799 999 997 019 767 242 752;
  • 16) 0.799 999 997 019 767 242 752 × 2 = 1 + 0.599 999 994 039 534 485 504;
  • 17) 0.599 999 994 039 534 485 504 × 2 = 1 + 0.199 999 988 079 068 971 008;
  • 18) 0.199 999 988 079 068 971 008 × 2 = 0 + 0.399 999 976 158 137 942 016;
  • 19) 0.399 999 976 158 137 942 016 × 2 = 0 + 0.799 999 952 316 275 884 032;
  • 20) 0.799 999 952 316 275 884 032 × 2 = 1 + 0.599 999 904 632 551 768 064;
  • 21) 0.599 999 904 632 551 768 064 × 2 = 1 + 0.199 999 809 265 103 536 128;
  • 22) 0.199 999 809 265 103 536 128 × 2 = 0 + 0.399 999 618 530 207 072 256;
  • 23) 0.399 999 618 530 207 072 256 × 2 = 0 + 0.799 999 237 060 414 144 512;
  • 24) 0.799 999 237 060 414 144 512 × 2 = 1 + 0.599 998 474 120 828 289 024;
  • 25) 0.599 998 474 120 828 289 024 × 2 = 1 + 0.199 996 948 241 656 578 048;
  • 26) 0.199 996 948 241 656 578 048 × 2 = 0 + 0.399 993 896 483 313 156 096;
  • 27) 0.399 993 896 483 313 156 096 × 2 = 0 + 0.799 987 792 966 626 312 192;
  • 28) 0.799 987 792 966 626 312 192 × 2 = 1 + 0.599 975 585 933 252 624 384;
  • 29) 0.599 975 585 933 252 624 384 × 2 = 1 + 0.199 951 171 866 505 248 768;
  • 30) 0.199 951 171 866 505 248 768 × 2 = 0 + 0.399 902 343 733 010 497 536;
  • 31) 0.399 902 343 733 010 497 536 × 2 = 0 + 0.799 804 687 466 020 995 072;
  • 32) 0.799 804 687 466 020 995 072 × 2 = 1 + 0.599 609 374 932 041 990 144;
  • 33) 0.599 609 374 932 041 990 144 × 2 = 1 + 0.199 218 749 864 083 980 288;
  • 34) 0.199 218 749 864 083 980 288 × 2 = 0 + 0.398 437 499 728 167 960 576;
  • 35) 0.398 437 499 728 167 960 576 × 2 = 0 + 0.796 874 999 456 335 921 152;
  • 36) 0.796 874 999 456 335 921 152 × 2 = 1 + 0.593 749 998 912 671 842 304;
  • 37) 0.593 749 998 912 671 842 304 × 2 = 1 + 0.187 499 997 825 343 684 608;
  • 38) 0.187 499 997 825 343 684 608 × 2 = 0 + 0.374 999 995 650 687 369 216;
  • 39) 0.374 999 995 650 687 369 216 × 2 = 0 + 0.749 999 991 301 374 738 432;
  • 40) 0.749 999 991 301 374 738 432 × 2 = 1 + 0.499 999 982 602 749 476 864;
  • 41) 0.499 999 982 602 749 476 864 × 2 = 0 + 0.999 999 965 205 498 953 728;
  • 42) 0.999 999 965 205 498 953 728 × 2 = 1 + 0.999 999 930 410 997 907 456;
  • 43) 0.999 999 930 410 997 907 456 × 2 = 1 + 0.999 999 860 821 995 814 912;
  • 44) 0.999 999 860 821 995 814 912 × 2 = 1 + 0.999 999 721 643 991 629 824;
  • 45) 0.999 999 721 643 991 629 824 × 2 = 1 + 0.999 999 443 287 983 259 648;
  • 46) 0.999 999 443 287 983 259 648 × 2 = 1 + 0.999 998 886 575 966 519 296;
  • 47) 0.999 998 886 575 966 519 296 × 2 = 1 + 0.999 997 773 151 933 038 592;
  • 48) 0.999 997 773 151 933 038 592 × 2 = 1 + 0.999 995 546 303 866 077 184;
  • 49) 0.999 995 546 303 866 077 184 × 2 = 1 + 0.999 991 092 607 732 154 368;
  • 50) 0.999 991 092 607 732 154 368 × 2 = 1 + 0.999 982 185 215 464 308 736;
  • 51) 0.999 982 185 215 464 308 736 × 2 = 1 + 0.999 964 370 430 928 617 472;
  • 52) 0.999 964 370 430 928 617 472 × 2 = 1 + 0.999 928 740 861 857 234 944;
  • 53) 0.999 928 740 861 857 234 944 × 2 = 1 + 0.999 857 481 723 714 469 888;

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).


4. 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:


0.599 999 999 999 909 050 514(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2)

5. Positive number before normalization:

654.599 999 999 999 909 050 514(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left, so that only one non zero digit remains to the left of it:


654.599 999 999 999 909 050 514(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2) × 20 =

1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 11(2) × 29


7. 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 0 (a positive number)

Exponent (unadjusted): 9

Mantissa (not normalized):
1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

9 + 2(11-1) - 1 =

(9 + 1 023)(10) =

1 032(10)


9. 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;
  • 1 032 ÷ 2 = 516 + 0;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 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;

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

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


Exponent (adjusted) =

1032(10) =

100 0000 1000(2)


11. 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. 0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 11 1111 1111 =

0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011


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

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 1000

Mantissa (52 bits) =
0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011


Decimal number 654.599 999 999 999 909 050 514 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1000 - 0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011

Share

» Convert decimal 654.599 999 999 999 909 050 524 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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