654.599 999 999 999 909 050 588 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 588(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
654.599 999 999 999 909 050 588(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: 0.599 999 999 999 909 050 588.

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) 0.599 999 999 999 909 050 588 × 2 = 1 + 0.199 999 999 999 818 101 176;
  • 2) 0.199 999 999 999 818 101 176 × 2 = 0 + 0.399 999 999 999 636 202 352;
  • 3) 0.399 999 999 999 636 202 352 × 2 = 0 + 0.799 999 999 999 272 404 704;
  • 4) 0.799 999 999 999 272 404 704 × 2 = 1 + 0.599 999 999 998 544 809 408;
  • 5) 0.599 999 999 998 544 809 408 × 2 = 1 + 0.199 999 999 997 089 618 816;
  • 6) 0.199 999 999 997 089 618 816 × 2 = 0 + 0.399 999 999 994 179 237 632;
  • 7) 0.399 999 999 994 179 237 632 × 2 = 0 + 0.799 999 999 988 358 475 264;
  • 8) 0.799 999 999 988 358 475 264 × 2 = 1 + 0.599 999 999 976 716 950 528;
  • 9) 0.599 999 999 976 716 950 528 × 2 = 1 + 0.199 999 999 953 433 901 056;
  • 10) 0.199 999 999 953 433 901 056 × 2 = 0 + 0.399 999 999 906 867 802 112;
  • 11) 0.399 999 999 906 867 802 112 × 2 = 0 + 0.799 999 999 813 735 604 224;
  • 12) 0.799 999 999 813 735 604 224 × 2 = 1 + 0.599 999 999 627 471 208 448;
  • 13) 0.599 999 999 627 471 208 448 × 2 = 1 + 0.199 999 999 254 942 416 896;
  • 14) 0.199 999 999 254 942 416 896 × 2 = 0 + 0.399 999 998 509 884 833 792;
  • 15) 0.399 999 998 509 884 833 792 × 2 = 0 + 0.799 999 997 019 769 667 584;
  • 16) 0.799 999 997 019 769 667 584 × 2 = 1 + 0.599 999 994 039 539 335 168;
  • 17) 0.599 999 994 039 539 335 168 × 2 = 1 + 0.199 999 988 079 078 670 336;
  • 18) 0.199 999 988 079 078 670 336 × 2 = 0 + 0.399 999 976 158 157 340 672;
  • 19) 0.399 999 976 158 157 340 672 × 2 = 0 + 0.799 999 952 316 314 681 344;
  • 20) 0.799 999 952 316 314 681 344 × 2 = 1 + 0.599 999 904 632 629 362 688;
  • 21) 0.599 999 904 632 629 362 688 × 2 = 1 + 0.199 999 809 265 258 725 376;
  • 22) 0.199 999 809 265 258 725 376 × 2 = 0 + 0.399 999 618 530 517 450 752;
  • 23) 0.399 999 618 530 517 450 752 × 2 = 0 + 0.799 999 237 061 034 901 504;
  • 24) 0.799 999 237 061 034 901 504 × 2 = 1 + 0.599 998 474 122 069 803 008;
  • 25) 0.599 998 474 122 069 803 008 × 2 = 1 + 0.199 996 948 244 139 606 016;
  • 26) 0.199 996 948 244 139 606 016 × 2 = 0 + 0.399 993 896 488 279 212 032;
  • 27) 0.399 993 896 488 279 212 032 × 2 = 0 + 0.799 987 792 976 558 424 064;
  • 28) 0.799 987 792 976 558 424 064 × 2 = 1 + 0.599 975 585 953 116 848 128;
  • 29) 0.599 975 585 953 116 848 128 × 2 = 1 + 0.199 951 171 906 233 696 256;
  • 30) 0.199 951 171 906 233 696 256 × 2 = 0 + 0.399 902 343 812 467 392 512;
  • 31) 0.399 902 343 812 467 392 512 × 2 = 0 + 0.799 804 687 624 934 785 024;
  • 32) 0.799 804 687 624 934 785 024 × 2 = 1 + 0.599 609 375 249 869 570 048;
  • 33) 0.599 609 375 249 869 570 048 × 2 = 1 + 0.199 218 750 499 739 140 096;
  • 34) 0.199 218 750 499 739 140 096 × 2 = 0 + 0.398 437 500 999 478 280 192;
  • 35) 0.398 437 500 999 478 280 192 × 2 = 0 + 0.796 875 001 998 956 560 384;
  • 36) 0.796 875 001 998 956 560 384 × 2 = 1 + 0.593 750 003 997 913 120 768;
  • 37) 0.593 750 003 997 913 120 768 × 2 = 1 + 0.187 500 007 995 826 241 536;
  • 38) 0.187 500 007 995 826 241 536 × 2 = 0 + 0.375 000 015 991 652 483 072;
  • 39) 0.375 000 015 991 652 483 072 × 2 = 0 + 0.750 000 031 983 304 966 144;
  • 40) 0.750 000 031 983 304 966 144 × 2 = 1 + 0.500 000 063 966 609 932 288;
  • 41) 0.500 000 063 966 609 932 288 × 2 = 1 + 0.000 000 127 933 219 864 576;
  • 42) 0.000 000 127 933 219 864 576 × 2 = 0 + 0.000 000 255 866 439 729 152;
  • 43) 0.000 000 255 866 439 729 152 × 2 = 0 + 0.000 000 511 732 879 458 304;
  • 44) 0.000 000 511 732 879 458 304 × 2 = 0 + 0.000 001 023 465 758 916 608;
  • 45) 0.000 001 023 465 758 916 608 × 2 = 0 + 0.000 002 046 931 517 833 216;
  • 46) 0.000 002 046 931 517 833 216 × 2 = 0 + 0.000 004 093 863 035 666 432;
  • 47) 0.000 004 093 863 035 666 432 × 2 = 0 + 0.000 008 187 726 071 332 864;
  • 48) 0.000 008 187 726 071 332 864 × 2 = 0 + 0.000 016 375 452 142 665 728;
  • 49) 0.000 016 375 452 142 665 728 × 2 = 0 + 0.000 032 750 904 285 331 456;
  • 50) 0.000 032 750 904 285 331 456 × 2 = 0 + 0.000 065 501 808 570 662 912;
  • 51) 0.000 065 501 808 570 662 912 × 2 = 0 + 0.000 131 003 617 141 325 824;
  • 52) 0.000 131 003 617 141 325 824 × 2 = 0 + 0.000 262 007 234 282 651 648;
  • 53) 0.000 262 007 234 282 651 648 × 2 = 0 + 0.000 524 014 468 565 303 296;

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:


0.599 999 999 999 909 050 588(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2)

5. Positive number before normalization:

654.599 999 999 999 909 050 588(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(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:


654.599 999 999 999 909 050 588(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) × 20 =

1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 00(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 1100 0000 0000 00


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 1100 00 0000 0000 =

0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


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 1100


Decimal number 654.599 999 999 999 909 050 588 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 1100

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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