654.599 999 999 999 909 050 626 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 626(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
654.599 999 999 999 909 050 626(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: 0.599 999 999 999 909 050 626.

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) 0.599 999 999 999 909 050 626 × 2 = 1 + 0.199 999 999 999 818 101 252;
  • 2) 0.199 999 999 999 818 101 252 × 2 = 0 + 0.399 999 999 999 636 202 504;
  • 3) 0.399 999 999 999 636 202 504 × 2 = 0 + 0.799 999 999 999 272 405 008;
  • 4) 0.799 999 999 999 272 405 008 × 2 = 1 + 0.599 999 999 998 544 810 016;
  • 5) 0.599 999 999 998 544 810 016 × 2 = 1 + 0.199 999 999 997 089 620 032;
  • 6) 0.199 999 999 997 089 620 032 × 2 = 0 + 0.399 999 999 994 179 240 064;
  • 7) 0.399 999 999 994 179 240 064 × 2 = 0 + 0.799 999 999 988 358 480 128;
  • 8) 0.799 999 999 988 358 480 128 × 2 = 1 + 0.599 999 999 976 716 960 256;
  • 9) 0.599 999 999 976 716 960 256 × 2 = 1 + 0.199 999 999 953 433 920 512;
  • 10) 0.199 999 999 953 433 920 512 × 2 = 0 + 0.399 999 999 906 867 841 024;
  • 11) 0.399 999 999 906 867 841 024 × 2 = 0 + 0.799 999 999 813 735 682 048;
  • 12) 0.799 999 999 813 735 682 048 × 2 = 1 + 0.599 999 999 627 471 364 096;
  • 13) 0.599 999 999 627 471 364 096 × 2 = 1 + 0.199 999 999 254 942 728 192;
  • 14) 0.199 999 999 254 942 728 192 × 2 = 0 + 0.399 999 998 509 885 456 384;
  • 15) 0.399 999 998 509 885 456 384 × 2 = 0 + 0.799 999 997 019 770 912 768;
  • 16) 0.799 999 997 019 770 912 768 × 2 = 1 + 0.599 999 994 039 541 825 536;
  • 17) 0.599 999 994 039 541 825 536 × 2 = 1 + 0.199 999 988 079 083 651 072;
  • 18) 0.199 999 988 079 083 651 072 × 2 = 0 + 0.399 999 976 158 167 302 144;
  • 19) 0.399 999 976 158 167 302 144 × 2 = 0 + 0.799 999 952 316 334 604 288;
  • 20) 0.799 999 952 316 334 604 288 × 2 = 1 + 0.599 999 904 632 669 208 576;
  • 21) 0.599 999 904 632 669 208 576 × 2 = 1 + 0.199 999 809 265 338 417 152;
  • 22) 0.199 999 809 265 338 417 152 × 2 = 0 + 0.399 999 618 530 676 834 304;
  • 23) 0.399 999 618 530 676 834 304 × 2 = 0 + 0.799 999 237 061 353 668 608;
  • 24) 0.799 999 237 061 353 668 608 × 2 = 1 + 0.599 998 474 122 707 337 216;
  • 25) 0.599 998 474 122 707 337 216 × 2 = 1 + 0.199 996 948 245 414 674 432;
  • 26) 0.199 996 948 245 414 674 432 × 2 = 0 + 0.399 993 896 490 829 348 864;
  • 27) 0.399 993 896 490 829 348 864 × 2 = 0 + 0.799 987 792 981 658 697 728;
  • 28) 0.799 987 792 981 658 697 728 × 2 = 1 + 0.599 975 585 963 317 395 456;
  • 29) 0.599 975 585 963 317 395 456 × 2 = 1 + 0.199 951 171 926 634 790 912;
  • 30) 0.199 951 171 926 634 790 912 × 2 = 0 + 0.399 902 343 853 269 581 824;
  • 31) 0.399 902 343 853 269 581 824 × 2 = 0 + 0.799 804 687 706 539 163 648;
  • 32) 0.799 804 687 706 539 163 648 × 2 = 1 + 0.599 609 375 413 078 327 296;
  • 33) 0.599 609 375 413 078 327 296 × 2 = 1 + 0.199 218 750 826 156 654 592;
  • 34) 0.199 218 750 826 156 654 592 × 2 = 0 + 0.398 437 501 652 313 309 184;
  • 35) 0.398 437 501 652 313 309 184 × 2 = 0 + 0.796 875 003 304 626 618 368;
  • 36) 0.796 875 003 304 626 618 368 × 2 = 1 + 0.593 750 006 609 253 236 736;
  • 37) 0.593 750 006 609 253 236 736 × 2 = 1 + 0.187 500 013 218 506 473 472;
  • 38) 0.187 500 013 218 506 473 472 × 2 = 0 + 0.375 000 026 437 012 946 944;
  • 39) 0.375 000 026 437 012 946 944 × 2 = 0 + 0.750 000 052 874 025 893 888;
  • 40) 0.750 000 052 874 025 893 888 × 2 = 1 + 0.500 000 105 748 051 787 776;
  • 41) 0.500 000 105 748 051 787 776 × 2 = 1 + 0.000 000 211 496 103 575 552;
  • 42) 0.000 000 211 496 103 575 552 × 2 = 0 + 0.000 000 422 992 207 151 104;
  • 43) 0.000 000 422 992 207 151 104 × 2 = 0 + 0.000 000 845 984 414 302 208;
  • 44) 0.000 000 845 984 414 302 208 × 2 = 0 + 0.000 001 691 968 828 604 416;
  • 45) 0.000 001 691 968 828 604 416 × 2 = 0 + 0.000 003 383 937 657 208 832;
  • 46) 0.000 003 383 937 657 208 832 × 2 = 0 + 0.000 006 767 875 314 417 664;
  • 47) 0.000 006 767 875 314 417 664 × 2 = 0 + 0.000 013 535 750 628 835 328;
  • 48) 0.000 013 535 750 628 835 328 × 2 = 0 + 0.000 027 071 501 257 670 656;
  • 49) 0.000 027 071 501 257 670 656 × 2 = 0 + 0.000 054 143 002 515 341 312;
  • 50) 0.000 054 143 002 515 341 312 × 2 = 0 + 0.000 108 286 005 030 682 624;
  • 51) 0.000 108 286 005 030 682 624 × 2 = 0 + 0.000 216 572 010 061 365 248;
  • 52) 0.000 216 572 010 061 365 248 × 2 = 0 + 0.000 433 144 020 122 730 496;
  • 53) 0.000 433 144 020 122 730 496 × 2 = 0 + 0.000 866 288 040 245 460 992;

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:


0.599 999 999 999 909 050 626(10) =

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

5. Positive number before normalization:

654.599 999 999 999 909 050 626(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:


654.599 999 999 999 909 050 626(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 654.599 999 999 999 909 050 626 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