654.599 999 999 999 909 050 540 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 540 4(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
654.599 999 999 999 909 050 540 4(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: 0.599 999 999 999 909 050 540 4.

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) 0.599 999 999 999 909 050 540 4 × 2 = 1 + 0.199 999 999 999 818 101 080 8;
  • 2) 0.199 999 999 999 818 101 080 8 × 2 = 0 + 0.399 999 999 999 636 202 161 6;
  • 3) 0.399 999 999 999 636 202 161 6 × 2 = 0 + 0.799 999 999 999 272 404 323 2;
  • 4) 0.799 999 999 999 272 404 323 2 × 2 = 1 + 0.599 999 999 998 544 808 646 4;
  • 5) 0.599 999 999 998 544 808 646 4 × 2 = 1 + 0.199 999 999 997 089 617 292 8;
  • 6) 0.199 999 999 997 089 617 292 8 × 2 = 0 + 0.399 999 999 994 179 234 585 6;
  • 7) 0.399 999 999 994 179 234 585 6 × 2 = 0 + 0.799 999 999 988 358 469 171 2;
  • 8) 0.799 999 999 988 358 469 171 2 × 2 = 1 + 0.599 999 999 976 716 938 342 4;
  • 9) 0.599 999 999 976 716 938 342 4 × 2 = 1 + 0.199 999 999 953 433 876 684 8;
  • 10) 0.199 999 999 953 433 876 684 8 × 2 = 0 + 0.399 999 999 906 867 753 369 6;
  • 11) 0.399 999 999 906 867 753 369 6 × 2 = 0 + 0.799 999 999 813 735 506 739 2;
  • 12) 0.799 999 999 813 735 506 739 2 × 2 = 1 + 0.599 999 999 627 471 013 478 4;
  • 13) 0.599 999 999 627 471 013 478 4 × 2 = 1 + 0.199 999 999 254 942 026 956 8;
  • 14) 0.199 999 999 254 942 026 956 8 × 2 = 0 + 0.399 999 998 509 884 053 913 6;
  • 15) 0.399 999 998 509 884 053 913 6 × 2 = 0 + 0.799 999 997 019 768 107 827 2;
  • 16) 0.799 999 997 019 768 107 827 2 × 2 = 1 + 0.599 999 994 039 536 215 654 4;
  • 17) 0.599 999 994 039 536 215 654 4 × 2 = 1 + 0.199 999 988 079 072 431 308 8;
  • 18) 0.199 999 988 079 072 431 308 8 × 2 = 0 + 0.399 999 976 158 144 862 617 6;
  • 19) 0.399 999 976 158 144 862 617 6 × 2 = 0 + 0.799 999 952 316 289 725 235 2;
  • 20) 0.799 999 952 316 289 725 235 2 × 2 = 1 + 0.599 999 904 632 579 450 470 4;
  • 21) 0.599 999 904 632 579 450 470 4 × 2 = 1 + 0.199 999 809 265 158 900 940 8;
  • 22) 0.199 999 809 265 158 900 940 8 × 2 = 0 + 0.399 999 618 530 317 801 881 6;
  • 23) 0.399 999 618 530 317 801 881 6 × 2 = 0 + 0.799 999 237 060 635 603 763 2;
  • 24) 0.799 999 237 060 635 603 763 2 × 2 = 1 + 0.599 998 474 121 271 207 526 4;
  • 25) 0.599 998 474 121 271 207 526 4 × 2 = 1 + 0.199 996 948 242 542 415 052 8;
  • 26) 0.199 996 948 242 542 415 052 8 × 2 = 0 + 0.399 993 896 485 084 830 105 6;
  • 27) 0.399 993 896 485 084 830 105 6 × 2 = 0 + 0.799 987 792 970 169 660 211 2;
  • 28) 0.799 987 792 970 169 660 211 2 × 2 = 1 + 0.599 975 585 940 339 320 422 4;
  • 29) 0.599 975 585 940 339 320 422 4 × 2 = 1 + 0.199 951 171 880 678 640 844 8;
  • 30) 0.199 951 171 880 678 640 844 8 × 2 = 0 + 0.399 902 343 761 357 281 689 6;
  • 31) 0.399 902 343 761 357 281 689 6 × 2 = 0 + 0.799 804 687 522 714 563 379 2;
  • 32) 0.799 804 687 522 714 563 379 2 × 2 = 1 + 0.599 609 375 045 429 126 758 4;
  • 33) 0.599 609 375 045 429 126 758 4 × 2 = 1 + 0.199 218 750 090 858 253 516 8;
  • 34) 0.199 218 750 090 858 253 516 8 × 2 = 0 + 0.398 437 500 181 716 507 033 6;
  • 35) 0.398 437 500 181 716 507 033 6 × 2 = 0 + 0.796 875 000 363 433 014 067 2;
  • 36) 0.796 875 000 363 433 014 067 2 × 2 = 1 + 0.593 750 000 726 866 028 134 4;
  • 37) 0.593 750 000 726 866 028 134 4 × 2 = 1 + 0.187 500 001 453 732 056 268 8;
  • 38) 0.187 500 001 453 732 056 268 8 × 2 = 0 + 0.375 000 002 907 464 112 537 6;
  • 39) 0.375 000 002 907 464 112 537 6 × 2 = 0 + 0.750 000 005 814 928 225 075 2;
  • 40) 0.750 000 005 814 928 225 075 2 × 2 = 1 + 0.500 000 011 629 856 450 150 4;
  • 41) 0.500 000 011 629 856 450 150 4 × 2 = 1 + 0.000 000 023 259 712 900 300 8;
  • 42) 0.000 000 023 259 712 900 300 8 × 2 = 0 + 0.000 000 046 519 425 800 601 6;
  • 43) 0.000 000 046 519 425 800 601 6 × 2 = 0 + 0.000 000 093 038 851 601 203 2;
  • 44) 0.000 000 093 038 851 601 203 2 × 2 = 0 + 0.000 000 186 077 703 202 406 4;
  • 45) 0.000 000 186 077 703 202 406 4 × 2 = 0 + 0.000 000 372 155 406 404 812 8;
  • 46) 0.000 000 372 155 406 404 812 8 × 2 = 0 + 0.000 000 744 310 812 809 625 6;
  • 47) 0.000 000 744 310 812 809 625 6 × 2 = 0 + 0.000 001 488 621 625 619 251 2;
  • 48) 0.000 001 488 621 625 619 251 2 × 2 = 0 + 0.000 002 977 243 251 238 502 4;
  • 49) 0.000 002 977 243 251 238 502 4 × 2 = 0 + 0.000 005 954 486 502 477 004 8;
  • 50) 0.000 005 954 486 502 477 004 8 × 2 = 0 + 0.000 011 908 973 004 954 009 6;
  • 51) 0.000 011 908 973 004 954 009 6 × 2 = 0 + 0.000 023 817 946 009 908 019 2;
  • 52) 0.000 023 817 946 009 908 019 2 × 2 = 0 + 0.000 047 635 892 019 816 038 4;
  • 53) 0.000 047 635 892 019 816 038 4 × 2 = 0 + 0.000 095 271 784 039 632 076 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).


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:


0.599 999 999 999 909 050 540 4(10) =

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

5. Positive number before normalization:

654.599 999 999 999 909 050 540 4(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:


654.599 999 999 999 909 050 540 4(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 654.599 999 999 999 909 050 540 4 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