0.000 043 757 252 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 757 252 7(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
0.000 043 757 252 7(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 043 757 252 7.

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) 0.000 043 757 252 7 × 2 = 0 + 0.000 087 514 505 4;
  • 2) 0.000 087 514 505 4 × 2 = 0 + 0.000 175 029 010 8;
  • 3) 0.000 175 029 010 8 × 2 = 0 + 0.000 350 058 021 6;
  • 4) 0.000 350 058 021 6 × 2 = 0 + 0.000 700 116 043 2;
  • 5) 0.000 700 116 043 2 × 2 = 0 + 0.001 400 232 086 4;
  • 6) 0.001 400 232 086 4 × 2 = 0 + 0.002 800 464 172 8;
  • 7) 0.002 800 464 172 8 × 2 = 0 + 0.005 600 928 345 6;
  • 8) 0.005 600 928 345 6 × 2 = 0 + 0.011 201 856 691 2;
  • 9) 0.011 201 856 691 2 × 2 = 0 + 0.022 403 713 382 4;
  • 10) 0.022 403 713 382 4 × 2 = 0 + 0.044 807 426 764 8;
  • 11) 0.044 807 426 764 8 × 2 = 0 + 0.089 614 853 529 6;
  • 12) 0.089 614 853 529 6 × 2 = 0 + 0.179 229 707 059 2;
  • 13) 0.179 229 707 059 2 × 2 = 0 + 0.358 459 414 118 4;
  • 14) 0.358 459 414 118 4 × 2 = 0 + 0.716 918 828 236 8;
  • 15) 0.716 918 828 236 8 × 2 = 1 + 0.433 837 656 473 6;
  • 16) 0.433 837 656 473 6 × 2 = 0 + 0.867 675 312 947 2;
  • 17) 0.867 675 312 947 2 × 2 = 1 + 0.735 350 625 894 4;
  • 18) 0.735 350 625 894 4 × 2 = 1 + 0.470 701 251 788 8;
  • 19) 0.470 701 251 788 8 × 2 = 0 + 0.941 402 503 577 6;
  • 20) 0.941 402 503 577 6 × 2 = 1 + 0.882 805 007 155 2;
  • 21) 0.882 805 007 155 2 × 2 = 1 + 0.765 610 014 310 4;
  • 22) 0.765 610 014 310 4 × 2 = 1 + 0.531 220 028 620 8;
  • 23) 0.531 220 028 620 8 × 2 = 1 + 0.062 440 057 241 6;
  • 24) 0.062 440 057 241 6 × 2 = 0 + 0.124 880 114 483 2;
  • 25) 0.124 880 114 483 2 × 2 = 0 + 0.249 760 228 966 4;
  • 26) 0.249 760 228 966 4 × 2 = 0 + 0.499 520 457 932 8;
  • 27) 0.499 520 457 932 8 × 2 = 0 + 0.999 040 915 865 6;
  • 28) 0.999 040 915 865 6 × 2 = 1 + 0.998 081 831 731 2;
  • 29) 0.998 081 831 731 2 × 2 = 1 + 0.996 163 663 462 4;
  • 30) 0.996 163 663 462 4 × 2 = 1 + 0.992 327 326 924 8;
  • 31) 0.992 327 326 924 8 × 2 = 1 + 0.984 654 653 849 6;
  • 32) 0.984 654 653 849 6 × 2 = 1 + 0.969 309 307 699 2;
  • 33) 0.969 309 307 699 2 × 2 = 1 + 0.938 618 615 398 4;
  • 34) 0.938 618 615 398 4 × 2 = 1 + 0.877 237 230 796 8;
  • 35) 0.877 237 230 796 8 × 2 = 1 + 0.754 474 461 593 6;
  • 36) 0.754 474 461 593 6 × 2 = 1 + 0.508 948 923 187 2;
  • 37) 0.508 948 923 187 2 × 2 = 1 + 0.017 897 846 374 4;
  • 38) 0.017 897 846 374 4 × 2 = 0 + 0.035 795 692 748 8;
  • 39) 0.035 795 692 748 8 × 2 = 0 + 0.071 591 385 497 6;
  • 40) 0.071 591 385 497 6 × 2 = 0 + 0.143 182 770 995 2;
  • 41) 0.143 182 770 995 2 × 2 = 0 + 0.286 365 541 990 4;
  • 42) 0.286 365 541 990 4 × 2 = 0 + 0.572 731 083 980 8;
  • 43) 0.572 731 083 980 8 × 2 = 1 + 0.145 462 167 961 6;
  • 44) 0.145 462 167 961 6 × 2 = 0 + 0.290 924 335 923 2;
  • 45) 0.290 924 335 923 2 × 2 = 0 + 0.581 848 671 846 4;
  • 46) 0.581 848 671 846 4 × 2 = 1 + 0.163 697 343 692 8;
  • 47) 0.163 697 343 692 8 × 2 = 0 + 0.327 394 687 385 6;
  • 48) 0.327 394 687 385 6 × 2 = 0 + 0.654 789 374 771 2;
  • 49) 0.654 789 374 771 2 × 2 = 1 + 0.309 578 749 542 4;
  • 50) 0.309 578 749 542 4 × 2 = 0 + 0.619 157 499 084 8;
  • 51) 0.619 157 499 084 8 × 2 = 1 + 0.238 314 998 169 6;
  • 52) 0.238 314 998 169 6 × 2 = 0 + 0.476 629 996 339 2;
  • 53) 0.476 629 996 339 2 × 2 = 0 + 0.953 259 992 678 4;
  • 54) 0.953 259 992 678 4 × 2 = 1 + 0.906 519 985 356 8;
  • 55) 0.906 519 985 356 8 × 2 = 1 + 0.813 039 970 713 6;
  • 56) 0.813 039 970 713 6 × 2 = 1 + 0.626 079 941 427 2;
  • 57) 0.626 079 941 427 2 × 2 = 1 + 0.252 159 882 854 4;
  • 58) 0.252 159 882 854 4 × 2 = 0 + 0.504 319 765 708 8;
  • 59) 0.504 319 765 708 8 × 2 = 1 + 0.008 639 531 417 6;
  • 60) 0.008 639 531 417 6 × 2 = 0 + 0.017 279 062 835 2;
  • 61) 0.017 279 062 835 2 × 2 = 0 + 0.034 558 125 670 4;
  • 62) 0.034 558 125 670 4 × 2 = 0 + 0.069 116 251 340 8;
  • 63) 0.069 116 251 340 8 × 2 = 0 + 0.138 232 502 681 6;
  • 64) 0.138 232 502 681 6 × 2 = 0 + 0.276 465 005 363 2;
  • 65) 0.276 465 005 363 2 × 2 = 0 + 0.552 930 010 726 4;
  • 66) 0.552 930 010 726 4 × 2 = 1 + 0.105 860 021 452 8;
  • 67) 0.105 860 021 452 8 × 2 = 0 + 0.211 720 042 905 6;

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:


0.000 043 757 252 7(10) =

0.0000 0000 0000 0010 1101 1110 0001 1111 1111 1000 0010 0100 1010 0111 1010 0000 010(2)

5. Positive number before normalization:

0.000 043 757 252 7(10) =

0.0000 0000 0000 0010 1101 1110 0001 1111 1111 1000 0010 0100 1010 0111 1010 0000 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 043 757 252 7(10) =

0.0000 0000 0000 0010 1101 1110 0001 1111 1111 1000 0010 0100 1010 0111 1010 0000 010(2) =

0.0000 0000 0000 0010 1101 1110 0001 1111 1111 1000 0010 0100 1010 0111 1010 0000 010(2) × 20 =

1.0110 1111 0000 1111 1111 1100 0001 0010 0101 0011 1101 0000 0010(2) × 2-15


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): -15

Mantissa (not normalized):
1.0110 1111 0000 1111 1111 1100 0001 0010 0101 0011 1101 0000 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1008(10) =

011 1111 0000(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0110 1111 0000 1111 1111 1100 0001 0010 0101 0011 1101 0000 0010 =

0110 1111 0000 1111 1111 1100 0001 0010 0101 0011 1101 0000 0010


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) =
011 1111 0000

Mantissa (52 bits) =
0110 1111 0000 1111 1111 1100 0001 0010 0101 0011 1101 0000 0010


Decimal number 0.000 043 757 252 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0000 1111 1111 1100 0001 0010 0101 0011 1101 0000 0010

» Convert decimal 0.000 043 757 252 9 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