0.000 085 532 101 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 085 532 101 9(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
0.000 085 532 101 9(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: 0.000 085 532 101 9.

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) 0.000 085 532 101 9 × 2 = 0 + 0.000 171 064 203 8;
  • 2) 0.000 171 064 203 8 × 2 = 0 + 0.000 342 128 407 6;
  • 3) 0.000 342 128 407 6 × 2 = 0 + 0.000 684 256 815 2;
  • 4) 0.000 684 256 815 2 × 2 = 0 + 0.001 368 513 630 4;
  • 5) 0.001 368 513 630 4 × 2 = 0 + 0.002 737 027 260 8;
  • 6) 0.002 737 027 260 8 × 2 = 0 + 0.005 474 054 521 6;
  • 7) 0.005 474 054 521 6 × 2 = 0 + 0.010 948 109 043 2;
  • 8) 0.010 948 109 043 2 × 2 = 0 + 0.021 896 218 086 4;
  • 9) 0.021 896 218 086 4 × 2 = 0 + 0.043 792 436 172 8;
  • 10) 0.043 792 436 172 8 × 2 = 0 + 0.087 584 872 345 6;
  • 11) 0.087 584 872 345 6 × 2 = 0 + 0.175 169 744 691 2;
  • 12) 0.175 169 744 691 2 × 2 = 0 + 0.350 339 489 382 4;
  • 13) 0.350 339 489 382 4 × 2 = 0 + 0.700 678 978 764 8;
  • 14) 0.700 678 978 764 8 × 2 = 1 + 0.401 357 957 529 6;
  • 15) 0.401 357 957 529 6 × 2 = 0 + 0.802 715 915 059 2;
  • 16) 0.802 715 915 059 2 × 2 = 1 + 0.605 431 830 118 4;
  • 17) 0.605 431 830 118 4 × 2 = 1 + 0.210 863 660 236 8;
  • 18) 0.210 863 660 236 8 × 2 = 0 + 0.421 727 320 473 6;
  • 19) 0.421 727 320 473 6 × 2 = 0 + 0.843 454 640 947 2;
  • 20) 0.843 454 640 947 2 × 2 = 1 + 0.686 909 281 894 4;
  • 21) 0.686 909 281 894 4 × 2 = 1 + 0.373 818 563 788 8;
  • 22) 0.373 818 563 788 8 × 2 = 0 + 0.747 637 127 577 6;
  • 23) 0.747 637 127 577 6 × 2 = 1 + 0.495 274 255 155 2;
  • 24) 0.495 274 255 155 2 × 2 = 0 + 0.990 548 510 310 4;
  • 25) 0.990 548 510 310 4 × 2 = 1 + 0.981 097 020 620 8;
  • 26) 0.981 097 020 620 8 × 2 = 1 + 0.962 194 041 241 6;
  • 27) 0.962 194 041 241 6 × 2 = 1 + 0.924 388 082 483 2;
  • 28) 0.924 388 082 483 2 × 2 = 1 + 0.848 776 164 966 4;
  • 29) 0.848 776 164 966 4 × 2 = 1 + 0.697 552 329 932 8;
  • 30) 0.697 552 329 932 8 × 2 = 1 + 0.395 104 659 865 6;
  • 31) 0.395 104 659 865 6 × 2 = 0 + 0.790 209 319 731 2;
  • 32) 0.790 209 319 731 2 × 2 = 1 + 0.580 418 639 462 4;
  • 33) 0.580 418 639 462 4 × 2 = 1 + 0.160 837 278 924 8;
  • 34) 0.160 837 278 924 8 × 2 = 0 + 0.321 674 557 849 6;
  • 35) 0.321 674 557 849 6 × 2 = 0 + 0.643 349 115 699 2;
  • 36) 0.643 349 115 699 2 × 2 = 1 + 0.286 698 231 398 4;
  • 37) 0.286 698 231 398 4 × 2 = 0 + 0.573 396 462 796 8;
  • 38) 0.573 396 462 796 8 × 2 = 1 + 0.146 792 925 593 6;
  • 39) 0.146 792 925 593 6 × 2 = 0 + 0.293 585 851 187 2;
  • 40) 0.293 585 851 187 2 × 2 = 0 + 0.587 171 702 374 4;
  • 41) 0.587 171 702 374 4 × 2 = 1 + 0.174 343 404 748 8;
  • 42) 0.174 343 404 748 8 × 2 = 0 + 0.348 686 809 497 6;
  • 43) 0.348 686 809 497 6 × 2 = 0 + 0.697 373 618 995 2;
  • 44) 0.697 373 618 995 2 × 2 = 1 + 0.394 747 237 990 4;
  • 45) 0.394 747 237 990 4 × 2 = 0 + 0.789 494 475 980 8;
  • 46) 0.789 494 475 980 8 × 2 = 1 + 0.578 988 951 961 6;
  • 47) 0.578 988 951 961 6 × 2 = 1 + 0.157 977 903 923 2;
  • 48) 0.157 977 903 923 2 × 2 = 0 + 0.315 955 807 846 4;
  • 49) 0.315 955 807 846 4 × 2 = 0 + 0.631 911 615 692 8;
  • 50) 0.631 911 615 692 8 × 2 = 1 + 0.263 823 231 385 6;
  • 51) 0.263 823 231 385 6 × 2 = 0 + 0.527 646 462 771 2;
  • 52) 0.527 646 462 771 2 × 2 = 1 + 0.055 292 925 542 4;
  • 53) 0.055 292 925 542 4 × 2 = 0 + 0.110 585 851 084 8;
  • 54) 0.110 585 851 084 8 × 2 = 0 + 0.221 171 702 169 6;
  • 55) 0.221 171 702 169 6 × 2 = 0 + 0.442 343 404 339 2;
  • 56) 0.442 343 404 339 2 × 2 = 0 + 0.884 686 808 678 4;
  • 57) 0.884 686 808 678 4 × 2 = 1 + 0.769 373 617 356 8;
  • 58) 0.769 373 617 356 8 × 2 = 1 + 0.538 747 234 713 6;
  • 59) 0.538 747 234 713 6 × 2 = 1 + 0.077 494 469 427 2;
  • 60) 0.077 494 469 427 2 × 2 = 0 + 0.154 988 938 854 4;
  • 61) 0.154 988 938 854 4 × 2 = 0 + 0.309 977 877 708 8;
  • 62) 0.309 977 877 708 8 × 2 = 0 + 0.619 955 755 417 6;
  • 63) 0.619 955 755 417 6 × 2 = 1 + 0.239 911 510 835 2;
  • 64) 0.239 911 510 835 2 × 2 = 0 + 0.479 823 021 670 4;
  • 65) 0.479 823 021 670 4 × 2 = 0 + 0.959 646 043 340 8;
  • 66) 0.959 646 043 340 8 × 2 = 1 + 0.919 292 086 681 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:


0.000 085 532 101 9(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1001 0100 1001 0110 0101 0000 1110 0010 01(2)

5. Positive number before normalization:

0.000 085 532 101 9(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1001 0100 1001 0110 0101 0000 1110 0010 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 085 532 101 9(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1001 0100 1001 0110 0101 0000 1110 0010 01(2) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1001 0100 1001 0110 0101 0000 1110 0010 01(2) × 20 =

1.0110 0110 1011 1111 0110 0101 0010 0101 1001 0100 0011 1000 1001(2) × 2-14


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): -14

Mantissa (not normalized):
1.0110 0110 1011 1111 0110 0101 0010 0101 1001 0100 0011 1000 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 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) =

1009(10) =

011 1111 0001(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 0110 1011 1111 0110 0101 0010 0101 1001 0100 0011 1000 1001 =

0110 0110 1011 1111 0110 0101 0010 0101 1001 0100 0011 1000 1001


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 0001

Mantissa (52 bits) =
0110 0110 1011 1111 0110 0101 0010 0101 1001 0100 0011 1000 1001


Decimal number 0.000 085 532 101 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0001 - 0110 0110 1011 1111 0110 0101 0010 0101 1001 0100 0011 1000 1001

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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