0.995 404 958 852 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.995 404 958 852(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
0.995 404 958 852(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: 0.995 404 958 852.

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) 0.995 404 958 852 × 2 = 1 + 0.990 809 917 704;
  • 2) 0.990 809 917 704 × 2 = 1 + 0.981 619 835 408;
  • 3) 0.981 619 835 408 × 2 = 1 + 0.963 239 670 816;
  • 4) 0.963 239 670 816 × 2 = 1 + 0.926 479 341 632;
  • 5) 0.926 479 341 632 × 2 = 1 + 0.852 958 683 264;
  • 6) 0.852 958 683 264 × 2 = 1 + 0.705 917 366 528;
  • 7) 0.705 917 366 528 × 2 = 1 + 0.411 834 733 056;
  • 8) 0.411 834 733 056 × 2 = 0 + 0.823 669 466 112;
  • 9) 0.823 669 466 112 × 2 = 1 + 0.647 338 932 224;
  • 10) 0.647 338 932 224 × 2 = 1 + 0.294 677 864 448;
  • 11) 0.294 677 864 448 × 2 = 0 + 0.589 355 728 896;
  • 12) 0.589 355 728 896 × 2 = 1 + 0.178 711 457 792;
  • 13) 0.178 711 457 792 × 2 = 0 + 0.357 422 915 584;
  • 14) 0.357 422 915 584 × 2 = 0 + 0.714 845 831 168;
  • 15) 0.714 845 831 168 × 2 = 1 + 0.429 691 662 336;
  • 16) 0.429 691 662 336 × 2 = 0 + 0.859 383 324 672;
  • 17) 0.859 383 324 672 × 2 = 1 + 0.718 766 649 344;
  • 18) 0.718 766 649 344 × 2 = 1 + 0.437 533 298 688;
  • 19) 0.437 533 298 688 × 2 = 0 + 0.875 066 597 376;
  • 20) 0.875 066 597 376 × 2 = 1 + 0.750 133 194 752;
  • 21) 0.750 133 194 752 × 2 = 1 + 0.500 266 389 504;
  • 22) 0.500 266 389 504 × 2 = 1 + 0.000 532 779 008;
  • 23) 0.000 532 779 008 × 2 = 0 + 0.001 065 558 016;
  • 24) 0.001 065 558 016 × 2 = 0 + 0.002 131 116 032;
  • 25) 0.002 131 116 032 × 2 = 0 + 0.004 262 232 064;
  • 26) 0.004 262 232 064 × 2 = 0 + 0.008 524 464 128;
  • 27) 0.008 524 464 128 × 2 = 0 + 0.017 048 928 256;
  • 28) 0.017 048 928 256 × 2 = 0 + 0.034 097 856 512;
  • 29) 0.034 097 856 512 × 2 = 0 + 0.068 195 713 024;
  • 30) 0.068 195 713 024 × 2 = 0 + 0.136 391 426 048;
  • 31) 0.136 391 426 048 × 2 = 0 + 0.272 782 852 096;
  • 32) 0.272 782 852 096 × 2 = 0 + 0.545 565 704 192;
  • 33) 0.545 565 704 192 × 2 = 1 + 0.091 131 408 384;
  • 34) 0.091 131 408 384 × 2 = 0 + 0.182 262 816 768;
  • 35) 0.182 262 816 768 × 2 = 0 + 0.364 525 633 536;
  • 36) 0.364 525 633 536 × 2 = 0 + 0.729 051 267 072;
  • 37) 0.729 051 267 072 × 2 = 1 + 0.458 102 534 144;
  • 38) 0.458 102 534 144 × 2 = 0 + 0.916 205 068 288;
  • 39) 0.916 205 068 288 × 2 = 1 + 0.832 410 136 576;
  • 40) 0.832 410 136 576 × 2 = 1 + 0.664 820 273 152;
  • 41) 0.664 820 273 152 × 2 = 1 + 0.329 640 546 304;
  • 42) 0.329 640 546 304 × 2 = 0 + 0.659 281 092 608;
  • 43) 0.659 281 092 608 × 2 = 1 + 0.318 562 185 216;
  • 44) 0.318 562 185 216 × 2 = 0 + 0.637 124 370 432;
  • 45) 0.637 124 370 432 × 2 = 1 + 0.274 248 740 864;
  • 46) 0.274 248 740 864 × 2 = 0 + 0.548 497 481 728;
  • 47) 0.548 497 481 728 × 2 = 1 + 0.096 994 963 456;
  • 48) 0.096 994 963 456 × 2 = 0 + 0.193 989 926 912;
  • 49) 0.193 989 926 912 × 2 = 0 + 0.387 979 853 824;
  • 50) 0.387 979 853 824 × 2 = 0 + 0.775 959 707 648;
  • 51) 0.775 959 707 648 × 2 = 1 + 0.551 919 415 296;
  • 52) 0.551 919 415 296 × 2 = 1 + 0.103 838 830 592;
  • 53) 0.103 838 830 592 × 2 = 0 + 0.207 677 661 184;

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:


0.995 404 958 852(10) =

0.1111 1110 1101 0010 1101 1100 0000 0000 1000 1011 1010 1010 0011 0(2)

5. Positive number before normalization:

0.995 404 958 852(10) =

0.1111 1110 1101 0010 1101 1100 0000 0000 1000 1011 1010 1010 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.995 404 958 852(10) =

0.1111 1110 1101 0010 1101 1100 0000 0000 1000 1011 1010 1010 0011 0(2) =

0.1111 1110 1101 0010 1101 1100 0000 0000 1000 1011 1010 1010 0011 0(2) × 20 =

1.1111 1101 1010 0101 1011 1000 0000 0001 0001 0111 0101 0100 0110(2) × 2-1


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): -1

Mantissa (not normalized):
1.1111 1101 1010 0101 1011 1000 0000 0001 0001 0111 0101 0100 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 1111 1101 1010 0101 1011 1000 0000 0001 0001 0111 0101 0100 0110 =

1111 1101 1010 0101 1011 1000 0000 0001 0001 0111 0101 0100 0110


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 1110

Mantissa (52 bits) =
1111 1101 1010 0101 1011 1000 0000 0001 0001 0111 0101 0100 0110


Decimal number 0.995 404 958 852 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 1101 1010 0101 1011 1000 0000 0001 0001 0111 0101 0100 0110

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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