0.995 404 958 724 974 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.995 404 958 724 974(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
0.995 404 958 724 974(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: 0.995 404 958 724 974.

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) 0.995 404 958 724 974 × 2 = 1 + 0.990 809 917 449 948;
  • 2) 0.990 809 917 449 948 × 2 = 1 + 0.981 619 834 899 896;
  • 3) 0.981 619 834 899 896 × 2 = 1 + 0.963 239 669 799 792;
  • 4) 0.963 239 669 799 792 × 2 = 1 + 0.926 479 339 599 584;
  • 5) 0.926 479 339 599 584 × 2 = 1 + 0.852 958 679 199 168;
  • 6) 0.852 958 679 199 168 × 2 = 1 + 0.705 917 358 398 336;
  • 7) 0.705 917 358 398 336 × 2 = 1 + 0.411 834 716 796 672;
  • 8) 0.411 834 716 796 672 × 2 = 0 + 0.823 669 433 593 344;
  • 9) 0.823 669 433 593 344 × 2 = 1 + 0.647 338 867 186 688;
  • 10) 0.647 338 867 186 688 × 2 = 1 + 0.294 677 734 373 376;
  • 11) 0.294 677 734 373 376 × 2 = 0 + 0.589 355 468 746 752;
  • 12) 0.589 355 468 746 752 × 2 = 1 + 0.178 710 937 493 504;
  • 13) 0.178 710 937 493 504 × 2 = 0 + 0.357 421 874 987 008;
  • 14) 0.357 421 874 987 008 × 2 = 0 + 0.714 843 749 974 016;
  • 15) 0.714 843 749 974 016 × 2 = 1 + 0.429 687 499 948 032;
  • 16) 0.429 687 499 948 032 × 2 = 0 + 0.859 374 999 896 064;
  • 17) 0.859 374 999 896 064 × 2 = 1 + 0.718 749 999 792 128;
  • 18) 0.718 749 999 792 128 × 2 = 1 + 0.437 499 999 584 256;
  • 19) 0.437 499 999 584 256 × 2 = 0 + 0.874 999 999 168 512;
  • 20) 0.874 999 999 168 512 × 2 = 1 + 0.749 999 998 337 024;
  • 21) 0.749 999 998 337 024 × 2 = 1 + 0.499 999 996 674 048;
  • 22) 0.499 999 996 674 048 × 2 = 0 + 0.999 999 993 348 096;
  • 23) 0.999 999 993 348 096 × 2 = 1 + 0.999 999 986 696 192;
  • 24) 0.999 999 986 696 192 × 2 = 1 + 0.999 999 973 392 384;
  • 25) 0.999 999 973 392 384 × 2 = 1 + 0.999 999 946 784 768;
  • 26) 0.999 999 946 784 768 × 2 = 1 + 0.999 999 893 569 536;
  • 27) 0.999 999 893 569 536 × 2 = 1 + 0.999 999 787 139 072;
  • 28) 0.999 999 787 139 072 × 2 = 1 + 0.999 999 574 278 144;
  • 29) 0.999 999 574 278 144 × 2 = 1 + 0.999 999 148 556 288;
  • 30) 0.999 999 148 556 288 × 2 = 1 + 0.999 998 297 112 576;
  • 31) 0.999 998 297 112 576 × 2 = 1 + 0.999 996 594 225 152;
  • 32) 0.999 996 594 225 152 × 2 = 1 + 0.999 993 188 450 304;
  • 33) 0.999 993 188 450 304 × 2 = 1 + 0.999 986 376 900 608;
  • 34) 0.999 986 376 900 608 × 2 = 1 + 0.999 972 753 801 216;
  • 35) 0.999 972 753 801 216 × 2 = 1 + 0.999 945 507 602 432;
  • 36) 0.999 945 507 602 432 × 2 = 1 + 0.999 891 015 204 864;
  • 37) 0.999 891 015 204 864 × 2 = 1 + 0.999 782 030 409 728;
  • 38) 0.999 782 030 409 728 × 2 = 1 + 0.999 564 060 819 456;
  • 39) 0.999 564 060 819 456 × 2 = 1 + 0.999 128 121 638 912;
  • 40) 0.999 128 121 638 912 × 2 = 1 + 0.998 256 243 277 824;
  • 41) 0.998 256 243 277 824 × 2 = 1 + 0.996 512 486 555 648;
  • 42) 0.996 512 486 555 648 × 2 = 1 + 0.993 024 973 111 296;
  • 43) 0.993 024 973 111 296 × 2 = 1 + 0.986 049 946 222 592;
  • 44) 0.986 049 946 222 592 × 2 = 1 + 0.972 099 892 445 184;
  • 45) 0.972 099 892 445 184 × 2 = 1 + 0.944 199 784 890 368;
  • 46) 0.944 199 784 890 368 × 2 = 1 + 0.888 399 569 780 736;
  • 47) 0.888 399 569 780 736 × 2 = 1 + 0.776 799 139 561 472;
  • 48) 0.776 799 139 561 472 × 2 = 1 + 0.553 598 279 122 944;
  • 49) 0.553 598 279 122 944 × 2 = 1 + 0.107 196 558 245 888;
  • 50) 0.107 196 558 245 888 × 2 = 0 + 0.214 393 116 491 776;
  • 51) 0.214 393 116 491 776 × 2 = 0 + 0.428 786 232 983 552;
  • 52) 0.428 786 232 983 552 × 2 = 0 + 0.857 572 465 967 104;
  • 53) 0.857 572 465 967 104 × 2 = 1 + 0.715 144 931 934 208;

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:


0.995 404 958 724 974(10) =

0.1111 1110 1101 0010 1101 1011 1111 1111 1111 1111 1111 1111 1000 1(2)

5. Positive number before normalization:

0.995 404 958 724 974(10) =

0.1111 1110 1101 0010 1101 1011 1111 1111 1111 1111 1111 1111 1000 1(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:


0.995 404 958 724 974(10) =

0.1111 1110 1101 0010 1101 1011 1111 1111 1111 1111 1111 1111 1000 1(2) =

0.1111 1110 1101 0010 1101 1011 1111 1111 1111 1111 1111 1111 1000 1(2) × 20 =

1.1111 1101 1010 0101 1011 0111 1111 1111 1111 1111 1111 1111 0001(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 0111 1111 1111 1111 1111 1111 1111 0001


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 0111 1111 1111 1111 1111 1111 1111 0001 =

1111 1101 1010 0101 1011 0111 1111 1111 1111 1111 1111 1111 0001


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 0111 1111 1111 1111 1111 1111 1111 0001


Decimal number 0.995 404 958 724 974 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 1101 1010 0101 1011 0111 1111 1111 1111 1111 1111 1111 0001

» Convert decimal 0.995 404 958 724 922 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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