-0.042 419 142 317 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 317 5(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
-0.042 419 142 317 5(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. Start with the positive version of the number:

|-0.042 419 142 317 5| = 0.042 419 142 317 5


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.042 419 142 317 5.

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) 0.042 419 142 317 5 × 2 = 0 + 0.084 838 284 635;
  • 2) 0.084 838 284 635 × 2 = 0 + 0.169 676 569 27;
  • 3) 0.169 676 569 27 × 2 = 0 + 0.339 353 138 54;
  • 4) 0.339 353 138 54 × 2 = 0 + 0.678 706 277 08;
  • 5) 0.678 706 277 08 × 2 = 1 + 0.357 412 554 16;
  • 6) 0.357 412 554 16 × 2 = 0 + 0.714 825 108 32;
  • 7) 0.714 825 108 32 × 2 = 1 + 0.429 650 216 64;
  • 8) 0.429 650 216 64 × 2 = 0 + 0.859 300 433 28;
  • 9) 0.859 300 433 28 × 2 = 1 + 0.718 600 866 56;
  • 10) 0.718 600 866 56 × 2 = 1 + 0.437 201 733 12;
  • 11) 0.437 201 733 12 × 2 = 0 + 0.874 403 466 24;
  • 12) 0.874 403 466 24 × 2 = 1 + 0.748 806 932 48;
  • 13) 0.748 806 932 48 × 2 = 1 + 0.497 613 864 96;
  • 14) 0.497 613 864 96 × 2 = 0 + 0.995 227 729 92;
  • 15) 0.995 227 729 92 × 2 = 1 + 0.990 455 459 84;
  • 16) 0.990 455 459 84 × 2 = 1 + 0.980 910 919 68;
  • 17) 0.980 910 919 68 × 2 = 1 + 0.961 821 839 36;
  • 18) 0.961 821 839 36 × 2 = 1 + 0.923 643 678 72;
  • 19) 0.923 643 678 72 × 2 = 1 + 0.847 287 357 44;
  • 20) 0.847 287 357 44 × 2 = 1 + 0.694 574 714 88;
  • 21) 0.694 574 714 88 × 2 = 1 + 0.389 149 429 76;
  • 22) 0.389 149 429 76 × 2 = 0 + 0.778 298 859 52;
  • 23) 0.778 298 859 52 × 2 = 1 + 0.556 597 719 04;
  • 24) 0.556 597 719 04 × 2 = 1 + 0.113 195 438 08;
  • 25) 0.113 195 438 08 × 2 = 0 + 0.226 390 876 16;
  • 26) 0.226 390 876 16 × 2 = 0 + 0.452 781 752 32;
  • 27) 0.452 781 752 32 × 2 = 0 + 0.905 563 504 64;
  • 28) 0.905 563 504 64 × 2 = 1 + 0.811 127 009 28;
  • 29) 0.811 127 009 28 × 2 = 1 + 0.622 254 018 56;
  • 30) 0.622 254 018 56 × 2 = 1 + 0.244 508 037 12;
  • 31) 0.244 508 037 12 × 2 = 0 + 0.489 016 074 24;
  • 32) 0.489 016 074 24 × 2 = 0 + 0.978 032 148 48;
  • 33) 0.978 032 148 48 × 2 = 1 + 0.956 064 296 96;
  • 34) 0.956 064 296 96 × 2 = 1 + 0.912 128 593 92;
  • 35) 0.912 128 593 92 × 2 = 1 + 0.824 257 187 84;
  • 36) 0.824 257 187 84 × 2 = 1 + 0.648 514 375 68;
  • 37) 0.648 514 375 68 × 2 = 1 + 0.297 028 751 36;
  • 38) 0.297 028 751 36 × 2 = 0 + 0.594 057 502 72;
  • 39) 0.594 057 502 72 × 2 = 1 + 0.188 115 005 44;
  • 40) 0.188 115 005 44 × 2 = 0 + 0.376 230 010 88;
  • 41) 0.376 230 010 88 × 2 = 0 + 0.752 460 021 76;
  • 42) 0.752 460 021 76 × 2 = 1 + 0.504 920 043 52;
  • 43) 0.504 920 043 52 × 2 = 1 + 0.009 840 087 04;
  • 44) 0.009 840 087 04 × 2 = 0 + 0.019 680 174 08;
  • 45) 0.019 680 174 08 × 2 = 0 + 0.039 360 348 16;
  • 46) 0.039 360 348 16 × 2 = 0 + 0.078 720 696 32;
  • 47) 0.078 720 696 32 × 2 = 0 + 0.157 441 392 64;
  • 48) 0.157 441 392 64 × 2 = 0 + 0.314 882 785 28;
  • 49) 0.314 882 785 28 × 2 = 0 + 0.629 765 570 56;
  • 50) 0.629 765 570 56 × 2 = 1 + 0.259 531 141 12;
  • 51) 0.259 531 141 12 × 2 = 0 + 0.519 062 282 24;
  • 52) 0.519 062 282 24 × 2 = 1 + 0.038 124 564 48;
  • 53) 0.038 124 564 48 × 2 = 0 + 0.076 249 128 96;
  • 54) 0.076 249 128 96 × 2 = 0 + 0.152 498 257 92;
  • 55) 0.152 498 257 92 × 2 = 0 + 0.304 996 515 84;
  • 56) 0.304 996 515 84 × 2 = 0 + 0.609 993 031 68;
  • 57) 0.609 993 031 68 × 2 = 1 + 0.219 986 063 36;

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).


5. 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:


0.042 419 142 317 5(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1010 0110 0000 0101 0000 1(2)

6. Positive number before normalization:

0.042 419 142 317 5(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1010 0110 0000 0101 0000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.042 419 142 317 5(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1010 0110 0000 0101 0000 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1010 0110 0000 0101 0000 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1111 0100 1100 0000 1010 0001(2) × 2-5


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): -5

Mantissa (not normalized):
1.0101 1011 0111 1111 0110 0011 1001 1111 0100 1100 0000 1010 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(10)


10. 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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1018(10) =

011 1111 1010(2)


12. 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. 0101 1011 0111 1111 0110 0011 1001 1111 0100 1100 0000 1010 0001 =

0101 1011 0111 1111 0110 0011 1001 1111 0100 1100 0000 1010 0001


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1010

Mantissa (52 bits) =
0101 1011 0111 1111 0110 0011 1001 1111 0100 1100 0000 1010 0001


Decimal number -0.042 419 142 317 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1001 1111 0100 1100 0000 1010 0001

» Convert decimal -0.042 419 142 327 4 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