-0.042 419 142 318 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 318 8(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
-0.042 419 142 318 8(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:

|-0.042 419 142 318 8| = 0.042 419 142 318 8


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: 0.042 419 142 318 8.

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) 0.042 419 142 318 8 × 2 = 0 + 0.084 838 284 637 6;
  • 2) 0.084 838 284 637 6 × 2 = 0 + 0.169 676 569 275 2;
  • 3) 0.169 676 569 275 2 × 2 = 0 + 0.339 353 138 550 4;
  • 4) 0.339 353 138 550 4 × 2 = 0 + 0.678 706 277 100 8;
  • 5) 0.678 706 277 100 8 × 2 = 1 + 0.357 412 554 201 6;
  • 6) 0.357 412 554 201 6 × 2 = 0 + 0.714 825 108 403 2;
  • 7) 0.714 825 108 403 2 × 2 = 1 + 0.429 650 216 806 4;
  • 8) 0.429 650 216 806 4 × 2 = 0 + 0.859 300 433 612 8;
  • 9) 0.859 300 433 612 8 × 2 = 1 + 0.718 600 867 225 6;
  • 10) 0.718 600 867 225 6 × 2 = 1 + 0.437 201 734 451 2;
  • 11) 0.437 201 734 451 2 × 2 = 0 + 0.874 403 468 902 4;
  • 12) 0.874 403 468 902 4 × 2 = 1 + 0.748 806 937 804 8;
  • 13) 0.748 806 937 804 8 × 2 = 1 + 0.497 613 875 609 6;
  • 14) 0.497 613 875 609 6 × 2 = 0 + 0.995 227 751 219 2;
  • 15) 0.995 227 751 219 2 × 2 = 1 + 0.990 455 502 438 4;
  • 16) 0.990 455 502 438 4 × 2 = 1 + 0.980 911 004 876 8;
  • 17) 0.980 911 004 876 8 × 2 = 1 + 0.961 822 009 753 6;
  • 18) 0.961 822 009 753 6 × 2 = 1 + 0.923 644 019 507 2;
  • 19) 0.923 644 019 507 2 × 2 = 1 + 0.847 288 039 014 4;
  • 20) 0.847 288 039 014 4 × 2 = 1 + 0.694 576 078 028 8;
  • 21) 0.694 576 078 028 8 × 2 = 1 + 0.389 152 156 057 6;
  • 22) 0.389 152 156 057 6 × 2 = 0 + 0.778 304 312 115 2;
  • 23) 0.778 304 312 115 2 × 2 = 1 + 0.556 608 624 230 4;
  • 24) 0.556 608 624 230 4 × 2 = 1 + 0.113 217 248 460 8;
  • 25) 0.113 217 248 460 8 × 2 = 0 + 0.226 434 496 921 6;
  • 26) 0.226 434 496 921 6 × 2 = 0 + 0.452 868 993 843 2;
  • 27) 0.452 868 993 843 2 × 2 = 0 + 0.905 737 987 686 4;
  • 28) 0.905 737 987 686 4 × 2 = 1 + 0.811 475 975 372 8;
  • 29) 0.811 475 975 372 8 × 2 = 1 + 0.622 951 950 745 6;
  • 30) 0.622 951 950 745 6 × 2 = 1 + 0.245 903 901 491 2;
  • 31) 0.245 903 901 491 2 × 2 = 0 + 0.491 807 802 982 4;
  • 32) 0.491 807 802 982 4 × 2 = 0 + 0.983 615 605 964 8;
  • 33) 0.983 615 605 964 8 × 2 = 1 + 0.967 231 211 929 6;
  • 34) 0.967 231 211 929 6 × 2 = 1 + 0.934 462 423 859 2;
  • 35) 0.934 462 423 859 2 × 2 = 1 + 0.868 924 847 718 4;
  • 36) 0.868 924 847 718 4 × 2 = 1 + 0.737 849 695 436 8;
  • 37) 0.737 849 695 436 8 × 2 = 1 + 0.475 699 390 873 6;
  • 38) 0.475 699 390 873 6 × 2 = 0 + 0.951 398 781 747 2;
  • 39) 0.951 398 781 747 2 × 2 = 1 + 0.902 797 563 494 4;
  • 40) 0.902 797 563 494 4 × 2 = 1 + 0.805 595 126 988 8;
  • 41) 0.805 595 126 988 8 × 2 = 1 + 0.611 190 253 977 6;
  • 42) 0.611 190 253 977 6 × 2 = 1 + 0.222 380 507 955 2;
  • 43) 0.222 380 507 955 2 × 2 = 0 + 0.444 761 015 910 4;
  • 44) 0.444 761 015 910 4 × 2 = 0 + 0.889 522 031 820 8;
  • 45) 0.889 522 031 820 8 × 2 = 1 + 0.779 044 063 641 6;
  • 46) 0.779 044 063 641 6 × 2 = 1 + 0.558 088 127 283 2;
  • 47) 0.558 088 127 283 2 × 2 = 1 + 0.116 176 254 566 4;
  • 48) 0.116 176 254 566 4 × 2 = 0 + 0.232 352 509 132 8;
  • 49) 0.232 352 509 132 8 × 2 = 0 + 0.464 705 018 265 6;
  • 50) 0.464 705 018 265 6 × 2 = 0 + 0.929 410 036 531 2;
  • 51) 0.929 410 036 531 2 × 2 = 1 + 0.858 820 073 062 4;
  • 52) 0.858 820 073 062 4 × 2 = 1 + 0.717 640 146 124 8;
  • 53) 0.717 640 146 124 8 × 2 = 1 + 0.435 280 292 249 6;
  • 54) 0.435 280 292 249 6 × 2 = 0 + 0.870 560 584 499 2;
  • 55) 0.870 560 584 499 2 × 2 = 1 + 0.741 121 168 998 4;
  • 56) 0.741 121 168 998 4 × 2 = 1 + 0.482 242 337 996 8;
  • 57) 0.482 242 337 996 8 × 2 = 0 + 0.964 484 675 993 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).


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:


0.042 419 142 318 8(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1011 1100 1110 0011 1011 0(2)

6. Positive number before normalization:

0.042 419 142 318 8(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1011 1100 1110 0011 1011 0(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:


0.042 419 142 318 8(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1011 1100 1110 0011 1011 0(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 1011 1100 1110 0011 1011 0(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1111 0111 1001 1100 0111 0110(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 0111 1001 1100 0111 0110


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 0111 1001 1100 0111 0110 =

0101 1011 0111 1111 0110 0011 1001 1111 0111 1001 1100 0111 0110


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 0111 1001 1100 0111 0110


Decimal number -0.042 419 142 318 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

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