-0.057 313 943 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.057 313 943 7(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
-0.057 313 943 7(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:

|-0.057 313 943 7| = 0.057 313 943 7


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: 0.057 313 943 7.

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) 0.057 313 943 7 × 2 = 0 + 0.114 627 887 4;
  • 2) 0.114 627 887 4 × 2 = 0 + 0.229 255 774 8;
  • 3) 0.229 255 774 8 × 2 = 0 + 0.458 511 549 6;
  • 4) 0.458 511 549 6 × 2 = 0 + 0.917 023 099 2;
  • 5) 0.917 023 099 2 × 2 = 1 + 0.834 046 198 4;
  • 6) 0.834 046 198 4 × 2 = 1 + 0.668 092 396 8;
  • 7) 0.668 092 396 8 × 2 = 1 + 0.336 184 793 6;
  • 8) 0.336 184 793 6 × 2 = 0 + 0.672 369 587 2;
  • 9) 0.672 369 587 2 × 2 = 1 + 0.344 739 174 4;
  • 10) 0.344 739 174 4 × 2 = 0 + 0.689 478 348 8;
  • 11) 0.689 478 348 8 × 2 = 1 + 0.378 956 697 6;
  • 12) 0.378 956 697 6 × 2 = 0 + 0.757 913 395 2;
  • 13) 0.757 913 395 2 × 2 = 1 + 0.515 826 790 4;
  • 14) 0.515 826 790 4 × 2 = 1 + 0.031 653 580 8;
  • 15) 0.031 653 580 8 × 2 = 0 + 0.063 307 161 6;
  • 16) 0.063 307 161 6 × 2 = 0 + 0.126 614 323 2;
  • 17) 0.126 614 323 2 × 2 = 0 + 0.253 228 646 4;
  • 18) 0.253 228 646 4 × 2 = 0 + 0.506 457 292 8;
  • 19) 0.506 457 292 8 × 2 = 1 + 0.012 914 585 6;
  • 20) 0.012 914 585 6 × 2 = 0 + 0.025 829 171 2;
  • 21) 0.025 829 171 2 × 2 = 0 + 0.051 658 342 4;
  • 22) 0.051 658 342 4 × 2 = 0 + 0.103 316 684 8;
  • 23) 0.103 316 684 8 × 2 = 0 + 0.206 633 369 6;
  • 24) 0.206 633 369 6 × 2 = 0 + 0.413 266 739 2;
  • 25) 0.413 266 739 2 × 2 = 0 + 0.826 533 478 4;
  • 26) 0.826 533 478 4 × 2 = 1 + 0.653 066 956 8;
  • 27) 0.653 066 956 8 × 2 = 1 + 0.306 133 913 6;
  • 28) 0.306 133 913 6 × 2 = 0 + 0.612 267 827 2;
  • 29) 0.612 267 827 2 × 2 = 1 + 0.224 535 654 4;
  • 30) 0.224 535 654 4 × 2 = 0 + 0.449 071 308 8;
  • 31) 0.449 071 308 8 × 2 = 0 + 0.898 142 617 6;
  • 32) 0.898 142 617 6 × 2 = 1 + 0.796 285 235 2;
  • 33) 0.796 285 235 2 × 2 = 1 + 0.592 570 470 4;
  • 34) 0.592 570 470 4 × 2 = 1 + 0.185 140 940 8;
  • 35) 0.185 140 940 8 × 2 = 0 + 0.370 281 881 6;
  • 36) 0.370 281 881 6 × 2 = 0 + 0.740 563 763 2;
  • 37) 0.740 563 763 2 × 2 = 1 + 0.481 127 526 4;
  • 38) 0.481 127 526 4 × 2 = 0 + 0.962 255 052 8;
  • 39) 0.962 255 052 8 × 2 = 1 + 0.924 510 105 6;
  • 40) 0.924 510 105 6 × 2 = 1 + 0.849 020 211 2;
  • 41) 0.849 020 211 2 × 2 = 1 + 0.698 040 422 4;
  • 42) 0.698 040 422 4 × 2 = 1 + 0.396 080 844 8;
  • 43) 0.396 080 844 8 × 2 = 0 + 0.792 161 689 6;
  • 44) 0.792 161 689 6 × 2 = 1 + 0.584 323 379 2;
  • 45) 0.584 323 379 2 × 2 = 1 + 0.168 646 758 4;
  • 46) 0.168 646 758 4 × 2 = 0 + 0.337 293 516 8;
  • 47) 0.337 293 516 8 × 2 = 0 + 0.674 587 033 6;
  • 48) 0.674 587 033 6 × 2 = 1 + 0.349 174 067 2;
  • 49) 0.349 174 067 2 × 2 = 0 + 0.698 348 134 4;
  • 50) 0.698 348 134 4 × 2 = 1 + 0.396 696 268 8;
  • 51) 0.396 696 268 8 × 2 = 0 + 0.793 392 537 6;
  • 52) 0.793 392 537 6 × 2 = 1 + 0.586 785 075 2;
  • 53) 0.586 785 075 2 × 2 = 1 + 0.173 570 150 4;
  • 54) 0.173 570 150 4 × 2 = 0 + 0.347 140 300 8;
  • 55) 0.347 140 300 8 × 2 = 0 + 0.694 280 601 6;
  • 56) 0.694 280 601 6 × 2 = 1 + 0.388 561 203 2;
  • 57) 0.388 561 203 2 × 2 = 0 + 0.777 122 406 4;

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:


0.057 313 943 7(10) =

0.0000 1110 1010 1100 0010 0000 0110 1001 1100 1011 1101 1001 0101 1001 0(2)

6. Positive number before normalization:

0.057 313 943 7(10) =

0.0000 1110 1010 1100 0010 0000 0110 1001 1100 1011 1101 1001 0101 1001 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:


0.057 313 943 7(10) =

0.0000 1110 1010 1100 0010 0000 0110 1001 1100 1011 1101 1001 0101 1001 0(2) =

0.0000 1110 1010 1100 0010 0000 0110 1001 1100 1011 1101 1001 0101 1001 0(2) × 20 =

1.1101 0101 1000 0100 0000 1101 0011 1001 0111 1011 0010 1011 0010(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):
1.1101 0101 1000 0100 0000 1101 0011 1001 0111 1011 0010 1011 0010


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. 1101 0101 1000 0100 0000 1101 0011 1001 0111 1011 0010 1011 0010 =

1101 0101 1000 0100 0000 1101 0011 1001 0111 1011 0010 1011 0010


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) =
1101 0101 1000 0100 0000 1101 0011 1001 0111 1011 0010 1011 0010


Decimal number -0.057 313 943 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1101 0101 1000 0100 0000 1101 0011 1001 0111 1011 0010 1011 0010

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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