-0.014 657 790 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.014 657 790 12(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
-0.014 657 790 12(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:

|-0.014 657 790 12| = 0.014 657 790 12


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: 0.014 657 790 12.

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) 0.014 657 790 12 × 2 = 0 + 0.029 315 580 24;
  • 2) 0.029 315 580 24 × 2 = 0 + 0.058 631 160 48;
  • 3) 0.058 631 160 48 × 2 = 0 + 0.117 262 320 96;
  • 4) 0.117 262 320 96 × 2 = 0 + 0.234 524 641 92;
  • 5) 0.234 524 641 92 × 2 = 0 + 0.469 049 283 84;
  • 6) 0.469 049 283 84 × 2 = 0 + 0.938 098 567 68;
  • 7) 0.938 098 567 68 × 2 = 1 + 0.876 197 135 36;
  • 8) 0.876 197 135 36 × 2 = 1 + 0.752 394 270 72;
  • 9) 0.752 394 270 72 × 2 = 1 + 0.504 788 541 44;
  • 10) 0.504 788 541 44 × 2 = 1 + 0.009 577 082 88;
  • 11) 0.009 577 082 88 × 2 = 0 + 0.019 154 165 76;
  • 12) 0.019 154 165 76 × 2 = 0 + 0.038 308 331 52;
  • 13) 0.038 308 331 52 × 2 = 0 + 0.076 616 663 04;
  • 14) 0.076 616 663 04 × 2 = 0 + 0.153 233 326 08;
  • 15) 0.153 233 326 08 × 2 = 0 + 0.306 466 652 16;
  • 16) 0.306 466 652 16 × 2 = 0 + 0.612 933 304 32;
  • 17) 0.612 933 304 32 × 2 = 1 + 0.225 866 608 64;
  • 18) 0.225 866 608 64 × 2 = 0 + 0.451 733 217 28;
  • 19) 0.451 733 217 28 × 2 = 0 + 0.903 466 434 56;
  • 20) 0.903 466 434 56 × 2 = 1 + 0.806 932 869 12;
  • 21) 0.806 932 869 12 × 2 = 1 + 0.613 865 738 24;
  • 22) 0.613 865 738 24 × 2 = 1 + 0.227 731 476 48;
  • 23) 0.227 731 476 48 × 2 = 0 + 0.455 462 952 96;
  • 24) 0.455 462 952 96 × 2 = 0 + 0.910 925 905 92;
  • 25) 0.910 925 905 92 × 2 = 1 + 0.821 851 811 84;
  • 26) 0.821 851 811 84 × 2 = 1 + 0.643 703 623 68;
  • 27) 0.643 703 623 68 × 2 = 1 + 0.287 407 247 36;
  • 28) 0.287 407 247 36 × 2 = 0 + 0.574 814 494 72;
  • 29) 0.574 814 494 72 × 2 = 1 + 0.149 628 989 44;
  • 30) 0.149 628 989 44 × 2 = 0 + 0.299 257 978 88;
  • 31) 0.299 257 978 88 × 2 = 0 + 0.598 515 957 76;
  • 32) 0.598 515 957 76 × 2 = 1 + 0.197 031 915 52;
  • 33) 0.197 031 915 52 × 2 = 0 + 0.394 063 831 04;
  • 34) 0.394 063 831 04 × 2 = 0 + 0.788 127 662 08;
  • 35) 0.788 127 662 08 × 2 = 1 + 0.576 255 324 16;
  • 36) 0.576 255 324 16 × 2 = 1 + 0.152 510 648 32;
  • 37) 0.152 510 648 32 × 2 = 0 + 0.305 021 296 64;
  • 38) 0.305 021 296 64 × 2 = 0 + 0.610 042 593 28;
  • 39) 0.610 042 593 28 × 2 = 1 + 0.220 085 186 56;
  • 40) 0.220 085 186 56 × 2 = 0 + 0.440 170 373 12;
  • 41) 0.440 170 373 12 × 2 = 0 + 0.880 340 746 24;
  • 42) 0.880 340 746 24 × 2 = 1 + 0.760 681 492 48;
  • 43) 0.760 681 492 48 × 2 = 1 + 0.521 362 984 96;
  • 44) 0.521 362 984 96 × 2 = 1 + 0.042 725 969 92;
  • 45) 0.042 725 969 92 × 2 = 0 + 0.085 451 939 84;
  • 46) 0.085 451 939 84 × 2 = 0 + 0.170 903 879 68;
  • 47) 0.170 903 879 68 × 2 = 0 + 0.341 807 759 36;
  • 48) 0.341 807 759 36 × 2 = 0 + 0.683 615 518 72;
  • 49) 0.683 615 518 72 × 2 = 1 + 0.367 231 037 44;
  • 50) 0.367 231 037 44 × 2 = 0 + 0.734 462 074 88;
  • 51) 0.734 462 074 88 × 2 = 1 + 0.468 924 149 76;
  • 52) 0.468 924 149 76 × 2 = 0 + 0.937 848 299 52;
  • 53) 0.937 848 299 52 × 2 = 1 + 0.875 696 599 04;
  • 54) 0.875 696 599 04 × 2 = 1 + 0.751 393 198 08;
  • 55) 0.751 393 198 08 × 2 = 1 + 0.502 786 396 16;
  • 56) 0.502 786 396 16 × 2 = 1 + 0.005 572 792 32;
  • 57) 0.005 572 792 32 × 2 = 0 + 0.011 145 584 64;
  • 58) 0.011 145 584 64 × 2 = 0 + 0.022 291 169 28;
  • 59) 0.022 291 169 28 × 2 = 0 + 0.044 582 338 56;

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:


0.014 657 790 12(10) =

0.0000 0011 1100 0000 1001 1100 1110 1001 0011 0010 0111 0000 1010 1111 000(2)

6. Positive number before normalization:

0.014 657 790 12(10) =

0.0000 0011 1100 0000 1001 1100 1110 1001 0011 0010 0111 0000 1010 1111 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.014 657 790 12(10) =

0.0000 0011 1100 0000 1001 1100 1110 1001 0011 0010 0111 0000 1010 1111 000(2) =

0.0000 0011 1100 0000 1001 1100 1110 1001 0011 0010 0111 0000 1010 1111 000(2) × 20 =

1.1110 0000 0100 1110 0111 0100 1001 1001 0011 1000 0101 0111 1000(2) × 2-7


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): -7

Mantissa (not normalized):
1.1110 0000 0100 1110 0111 0100 1001 1001 0011 1000 0101 0111 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 1110 0000 0100 1110 0111 0100 1001 1001 0011 1000 0101 0111 1000 =

1110 0000 0100 1110 0111 0100 1001 1001 0011 1000 0101 0111 1000


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 1000

Mantissa (52 bits) =
1110 0000 0100 1110 0111 0100 1001 1001 0011 1000 0101 0111 1000


Decimal number -0.014 657 790 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 1110 0000 0100 1110 0111 0100 1001 1001 0011 1000 0101 0111 1000

» Convert decimal -0.014 657 789 75 to 64 bit double precision IEEE 754 binary floating point representation standard

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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