-0.014 657 791 96 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.014 657 791 96(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
-0.014 657 791 96(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:

|-0.014 657 791 96| = 0.014 657 791 96


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: 0.014 657 791 96.

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) 0.014 657 791 96 × 2 = 0 + 0.029 315 583 92;
  • 2) 0.029 315 583 92 × 2 = 0 + 0.058 631 167 84;
  • 3) 0.058 631 167 84 × 2 = 0 + 0.117 262 335 68;
  • 4) 0.117 262 335 68 × 2 = 0 + 0.234 524 671 36;
  • 5) 0.234 524 671 36 × 2 = 0 + 0.469 049 342 72;
  • 6) 0.469 049 342 72 × 2 = 0 + 0.938 098 685 44;
  • 7) 0.938 098 685 44 × 2 = 1 + 0.876 197 370 88;
  • 8) 0.876 197 370 88 × 2 = 1 + 0.752 394 741 76;
  • 9) 0.752 394 741 76 × 2 = 1 + 0.504 789 483 52;
  • 10) 0.504 789 483 52 × 2 = 1 + 0.009 578 967 04;
  • 11) 0.009 578 967 04 × 2 = 0 + 0.019 157 934 08;
  • 12) 0.019 157 934 08 × 2 = 0 + 0.038 315 868 16;
  • 13) 0.038 315 868 16 × 2 = 0 + 0.076 631 736 32;
  • 14) 0.076 631 736 32 × 2 = 0 + 0.153 263 472 64;
  • 15) 0.153 263 472 64 × 2 = 0 + 0.306 526 945 28;
  • 16) 0.306 526 945 28 × 2 = 0 + 0.613 053 890 56;
  • 17) 0.613 053 890 56 × 2 = 1 + 0.226 107 781 12;
  • 18) 0.226 107 781 12 × 2 = 0 + 0.452 215 562 24;
  • 19) 0.452 215 562 24 × 2 = 0 + 0.904 431 124 48;
  • 20) 0.904 431 124 48 × 2 = 1 + 0.808 862 248 96;
  • 21) 0.808 862 248 96 × 2 = 1 + 0.617 724 497 92;
  • 22) 0.617 724 497 92 × 2 = 1 + 0.235 448 995 84;
  • 23) 0.235 448 995 84 × 2 = 0 + 0.470 897 991 68;
  • 24) 0.470 897 991 68 × 2 = 0 + 0.941 795 983 36;
  • 25) 0.941 795 983 36 × 2 = 1 + 0.883 591 966 72;
  • 26) 0.883 591 966 72 × 2 = 1 + 0.767 183 933 44;
  • 27) 0.767 183 933 44 × 2 = 1 + 0.534 367 866 88;
  • 28) 0.534 367 866 88 × 2 = 1 + 0.068 735 733 76;
  • 29) 0.068 735 733 76 × 2 = 0 + 0.137 471 467 52;
  • 30) 0.137 471 467 52 × 2 = 0 + 0.274 942 935 04;
  • 31) 0.274 942 935 04 × 2 = 0 + 0.549 885 870 08;
  • 32) 0.549 885 870 08 × 2 = 1 + 0.099 771 740 16;
  • 33) 0.099 771 740 16 × 2 = 0 + 0.199 543 480 32;
  • 34) 0.199 543 480 32 × 2 = 0 + 0.399 086 960 64;
  • 35) 0.399 086 960 64 × 2 = 0 + 0.798 173 921 28;
  • 36) 0.798 173 921 28 × 2 = 1 + 0.596 347 842 56;
  • 37) 0.596 347 842 56 × 2 = 1 + 0.192 695 685 12;
  • 38) 0.192 695 685 12 × 2 = 0 + 0.385 391 370 24;
  • 39) 0.385 391 370 24 × 2 = 0 + 0.770 782 740 48;
  • 40) 0.770 782 740 48 × 2 = 1 + 0.541 565 480 96;
  • 41) 0.541 565 480 96 × 2 = 1 + 0.083 130 961 92;
  • 42) 0.083 130 961 92 × 2 = 0 + 0.166 261 923 84;
  • 43) 0.166 261 923 84 × 2 = 0 + 0.332 523 847 68;
  • 44) 0.332 523 847 68 × 2 = 0 + 0.665 047 695 36;
  • 45) 0.665 047 695 36 × 2 = 1 + 0.330 095 390 72;
  • 46) 0.330 095 390 72 × 2 = 0 + 0.660 190 781 44;
  • 47) 0.660 190 781 44 × 2 = 1 + 0.320 381 562 88;
  • 48) 0.320 381 562 88 × 2 = 0 + 0.640 763 125 76;
  • 49) 0.640 763 125 76 × 2 = 1 + 0.281 526 251 52;
  • 50) 0.281 526 251 52 × 2 = 0 + 0.563 052 503 04;
  • 51) 0.563 052 503 04 × 2 = 1 + 0.126 105 006 08;
  • 52) 0.126 105 006 08 × 2 = 0 + 0.252 210 012 16;
  • 53) 0.252 210 012 16 × 2 = 0 + 0.504 420 024 32;
  • 54) 0.504 420 024 32 × 2 = 1 + 0.008 840 048 64;
  • 55) 0.008 840 048 64 × 2 = 0 + 0.017 680 097 28;
  • 56) 0.017 680 097 28 × 2 = 0 + 0.035 360 194 56;
  • 57) 0.035 360 194 56 × 2 = 0 + 0.070 720 389 12;
  • 58) 0.070 720 389 12 × 2 = 0 + 0.141 440 778 24;
  • 59) 0.141 440 778 24 × 2 = 0 + 0.282 881 556 48;

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:


0.014 657 791 96(10) =

0.0000 0011 1100 0000 1001 1100 1111 0001 0001 1001 1000 1010 1010 0100 000(2)

6. Positive number before normalization:

0.014 657 791 96(10) =

0.0000 0011 1100 0000 1001 1100 1111 0001 0001 1001 1000 1010 1010 0100 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:


0.014 657 791 96(10) =

0.0000 0011 1100 0000 1001 1100 1111 0001 0001 1001 1000 1010 1010 0100 000(2) =

0.0000 0011 1100 0000 1001 1100 1111 0001 0001 1001 1000 1010 1010 0100 000(2) × 20 =

1.1110 0000 0100 1110 0111 1000 1000 1100 1100 0101 0101 0010 0000(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 1000 1000 1100 1100 0101 0101 0010 0000


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 1000 1000 1100 1100 0101 0101 0010 0000 =

1110 0000 0100 1110 0111 1000 1000 1100 1100 0101 0101 0010 0000


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 1000 1000 1100 1100 0101 0101 0010 0000


Decimal number -0.014 657 791 96 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 1110 0000 0100 1110 0111 1000 1000 1100 1100 0101 0101 0010 0000

» Convert decimal -0.014 657 791 1 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