-0.024 189 053 39 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.024 189 053 39(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
-0.024 189 053 39(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:

|-0.024 189 053 39| = 0.024 189 053 39


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: 0.024 189 053 39.

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) 0.024 189 053 39 × 2 = 0 + 0.048 378 106 78;
  • 2) 0.048 378 106 78 × 2 = 0 + 0.096 756 213 56;
  • 3) 0.096 756 213 56 × 2 = 0 + 0.193 512 427 12;
  • 4) 0.193 512 427 12 × 2 = 0 + 0.387 024 854 24;
  • 5) 0.387 024 854 24 × 2 = 0 + 0.774 049 708 48;
  • 6) 0.774 049 708 48 × 2 = 1 + 0.548 099 416 96;
  • 7) 0.548 099 416 96 × 2 = 1 + 0.096 198 833 92;
  • 8) 0.096 198 833 92 × 2 = 0 + 0.192 397 667 84;
  • 9) 0.192 397 667 84 × 2 = 0 + 0.384 795 335 68;
  • 10) 0.384 795 335 68 × 2 = 0 + 0.769 590 671 36;
  • 11) 0.769 590 671 36 × 2 = 1 + 0.539 181 342 72;
  • 12) 0.539 181 342 72 × 2 = 1 + 0.078 362 685 44;
  • 13) 0.078 362 685 44 × 2 = 0 + 0.156 725 370 88;
  • 14) 0.156 725 370 88 × 2 = 0 + 0.313 450 741 76;
  • 15) 0.313 450 741 76 × 2 = 0 + 0.626 901 483 52;
  • 16) 0.626 901 483 52 × 2 = 1 + 0.253 802 967 04;
  • 17) 0.253 802 967 04 × 2 = 0 + 0.507 605 934 08;
  • 18) 0.507 605 934 08 × 2 = 1 + 0.015 211 868 16;
  • 19) 0.015 211 868 16 × 2 = 0 + 0.030 423 736 32;
  • 20) 0.030 423 736 32 × 2 = 0 + 0.060 847 472 64;
  • 21) 0.060 847 472 64 × 2 = 0 + 0.121 694 945 28;
  • 22) 0.121 694 945 28 × 2 = 0 + 0.243 389 890 56;
  • 23) 0.243 389 890 56 × 2 = 0 + 0.486 779 781 12;
  • 24) 0.486 779 781 12 × 2 = 0 + 0.973 559 562 24;
  • 25) 0.973 559 562 24 × 2 = 1 + 0.947 119 124 48;
  • 26) 0.947 119 124 48 × 2 = 1 + 0.894 238 248 96;
  • 27) 0.894 238 248 96 × 2 = 1 + 0.788 476 497 92;
  • 28) 0.788 476 497 92 × 2 = 1 + 0.576 952 995 84;
  • 29) 0.576 952 995 84 × 2 = 1 + 0.153 905 991 68;
  • 30) 0.153 905 991 68 × 2 = 0 + 0.307 811 983 36;
  • 31) 0.307 811 983 36 × 2 = 0 + 0.615 623 966 72;
  • 32) 0.615 623 966 72 × 2 = 1 + 0.231 247 933 44;
  • 33) 0.231 247 933 44 × 2 = 0 + 0.462 495 866 88;
  • 34) 0.462 495 866 88 × 2 = 0 + 0.924 991 733 76;
  • 35) 0.924 991 733 76 × 2 = 1 + 0.849 983 467 52;
  • 36) 0.849 983 467 52 × 2 = 1 + 0.699 966 935 04;
  • 37) 0.699 966 935 04 × 2 = 1 + 0.399 933 870 08;
  • 38) 0.399 933 870 08 × 2 = 0 + 0.799 867 740 16;
  • 39) 0.799 867 740 16 × 2 = 1 + 0.599 735 480 32;
  • 40) 0.599 735 480 32 × 2 = 1 + 0.199 470 960 64;
  • 41) 0.199 470 960 64 × 2 = 0 + 0.398 941 921 28;
  • 42) 0.398 941 921 28 × 2 = 0 + 0.797 883 842 56;
  • 43) 0.797 883 842 56 × 2 = 1 + 0.595 767 685 12;
  • 44) 0.595 767 685 12 × 2 = 1 + 0.191 535 370 24;
  • 45) 0.191 535 370 24 × 2 = 0 + 0.383 070 740 48;
  • 46) 0.383 070 740 48 × 2 = 0 + 0.766 141 480 96;
  • 47) 0.766 141 480 96 × 2 = 1 + 0.532 282 961 92;
  • 48) 0.532 282 961 92 × 2 = 1 + 0.064 565 923 84;
  • 49) 0.064 565 923 84 × 2 = 0 + 0.129 131 847 68;
  • 50) 0.129 131 847 68 × 2 = 0 + 0.258 263 695 36;
  • 51) 0.258 263 695 36 × 2 = 0 + 0.516 527 390 72;
  • 52) 0.516 527 390 72 × 2 = 1 + 0.033 054 781 44;
  • 53) 0.033 054 781 44 × 2 = 0 + 0.066 109 562 88;
  • 54) 0.066 109 562 88 × 2 = 0 + 0.132 219 125 76;
  • 55) 0.132 219 125 76 × 2 = 0 + 0.264 438 251 52;
  • 56) 0.264 438 251 52 × 2 = 0 + 0.528 876 503 04;
  • 57) 0.528 876 503 04 × 2 = 1 + 0.057 753 006 08;
  • 58) 0.057 753 006 08 × 2 = 0 + 0.115 506 012 16;

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:


0.024 189 053 39(10) =

0.0000 0110 0011 0001 0100 0000 1111 1001 0011 1011 0011 0011 0001 0000 10(2)

6. Positive number before normalization:

0.024 189 053 39(10) =

0.0000 0110 0011 0001 0100 0000 1111 1001 0011 1011 0011 0011 0001 0000 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.024 189 053 39(10) =

0.0000 0110 0011 0001 0100 0000 1111 1001 0011 1011 0011 0011 0001 0000 10(2) =

0.0000 0110 0011 0001 0100 0000 1111 1001 0011 1011 0011 0011 0001 0000 10(2) × 20 =

1.1000 1100 0101 0000 0011 1110 0100 1110 1100 1100 1100 0100 0010(2) × 2-6


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): -6

Mantissa (not normalized):
1.1000 1100 0101 0000 0011 1110 0100 1110 1100 1100 1100 0100 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 1000 1100 0101 0000 0011 1110 0100 1110 1100 1100 1100 0100 0010 =

1000 1100 0101 0000 0011 1110 0100 1110 1100 1100 1100 0100 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 1001

Mantissa (52 bits) =
1000 1100 0101 0000 0011 1110 0100 1110 1100 1100 1100 0100 0010


Decimal number -0.024 189 053 39 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 1000 1100 0101 0000 0011 1110 0100 1110 1100 1100 1100 0100 0010

» Convert decimal -0.024 189 052 52 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