-0.089 999 619 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.089 999 619(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
-0.089 999 619(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:

|-0.089 999 619| = 0.089 999 619


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: 0.089 999 619.

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) 0.089 999 619 × 2 = 0 + 0.179 999 238;
  • 2) 0.179 999 238 × 2 = 0 + 0.359 998 476;
  • 3) 0.359 998 476 × 2 = 0 + 0.719 996 952;
  • 4) 0.719 996 952 × 2 = 1 + 0.439 993 904;
  • 5) 0.439 993 904 × 2 = 0 + 0.879 987 808;
  • 6) 0.879 987 808 × 2 = 1 + 0.759 975 616;
  • 7) 0.759 975 616 × 2 = 1 + 0.519 951 232;
  • 8) 0.519 951 232 × 2 = 1 + 0.039 902 464;
  • 9) 0.039 902 464 × 2 = 0 + 0.079 804 928;
  • 10) 0.079 804 928 × 2 = 0 + 0.159 609 856;
  • 11) 0.159 609 856 × 2 = 0 + 0.319 219 712;
  • 12) 0.319 219 712 × 2 = 0 + 0.638 439 424;
  • 13) 0.638 439 424 × 2 = 1 + 0.276 878 848;
  • 14) 0.276 878 848 × 2 = 0 + 0.553 757 696;
  • 15) 0.553 757 696 × 2 = 1 + 0.107 515 392;
  • 16) 0.107 515 392 × 2 = 0 + 0.215 030 784;
  • 17) 0.215 030 784 × 2 = 0 + 0.430 061 568;
  • 18) 0.430 061 568 × 2 = 0 + 0.860 123 136;
  • 19) 0.860 123 136 × 2 = 1 + 0.720 246 272;
  • 20) 0.720 246 272 × 2 = 1 + 0.440 492 544;
  • 21) 0.440 492 544 × 2 = 0 + 0.880 985 088;
  • 22) 0.880 985 088 × 2 = 1 + 0.761 970 176;
  • 23) 0.761 970 176 × 2 = 1 + 0.523 940 352;
  • 24) 0.523 940 352 × 2 = 1 + 0.047 880 704;
  • 25) 0.047 880 704 × 2 = 0 + 0.095 761 408;
  • 26) 0.095 761 408 × 2 = 0 + 0.191 522 816;
  • 27) 0.191 522 816 × 2 = 0 + 0.383 045 632;
  • 28) 0.383 045 632 × 2 = 0 + 0.766 091 264;
  • 29) 0.766 091 264 × 2 = 1 + 0.532 182 528;
  • 30) 0.532 182 528 × 2 = 1 + 0.064 365 056;
  • 31) 0.064 365 056 × 2 = 0 + 0.128 730 112;
  • 32) 0.128 730 112 × 2 = 0 + 0.257 460 224;
  • 33) 0.257 460 224 × 2 = 0 + 0.514 920 448;
  • 34) 0.514 920 448 × 2 = 1 + 0.029 840 896;
  • 35) 0.029 840 896 × 2 = 0 + 0.059 681 792;
  • 36) 0.059 681 792 × 2 = 0 + 0.119 363 584;
  • 37) 0.119 363 584 × 2 = 0 + 0.238 727 168;
  • 38) 0.238 727 168 × 2 = 0 + 0.477 454 336;
  • 39) 0.477 454 336 × 2 = 0 + 0.954 908 672;
  • 40) 0.954 908 672 × 2 = 1 + 0.909 817 344;
  • 41) 0.909 817 344 × 2 = 1 + 0.819 634 688;
  • 42) 0.819 634 688 × 2 = 1 + 0.639 269 376;
  • 43) 0.639 269 376 × 2 = 1 + 0.278 538 752;
  • 44) 0.278 538 752 × 2 = 0 + 0.557 077 504;
  • 45) 0.557 077 504 × 2 = 1 + 0.114 155 008;
  • 46) 0.114 155 008 × 2 = 0 + 0.228 310 016;
  • 47) 0.228 310 016 × 2 = 0 + 0.456 620 032;
  • 48) 0.456 620 032 × 2 = 0 + 0.913 240 064;
  • 49) 0.913 240 064 × 2 = 1 + 0.826 480 128;
  • 50) 0.826 480 128 × 2 = 1 + 0.652 960 256;
  • 51) 0.652 960 256 × 2 = 1 + 0.305 920 512;
  • 52) 0.305 920 512 × 2 = 0 + 0.611 841 024;
  • 53) 0.611 841 024 × 2 = 1 + 0.223 682 048;
  • 54) 0.223 682 048 × 2 = 0 + 0.447 364 096;
  • 55) 0.447 364 096 × 2 = 0 + 0.894 728 192;
  • 56) 0.894 728 192 × 2 = 1 + 0.789 456 384;

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:


0.089 999 619(10) =

0.0001 0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001(2)

6. Positive number before normalization:

0.089 999 619(10) =

0.0001 0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.089 999 619(10) =

0.0001 0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001(2) =

0.0001 0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001(2) × 20 =

1.0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001(2) × 2-4


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):
1.0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001 =

0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001


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 1011

Mantissa (52 bits) =
0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001


Decimal number -0.089 999 619 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0111 0000 1010 0011 0111 0000 1100 0100 0001 1110 1000 1110 1001

Share

» Convert decimal -0.089 999 67 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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