-0.000 010 469 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 010 469(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
-0.000 010 469(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:

|-0.000 010 469| = 0.000 010 469


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: 0.000 010 469.

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) 0.000 010 469 × 2 = 0 + 0.000 020 938;
  • 2) 0.000 020 938 × 2 = 0 + 0.000 041 876;
  • 3) 0.000 041 876 × 2 = 0 + 0.000 083 752;
  • 4) 0.000 083 752 × 2 = 0 + 0.000 167 504;
  • 5) 0.000 167 504 × 2 = 0 + 0.000 335 008;
  • 6) 0.000 335 008 × 2 = 0 + 0.000 670 016;
  • 7) 0.000 670 016 × 2 = 0 + 0.001 340 032;
  • 8) 0.001 340 032 × 2 = 0 + 0.002 680 064;
  • 9) 0.002 680 064 × 2 = 0 + 0.005 360 128;
  • 10) 0.005 360 128 × 2 = 0 + 0.010 720 256;
  • 11) 0.010 720 256 × 2 = 0 + 0.021 440 512;
  • 12) 0.021 440 512 × 2 = 0 + 0.042 881 024;
  • 13) 0.042 881 024 × 2 = 0 + 0.085 762 048;
  • 14) 0.085 762 048 × 2 = 0 + 0.171 524 096;
  • 15) 0.171 524 096 × 2 = 0 + 0.343 048 192;
  • 16) 0.343 048 192 × 2 = 0 + 0.686 096 384;
  • 17) 0.686 096 384 × 2 = 1 + 0.372 192 768;
  • 18) 0.372 192 768 × 2 = 0 + 0.744 385 536;
  • 19) 0.744 385 536 × 2 = 1 + 0.488 771 072;
  • 20) 0.488 771 072 × 2 = 0 + 0.977 542 144;
  • 21) 0.977 542 144 × 2 = 1 + 0.955 084 288;
  • 22) 0.955 084 288 × 2 = 1 + 0.910 168 576;
  • 23) 0.910 168 576 × 2 = 1 + 0.820 337 152;
  • 24) 0.820 337 152 × 2 = 1 + 0.640 674 304;
  • 25) 0.640 674 304 × 2 = 1 + 0.281 348 608;
  • 26) 0.281 348 608 × 2 = 0 + 0.562 697 216;
  • 27) 0.562 697 216 × 2 = 1 + 0.125 394 432;
  • 28) 0.125 394 432 × 2 = 0 + 0.250 788 864;
  • 29) 0.250 788 864 × 2 = 0 + 0.501 577 728;
  • 30) 0.501 577 728 × 2 = 1 + 0.003 155 456;
  • 31) 0.003 155 456 × 2 = 0 + 0.006 310 912;
  • 32) 0.006 310 912 × 2 = 0 + 0.012 621 824;
  • 33) 0.012 621 824 × 2 = 0 + 0.025 243 648;
  • 34) 0.025 243 648 × 2 = 0 + 0.050 487 296;
  • 35) 0.050 487 296 × 2 = 0 + 0.100 974 592;
  • 36) 0.100 974 592 × 2 = 0 + 0.201 949 184;
  • 37) 0.201 949 184 × 2 = 0 + 0.403 898 368;
  • 38) 0.403 898 368 × 2 = 0 + 0.807 796 736;
  • 39) 0.807 796 736 × 2 = 1 + 0.615 593 472;
  • 40) 0.615 593 472 × 2 = 1 + 0.231 186 944;
  • 41) 0.231 186 944 × 2 = 0 + 0.462 373 888;
  • 42) 0.462 373 888 × 2 = 0 + 0.924 747 776;
  • 43) 0.924 747 776 × 2 = 1 + 0.849 495 552;
  • 44) 0.849 495 552 × 2 = 1 + 0.698 991 104;
  • 45) 0.698 991 104 × 2 = 1 + 0.397 982 208;
  • 46) 0.397 982 208 × 2 = 0 + 0.795 964 416;
  • 47) 0.795 964 416 × 2 = 1 + 0.591 928 832;
  • 48) 0.591 928 832 × 2 = 1 + 0.183 857 664;
  • 49) 0.183 857 664 × 2 = 0 + 0.367 715 328;
  • 50) 0.367 715 328 × 2 = 0 + 0.735 430 656;
  • 51) 0.735 430 656 × 2 = 1 + 0.470 861 312;
  • 52) 0.470 861 312 × 2 = 0 + 0.941 722 624;
  • 53) 0.941 722 624 × 2 = 1 + 0.883 445 248;
  • 54) 0.883 445 248 × 2 = 1 + 0.766 890 496;
  • 55) 0.766 890 496 × 2 = 1 + 0.533 780 992;
  • 56) 0.533 780 992 × 2 = 1 + 0.067 561 984;
  • 57) 0.067 561 984 × 2 = 0 + 0.135 123 968;
  • 58) 0.135 123 968 × 2 = 0 + 0.270 247 936;
  • 59) 0.270 247 936 × 2 = 0 + 0.540 495 872;
  • 60) 0.540 495 872 × 2 = 1 + 0.080 991 744;
  • 61) 0.080 991 744 × 2 = 0 + 0.161 983 488;
  • 62) 0.161 983 488 × 2 = 0 + 0.323 966 976;
  • 63) 0.323 966 976 × 2 = 0 + 0.647 933 952;
  • 64) 0.647 933 952 × 2 = 1 + 0.295 867 904;
  • 65) 0.295 867 904 × 2 = 0 + 0.591 735 808;
  • 66) 0.591 735 808 × 2 = 1 + 0.183 471 616;
  • 67) 0.183 471 616 × 2 = 0 + 0.366 943 232;
  • 68) 0.366 943 232 × 2 = 0 + 0.733 886 464;
  • 69) 0.733 886 464 × 2 = 1 + 0.467 772 928;

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:


0.000 010 469(10) =

0.0000 0000 0000 0000 1010 1111 1010 0100 0000 0011 0011 1011 0010 1111 0001 0001 0100 1(2)

6. Positive number before normalization:

0.000 010 469(10) =

0.0000 0000 0000 0000 1010 1111 1010 0100 0000 0011 0011 1011 0010 1111 0001 0001 0100 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 010 469(10) =

0.0000 0000 0000 0000 1010 1111 1010 0100 0000 0011 0011 1011 0010 1111 0001 0001 0100 1(2) =

0.0000 0000 0000 0000 1010 1111 1010 0100 0000 0011 0011 1011 0010 1111 0001 0001 0100 1(2) × 20 =

1.0101 1111 0100 1000 0000 0110 0111 0110 0101 1110 0010 0010 1001(2) × 2-17


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): -17

Mantissa (not normalized):
1.0101 1111 0100 1000 0000 0110 0111 0110 0101 1110 0010 0010 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1006(10) =

011 1110 1110(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. 0101 1111 0100 1000 0000 0110 0111 0110 0101 1110 0010 0010 1001 =

0101 1111 0100 1000 0000 0110 0111 0110 0101 1110 0010 0010 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 1110 1110

Mantissa (52 bits) =
0101 1111 0100 1000 0000 0110 0111 0110 0101 1110 0010 0010 1001


Decimal number -0.000 010 469 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0101 1111 0100 1000 0000 0110 0111 0110 0101 1110 0010 0010 1001

» Convert decimal -0.000 010 459 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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