-0.000 003 300 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 003 300 5(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
-0.000 003 300 5(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:

|-0.000 003 300 5| = 0.000 003 300 5


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: 0.000 003 300 5.

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) 0.000 003 300 5 × 2 = 0 + 0.000 006 601;
  • 2) 0.000 006 601 × 2 = 0 + 0.000 013 202;
  • 3) 0.000 013 202 × 2 = 0 + 0.000 026 404;
  • 4) 0.000 026 404 × 2 = 0 + 0.000 052 808;
  • 5) 0.000 052 808 × 2 = 0 + 0.000 105 616;
  • 6) 0.000 105 616 × 2 = 0 + 0.000 211 232;
  • 7) 0.000 211 232 × 2 = 0 + 0.000 422 464;
  • 8) 0.000 422 464 × 2 = 0 + 0.000 844 928;
  • 9) 0.000 844 928 × 2 = 0 + 0.001 689 856;
  • 10) 0.001 689 856 × 2 = 0 + 0.003 379 712;
  • 11) 0.003 379 712 × 2 = 0 + 0.006 759 424;
  • 12) 0.006 759 424 × 2 = 0 + 0.013 518 848;
  • 13) 0.013 518 848 × 2 = 0 + 0.027 037 696;
  • 14) 0.027 037 696 × 2 = 0 + 0.054 075 392;
  • 15) 0.054 075 392 × 2 = 0 + 0.108 150 784;
  • 16) 0.108 150 784 × 2 = 0 + 0.216 301 568;
  • 17) 0.216 301 568 × 2 = 0 + 0.432 603 136;
  • 18) 0.432 603 136 × 2 = 0 + 0.865 206 272;
  • 19) 0.865 206 272 × 2 = 1 + 0.730 412 544;
  • 20) 0.730 412 544 × 2 = 1 + 0.460 825 088;
  • 21) 0.460 825 088 × 2 = 0 + 0.921 650 176;
  • 22) 0.921 650 176 × 2 = 1 + 0.843 300 352;
  • 23) 0.843 300 352 × 2 = 1 + 0.686 600 704;
  • 24) 0.686 600 704 × 2 = 1 + 0.373 201 408;
  • 25) 0.373 201 408 × 2 = 0 + 0.746 402 816;
  • 26) 0.746 402 816 × 2 = 1 + 0.492 805 632;
  • 27) 0.492 805 632 × 2 = 0 + 0.985 611 264;
  • 28) 0.985 611 264 × 2 = 1 + 0.971 222 528;
  • 29) 0.971 222 528 × 2 = 1 + 0.942 445 056;
  • 30) 0.942 445 056 × 2 = 1 + 0.884 890 112;
  • 31) 0.884 890 112 × 2 = 1 + 0.769 780 224;
  • 32) 0.769 780 224 × 2 = 1 + 0.539 560 448;
  • 33) 0.539 560 448 × 2 = 1 + 0.079 120 896;
  • 34) 0.079 120 896 × 2 = 0 + 0.158 241 792;
  • 35) 0.158 241 792 × 2 = 0 + 0.316 483 584;
  • 36) 0.316 483 584 × 2 = 0 + 0.632 967 168;
  • 37) 0.632 967 168 × 2 = 1 + 0.265 934 336;
  • 38) 0.265 934 336 × 2 = 0 + 0.531 868 672;
  • 39) 0.531 868 672 × 2 = 1 + 0.063 737 344;
  • 40) 0.063 737 344 × 2 = 0 + 0.127 474 688;
  • 41) 0.127 474 688 × 2 = 0 + 0.254 949 376;
  • 42) 0.254 949 376 × 2 = 0 + 0.509 898 752;
  • 43) 0.509 898 752 × 2 = 1 + 0.019 797 504;
  • 44) 0.019 797 504 × 2 = 0 + 0.039 595 008;
  • 45) 0.039 595 008 × 2 = 0 + 0.079 190 016;
  • 46) 0.079 190 016 × 2 = 0 + 0.158 380 032;
  • 47) 0.158 380 032 × 2 = 0 + 0.316 760 064;
  • 48) 0.316 760 064 × 2 = 0 + 0.633 520 128;
  • 49) 0.633 520 128 × 2 = 1 + 0.267 040 256;
  • 50) 0.267 040 256 × 2 = 0 + 0.534 080 512;
  • 51) 0.534 080 512 × 2 = 1 + 0.068 161 024;
  • 52) 0.068 161 024 × 2 = 0 + 0.136 322 048;
  • 53) 0.136 322 048 × 2 = 0 + 0.272 644 096;
  • 54) 0.272 644 096 × 2 = 0 + 0.545 288 192;
  • 55) 0.545 288 192 × 2 = 1 + 0.090 576 384;
  • 56) 0.090 576 384 × 2 = 0 + 0.181 152 768;
  • 57) 0.181 152 768 × 2 = 0 + 0.362 305 536;
  • 58) 0.362 305 536 × 2 = 0 + 0.724 611 072;
  • 59) 0.724 611 072 × 2 = 1 + 0.449 222 144;
  • 60) 0.449 222 144 × 2 = 0 + 0.898 444 288;
  • 61) 0.898 444 288 × 2 = 1 + 0.796 888 576;
  • 62) 0.796 888 576 × 2 = 1 + 0.593 777 152;
  • 63) 0.593 777 152 × 2 = 1 + 0.187 554 304;
  • 64) 0.187 554 304 × 2 = 0 + 0.375 108 608;
  • 65) 0.375 108 608 × 2 = 0 + 0.750 217 216;
  • 66) 0.750 217 216 × 2 = 1 + 0.500 434 432;
  • 67) 0.500 434 432 × 2 = 1 + 0.000 868 864;
  • 68) 0.000 868 864 × 2 = 0 + 0.001 737 728;
  • 69) 0.001 737 728 × 2 = 0 + 0.003 475 456;
  • 70) 0.003 475 456 × 2 = 0 + 0.006 950 912;
  • 71) 0.006 950 912 × 2 = 0 + 0.013 901 824;

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:


0.000 003 300 5(10) =

0.0000 0000 0000 0000 0011 0111 0101 1111 1000 1010 0010 0000 1010 0010 0010 1110 0110 000(2)

6. Positive number before normalization:

0.000 003 300 5(10) =

0.0000 0000 0000 0000 0011 0111 0101 1111 1000 1010 0010 0000 1010 0010 0010 1110 0110 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 300 5(10) =

0.0000 0000 0000 0000 0011 0111 0101 1111 1000 1010 0010 0000 1010 0010 0010 1110 0110 000(2) =

0.0000 0000 0000 0000 0011 0111 0101 1111 1000 1010 0010 0000 1010 0010 0010 1110 0110 000(2) × 20 =

1.1011 1010 1111 1100 0101 0001 0000 0101 0001 0001 0111 0011 0000(2) × 2-19


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): -19

Mantissa (not normalized):
1.1011 1010 1111 1100 0101 0001 0000 0101 0001 0001 0111 0011 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(11-1) - 1 =

(-19 + 1 023)(10) =

1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 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) =

1004(10) =

011 1110 1100(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. 1011 1010 1111 1100 0101 0001 0000 0101 0001 0001 0111 0011 0000 =

1011 1010 1111 1100 0101 0001 0000 0101 0001 0001 0111 0011 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 1110 1100

Mantissa (52 bits) =
1011 1010 1111 1100 0101 0001 0000 0101 0001 0001 0111 0011 0000


Decimal number -0.000 003 300 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1100 - 1011 1010 1111 1100 0101 0001 0000 0101 0001 0001 0111 0011 0000

Share

» Convert decimal -0.000 003 304 1 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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